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INSTRUCTOR SOLUTIONS MANUAL

Chapter 1 1. THINK In this problem we’re given the radius of Earth, and asked to compute its circumference, surface area and volume. EXPRESS Assuming Earth to be a sphere of radius RE   6.37  106 m 103 km m   6.37  103 km,

the corresponding circumference, surface area and volume are: 4 3 C  2 RE , A  4 RE2 , V RE . 3 The geometric formulas are given in Appendix E. ANALYZE (a) Using the formulas given above, we find the circumference to be C  2 RE  2 (6.37  103 km)  4.00 104 km.

(b) Similarly, the surface area of Earth is



A  4 RE2  4 6.37  103 km

  5.10  10 km , 2

3 4 3 4 RE  6.37  103 km  1.08  1012 km3. 3 3





LEARN From the formulas given, we see that C RE , A RE2 , and V RE3 . The ratios of volume to surface area, and surface area to circumference are V / A  RE / 3 and A / C  2RE . 2. The conversion factors are: 1 gry  1/10 line , 1 line  1/12 inch and 1 point = 1/72 inch. The factors imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry2 = 0.18 point 2 . 3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2).

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2 (a) Since 1 km = 1  103 m and 1 m = 1  106 m,







1km  103 m  103 m 106  m m  109  m.

The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0  109 m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,







1cm = 102 m = 102m 106  m m  104 m.

We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,





1.0 yd =  0.91m  106  m m  9.1  105  m.

4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain  1 inch  6 picas  0.80 cm =  0.80 cm      1.9 picas.  2.54 cm  1 inch  (b) With 12 points = 1 pica, we have

 1 inch  6 picas  12 points  0.80 cm =  0.80 cm       23 points.  2.54 cm  1 inch  1 pica  5. THINK This problem deals with conversion of furlongs to rods and chains, all of which are units for distance. EXPRESS Given that 1 furlong  201.168 m, 1 rod  5.0292 m and 1chain  20.117 m , the relevant conversion factors are 1 rod 1.0 furlong  201.168 m  (201.168 m )  40 rods, 5.0292 m and 1 chain 1.0 furlong  201.168 m  (201.168 m ) 10 chains . 20.117 m Note the cancellation of m (meters), the unwanted unit. ANALYZE Using the above conversion factors, we find (a) the distance d in rods to be d  4.0 furlongs   4.0 furlongs 

40 rods  160 rods, 1 furlong

3 (b) and in chains to be d  4.0 furlongs   4.0 furlongs 

10 chains  40 chains. 1 furlong

LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160 rods ( 1 rod  5 m ) and 40 chains ( 1 chain  20 m ). So our results make sense. 6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. 1 Thus, 1 fanega = 12 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the 1

already completed part) implies that 1 cuartilla = 48 cahiz, or 2.08  102 cahiz.

Continuing in this way, the remaining entries in the first column are 6.94  103 and 3.47 103 . (b) In the second (“fanega”) column, we find 0.250, 8.33  102, and 4.17  102 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. 1

(d) Finally, in the fourth (“almude”) column, we get 2 = 0.500 for the last entry. (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86  102 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 7.00 7.00 m3 or 55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24  104 cm3. 7. We use the conversion factors found in Appendix D. 1 acre  ft = (43,560 ft 2 )  ft = 43,560 ft 3

Since 2 in. = (1/6) ft, the volume of water that fell during the storm is V  (26 km2 )(1/6 ft)  (26 km2 )(3281ft/km) 2 (1/6 ft )  4.66 107 ft 3.

Thus, 4.66  107 ft 3 V   11 .  103 acre  ft. 4 3 4.3560  10 ft acre  ft

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8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have

 258 W  50.0 S   50.0 S    60.8 W  212 S  (b) In units of Z, we have

 156 Z  50.0 S   50.0 S    43.3 Z  180 S  9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = r2/2, where r is the radius. Therefore, the volume is  V  r2 z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

 103 m   102 cm  5 r   2000 km       2000  10 cm.  1km   1m  In these units, the thickness becomes

 102 cm  2 z  3000 m   3000 m     3000  10 cm  1m  which yields V 

2  2000  105 cm 3000  102 cm  1.9  1022 cm3. 2



 



10. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change longitude by 360 / 24 15 before resetting one's watch by 1.0 h. 11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so

b3.7 mgc10  m mh  31.  m s. b14 daygb86400 s dayg 6

13. The time on any of these clocks is a straight-line function of that on another, with slopes  1 and y-intercepts  0. From the data in the figure we deduce

tC 

2 594 tB  , 7 7

tB 

33 662 tA  . 40 5

These are used in obtaining the following results. (a) We find tB  tB 

when t'A  tA = 600 s. (b) We obtain tC  tC 

33  t A  t A   495 s 40

2 2 t B  t B  495  141 s. 7 7

b g

(c) Clock B reads tB = (33/40)(400) (662/5)  198 s when clock A reads tA = 400 s. (d) From tC = 15 = (2/7)tB + (594/7), we get tB  245 s. 14. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also Table 1–2).

 100 y   365 day   24 h   60 min  (a) 1  century  106 century       52.6 min.  1 century   1 y   1 day   1 h 





(b) The percent difference is therefore 52.6 min  50 min  4.9%. 52.6 min

15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21  1012 s. 16. We denote the pulsar rotation rate f (for frequency). f 

1 rotation 1.55780644887275  103 s

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(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations:   1 rotation N    604800 s   388238218.4 3  1.55780644887275  10 s  which should now be rounded to 3.88  108 rotations since the time-interval was specified in the problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft, or

  1 rotation 1  106   t 3  1.55780644887275  10 s  which yields the result t = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). (c) Careful reading of the problem shows that the time-uncertainty per revolution is  3 1017s . We therefore expect that as a result of one million revolutions, the uncertainty should be (  3 1017 )(1106 )=  3 1011 s . 17. THINK In this problem we are asked to rank 5 clocks, based on their performance as timekeepers. EXPRESS We first note that none of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important here is that the clock advance by the same (or nearly the same) amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. ANALYZE The chart below gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from 5 s to +10 s, for clock E it is in the range from 70 s to 2 s. After C and D, A has

7 the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. CLOCK A B C D E

Sun. -Mon. 16 3 58 +67 +70

Mon. -Tues. 16 +5 58 +67 +55

Tues. -Wed. 15 10 58 +67 +2

Wed. -Thurs. 17 +5 58 +67 +20

Thurs. -Fri. 15 +6 58 +67 +10

Fri. -Sat. 15 7 58 +67 +10

LEARN Of the five clocks, the readings in clocks A, B and E jump around from one 24h period to another, making it difficult to correct them. 18. The last day of the 20 centuries is longer than the first day by

 20 century   0.001 s century   0.02 s. The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is

T   average increase in length of a day  number of days   0.01 s   365.25 day      2000 y  y  day     7305 s or roughly two hours. 19. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B. Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have d 2  r 2  (r  h)2  r 2  2rh  h2

CHAPTER 1

8 or d 2  2rh  h2 , where r is the radius of the Earth. Since r

h , the second term can be

dropped, leading to d  2rh . Now the angle between the two radii to the two tangent points A and B is , which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of  can be obtained by using

 360 

This yields





t . 24 h

(360)(11.1 s)  0.04625. (24 h)(60 min/h)(60 s/min)

Using d  r tan  , we have d 2  r 2 tan 2   2rh , or r

2h tan 2 

Using the above value for  and h = 1.7 m, we have r  5.2 106 m. 20. (a) We find the volume in cubic centimeters 3

 231 in 3   2.54 cm  5 3 193 gal = 193 gal      7.31  10 cm  1 gal   1in 

and subtract this from 1  106 cm3 to obtain 2.69  105 cm3. The conversion gal  in3 is given in Appendix D (immediately below the table of Volume conversions). (b) The volume found in part (a) is converted (by dividing by (100 cm/m) 3) to 0.731 m3, which corresponds to a mass of

c1000 kg m h c0.731 m h = 731 kg 3

using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731kg  4.06  105 min = 0.77 y 0.0018 kg min after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 21. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661  1027 kg). Thus,

N 

5.98  1024 kg ME   9.0  1049 . 27 m 40 u 1661 .  10 kg u

b gc

22. The density of gold is



m 19.32 g   19.32 g/cm3 . V 1 cm3

(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density  = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be V 



 1430 . cm3 .

We convert the volume to SI units: 3

 1m  6 3 V  1.430 cm     1.430  10 m . 100 cm   3

Since V = Az with z = 1  10-6 m (metric prefixes can be found in Table 1–2), we obtain 1430 .  106 m3 A  1430 . m2 . 6 1  10 m

(b) The volume of a cylinder of length  is V  A where the cross-section area is that of a circle: A = r2. Therefore, with r = 2.500  106 m and V = 1.430  106 m3, we obtain 

V  7.284  104 m  72.84 km. 2 r

23. THINK This problem consists of two parts: in the first part, we are asked to find the mass of water, given its volume and density; the second part deals with the mass flow rate of water, which is expressed as kg/s in SI units. EXPRESS From the definition of density:   m / V , we see that mass can be calculated as m  V , the product of the volume of water and its density. With 1 g = 1  103 kg and 1 cm3 = (1  102m)3 = 1  106m3, the density of water in SI units (kg/m3) is 3 3  1 g   10 kg   cm  3 3   6 3   1  10 kg m . 3    cm   g   10 m 

  1 g/cm3  

To obtain the flow rate, we simply divide the total mass of the water by the time taken to drain it. ANALYZE (a) Using m  V , the mass of a cubic meter of water is

CHAPTER 1

m  V  (1  103 kg/m3 )(1 m3 )  1000 kg.

(b) The total mass of water in the container is M  V  (1  103 kg m3 )(5700 m3 )  5.70  106 kg ,

and the time elapsed is t = (10 h)(3600 s/h) = 3.6  104 s. Thus, the mass flow rate R is R

M 5.70  106 kg   158 kg s. t 3.6  104 s

LEARN In terms of volume, the drain rate can be expressed as R 

V 5700 m3   0.158 m3 /s  42 gal/s. 4 t 3.6  10 s

The greater the flow rate, the less time required to drain a given amount of water. 24. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). The surface area A of each grain of sand of radius r = 50 m = 50  106 m is given by A = 4(50  106)2 = 3.14  108 m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of density,   m / V , so that the mass can be found from m = V, where  = 2600 kg/m3. Thus, using V = 4r3/3, the mass of each grain is

 4 r 3   kg  4  50  10 m  V     2600    m3  3  3  

6

m

 1.36  109 kg.

We observe that (because a cube has six equal faces) the indicated surface area is 6 m2. The number of spheres (the grains of sand) N that have a total surface area of 6 m2 is given by 6 m2 N   1.91  108. 3.14  108 m2 Therefore, the total mass M is M  Nm  1.91  108  1.36  109 kg   0.260 kg. 25. The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0  106 m3. Letting “d” stand for the thickness of the mud after it has (uniformly) distributed in the valley, then its volume there would be (400 m)(400 m)d. Requiring these two volumes to be equal, we can solve for d. Thus, d = 25 m. The volume of a small part of the mud over a patch of area of 4.0 m2 is (4.0)d = 100 m3. Since each cubic meter corresponds to a mass of

11 1900 kg (stated in the problem), then the mass of that small part of the mud is 1.9 105 kg . 26. (a) The volume of the cloud is (3000 m)(1000 m)2 = 9.4  109 m3. Since each cubic meter of the cloud contains from 50  106 to 500  106 water drops, then we conclude that the entire cloud contains from 4.7  1018 to 4.7  1019 drops. Since the volume of 4 each drop is 3 (10  106 m)3 = 4.2  1015 m3, then the total volume of water in a cloud is from 2 103 to 2 104 m3.

(b) Using the fact that 1 L  1103 cm3  1103 m3 , the amount of water estimated in part (a) would fill from 2 106 to 2 107 bottles. (c) At 1000 kg for every cubic meter, the mass of water is from 2 106 to 2 107 kg. The coincidence in numbers between the results of parts (b) and (c) of this problem is due to the fact that each liter has a mass of one kilogram when water is at its normal density (under standard conditions). 27. We introduce the notion of density,   m / V , and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density  of a sample of iron is

   7.87 g cm  3

 1 kg   100 cm  3     7870 kg/m . 1000 g 1 m   

If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then 9.27  1026 kg V    1.18  1029 m3 . 3 3  7.87  10 kg m M

(b) We set V = 4R3/3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find 13

 3V  R   4 





 3 1.18  1029 m3      4  

 1.41  1010 m.

The center-to-center distance between atoms is twice the radius, or 2.82  1010 m.

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28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom (in the cat) as 10 u  2  1026 kg, then there are roughly (10 kg)/( 2  1026 kg)  5  1026 atoms. This is close to being a factor of a thousand greater than Avogadro’s number. Thus this is roughly a kilomole of atoms. 29. The mass in kilograms is

gin I F 16 tahil I F 10 chee I F 10 hoon I F 0.3779 g I b28.9 piculsg FGH 100 JG JG JG JG J 1picul K H 1gin K H 1tahil K H 1 chee K H 1hoon K which yields 1.747  106 g or roughly 1.75 103 kg. 30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum. (a) Differentiating m(t )  5.00t 0.8  3.00t  20.00 with respect to t gives dm  4.00t 0.2  3.00. dt

The water mass is the greatest when dm / dt  0, or at t  (4.00 / 3.00)1/ 0.2  4.21s. (b) At t  4.21s, the water mass is m(t  4.21s)  5.00(4.21)0.8  3.00(4.21)  20.00  23.2 g.

(c) The rate of mass change at t  2.00 s is dm g 1 kg 60 s   4.00(2.00)0.2  3.00 g/s  0.48 g/s  0.48   dt t 2.00 s s 1000 g 1 min  2.89 102 kg/min.

(d) Similarly, the rate of mass change at t  5.00 s is dm g 1 kg 60 s   4.00(5.00)0.2  3.00 g/s  0.101 g/s  0.101   dt t 2.00 s s 1000 g 1 min  6.05 103 kg/min.

31. The mass density of the candy is



m 0.0200 g   4.00 104 g/mm3  4.00 104 kg/cm3 . 3 V 50.0 mm

If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is M   Ah, where A  (14.0 cm)(17.0 cm)  238 cm2 is the base area of the container that remains unchanged. Thus, the rate of mass change is given by dM d (  Ah) dh    A  (4.00 104 kg/cm3 )(238 cm 2 )(0.250 cm/s) dt dt dt  0.0238 kg/s  1.43 kg/min.

32. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20  12 = 240 m2) in addition to a rectangular box (height h´ = 6.0 m and same base). Therefore, 1 h  V  hA  hA    h  A  1800 m3 . 2 2  (a) Each dimension is reduced by a factor of 1/12, and we find

h FGH 121 IJK  10. m . 3

Vdoll  1800 m3

(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, 3 1 3 Vminiature  1800 m  6.0  104 m3 . 144

h FGH IJK

33. THINK In this problem we are asked to differentiate between three types of tons: displacement ton, freight ton and register ton, all of which are units of volume. EXPRESS The three different tons are defined in terms of barrel bulk, with 1 barrel bulk  0.1415 m3  4.0155 U.S. bushels (using 1 m3  28.378 U.S. bushels ). Thus, in terms of U.S. bushels, we have  4.0155 U.S. bushels  1 displacement ton  (7 barrels bulk)     28.108 U.S. bushels 1 barrel bulk    4.0155 U.S. bushels  1 freight ton  (8 barrels bulk)     32.124 U.S. bushels 1 barrel bulk  

 4.0155 U.S. bushels  1 register ton  (20 barrels bulk)     80.31 U.S. bushels 1 barrel bulk  

CHAPTER 1

ANALYZE (a) The difference between 73 “freight” tons and 73 “displacement” tons is V  73(freight tons  displacement tons)  73(32.124 U.S. bushels  28.108 U.S. bushels)  293.168 U.S. bushels  293 U.S. bushels

(b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is V  73(register tons  displacement tons)  73(80.31 U.S. bushels  28.108 U.S. bushels)  3810.746 U.S. bushels  3.81103 U.S. bushels

LEARN With 1 register ton  1 freight ton  1displacement ton, we expect the difference found in (b) to be greater than that in (a). This is indeed the case. 34. The customer expects a volume V1 = 20  7056 in3 and receives V2 = 20  5826 in.3, the difference being  V  V1  V2  24600 in.3 , or



V  24600 in.



 2.54cm   1L   403L    3   1 inch   1000 cm 

where Appendix D has been used. 35. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally:

 2 peck  (a) 11 tuffets = 11 tuffets     22 pecks .  1 tuffet   0.50 Imperial bushel  (b) 11 tuffets = 11 tuffets     5.5 Imperial bushels . 1 tuffet  

 36.3687 L  (c) 11 tuffets =  5.5 Imperial bushel     200 L .  1 Imperial bushel  36. Table 7 can be completed as follows: (a) It should be clear that the first column (under “wey”) is the reciprocal of the first 9 3 row – so that 10 = 0.900, 40 = 7.50  102, and so forth. Thus, 1 pottle = 1.56  103 wey and 1 gill = 8.32  106 wey are the last two entries in the first column.

(b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron (that is, the entries along the “diagonal” in the table must be 1’s). To find out how many

15 1

chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 12 1

chaldron = 1 bag. Thus, the next entry in that second column is 12 = 8.33  102. Similarly, 1 pottle = 1.74  103 chaldron and 1 gill = 9.24  106 chaldron.

(c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1 pottle = 2.08  102 bag, and 1 gill = 1.11  104 bag. (d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 pottle, 1 pottle = 1 pottle, and 1 gill = 5.32  103 pottle. (e) In the last column (under “gill”), we obtain 1 chaldron = 1.08  105 gill, 1 bag = 9.02  103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill. (f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3. 37. The volume of one unit is 1 cm3 = 1  106 m3, so the volume of a mole of them is 6.02  1023 cm3 = 6.02  1017 m3. The cube root of this number gives the edge length: 8.4 105 m3 . This is equivalent to roughly 8  102 km. 38. (a) Using the fact that the area A of a rectangle is (width) length), we find Atotal   3.00 acre    25.0 perch  4.00 perch    40 perch  4 perch   2   3.00 acre     100 perch 1acre   2  580 perch .

We multiply this by the perch2  rood conversion factor (1 rood/40 perch2) to obtain the answer: Atotal = 14.5 roods. (b) We convert our intermediate result in part (a):



Atotal  580 perch



 16.5ft  5 2    1.58  10 ft .  1perch 

Now, we use the feet  meters conversion given in Appendix D to obtain



Atotal  1.58  10 ft



 1m  4 2    1.47  10 m .  3.281ft 

CHAPTER 1

39. THINK This problem compares the U.K. gallon with U.S. gallon, two non-SI units for volume. The interpretation of the type of gallons, whether U.K. or U.S., affects the amount of gasoline one calculates for traveling a given distance. EXPRESS If the fuel consumption rate is R (in miles/gallon), then the amount of gasoline (in gallons) needed for a trip of distance d (in miles) would be V (gallon) 

d (miles) R (miles/gallon)

Since the car was manufactured in U.K., the fuel consumption rate is calibrated based on U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S. gallon. Note also that since 1 U.K. gallon  4.5460900 L and 1 U.S. gallon  3.7854118 L , the relationship between the two is  1 U.S. gallon  1 U.K. gallon  (4.5460900 L)    1.20095 U.S. gallons  3.7854118 L 

ANALYZE (a) The amount of gasoline actually required is V 

750 miles  18.75 U.K. gallons  18.8 U.K. gallons 40 miles/U.K. gallon

This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. (b) Using the conversion factor found above, this is equivalent to

 1.20095 U.S. gallons  V   18.75 U.K. gallons      22.5 U.S. gallons 1 U.K. gallon   LEARN One U.K. gallon is greater than one U.S gallon by roughly a factor of 1.2 in volume. Therefore, 40 mi/U.K. gallon is less fuel-efficient than 40 mi/U.S. gallon. 40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq. 1-9 and rounding off appropriately. Thus, the answer is 6.0  1026. 41. THINK This problem involves converting cord, a non-SI unit for volume, to SI unit. EXPRESS Using the (exact) conversion 1 in. = 2.54 cm = 0.0254 m for length, we have

 0.0254 m  1 ft  12 in  (12 in.)     0.3048 m .  1in  Thus, 1 ft 3  (0.3048 m)3  0.0283 m3 for volume (these results also can be found in Appendix D). ANALYZE The volume of a cord of wood is V  (8 ft)  (4 ft)  (4 ft)  128 ft 3 . Using the conversion factor found above, we obtain  0.0283 m3  3 V  1 cord  128 ft 3  (128 ft 3 )     3.625 m 3 1 ft    1  which implies that 1 m3    cord  0.276 cord  0.3 cord .  3.625 

LEARN The unwanted units ft3 all cancel out, as they should. In conversions, units obey the same algebraic rules as variables and numbers. 42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 1-9, we find  1.6605402  1027 kg  26 18u = 18u     3.0  10 kg. 1u   (b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules: 1.4  1021 N  5  1046. 3.0  10 26 43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3  106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3  106 mg/week)/(604800 s/week) = 3.8 mg/s. 44. The volume of the water that fell is

   2.0 in.   26 km    26  10 m   0.0508 m 

V  26 km

 1000 m     1 km 

 0.0254 m    1 in. 

 2.0 in. 

 1.3  106 m3 . We write the mass-per-unit-volume (density) of the water as:   The mass of the water that fell is therefore given by m = V:



m  1  103 kg m3

m  1  103 kg m3 . V

 1.3  10 m   1.3  10 kg. 6

CHAPTER 1

45. The number of seconds in a year is 3.156  107. This is listed in Appendix D and results from the product (365.25 day/y) (24 h/day) (60 min/h) (60 s/min). (a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by

106  104 u - day, 10 10

which may also be expressed as 104 u - day

u - sec I h FGH 86400 J  8.6 u - sec. 1 u - day K

46. The volume removed in one year is V = (75  104 m2 ) (26 m)  2  107 m3 ,

F 1 km IJ  0.020 km . which we convert to cubic kilometers: V  c2  10 m h G H 1000 mK 3

47. THINK This problem involves expressing the speed of light in astronomical units per minute. EXPRESS We first convert meters to astronomical units (AU), and seconds to minutes, using 1000 m 1 km, 1 AU 1.50 108 km, 60 s 1 min. ANALYZE Using the conversion factors above, the speed of light can be rewritten as

 3.0  108 m   1 km     60 s  AU c  3.0  108 m/s        0.12 AU min. 8 s    1000 m   1.50  10 km   min  LEARN When expressed the speed of light c in AU/min, we readily see that it takes about 8.3 (= 1/0.12) minutes for sunlight to reach the Earth (i.e., to travel a distance of 1 AU). 48. Since one atomic mass unit is 1 u  1.66 1024 g (see Appendix D), the mass of one mole of atoms is about m  (1.66 1024 g)(6.02 1023 )  1g. On the other hand, the mass of one mole of atoms in the common Eastern mole is

19 m 

75 g  10 g 7.5

Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is m 10 g   1.66 1023 g  10 u. 23 N A 6.02 10 49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain

(b) Similarly, we find

1 ken 2 1.972 m 2   3.88. 1 m2 1 m2 1 ken 3 197 . 3 m3   7.65. 1 m3 1 m3

 r 2 h    3.00  5.50  156 ken 3. 2

(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19  103 m3. 50. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km. 51. (a) For the minimum (43 cm) case, 9 cubits converts as follows:

 0.43m  9cubits   9cubits     3.9m.  1cubit   0.53m  And for the maximum (53 cm) case we have 9cubits   9cubits     4.8m.  1cubit  (b) Similarly, with 0.43 m  430 mm and 0.53 m  530 mm, we find 3.9  103 mm and 4.8  103 mm, respectively. (c) We can convert length and diameter first and then compute the volume, or first compute the volume and then convert. We proceed using the latter approach (where d is diameter and  is length).

CHAPTER 1

Vcylinder, min 

 4



d  28 cubit  28 cubit



 0.43m  3    2.2 m .  1 cubit 

Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max = 4.2 m3. 52. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors: acre 4047 m  25 wp   1001 wphide   110 1 hide   1 acre  2

11 barn  

1  1028 m2 1 barn



 1  1036.

53. THINK The objective of this problem is to convert the Earth-Sun distance (1 AU) to parsecs and light-years. EXPRESS To relate parsec (pc) to AU, we note that when  is measured in radians, it is equal to the arc length s divided by the radius R. For a very large radius circle and small value of , the arc may be approximated as the straight line-segment of length 1 AU. Thus,  1 arcmin    2 radian  1 6   1 arcsec  1 arcsec       4.85 10 rad .  60 arcsec  60 arcmin   360  Therefore, one parsec is s 1 AU 1 pc    2.06  105 AU .  4.85  106 Next, we relate AU to light-year (ly). Since a year is about 3.16  107 s,





1ly  186,000 mi s  3.16  107 s  5.9  1012 mi .

ANALYZE (a) Since 1 pc  2.06  105 AU , inverting the relation gives

  1 pc 6 1 AU  1 AU     4.9  10 pc. 5  2.06  10 AU  (b) Given that 1 AU together lead to

92.9 106 mi and 1 ly  5.9  1012 mi , the two expressions

  1 ly 5 1 AU  92.9 106 mi  (92.9 106 mi)    1.57 10 ly . 12 5.9  10 mi  

21 LEARN Our results can be further combined to give 1 pc  3.2 ly. From the above expression, we readily see that it takes 1.57 10 5 y , or about 8.3 min, for Sunlight to travel a distance of 1 AU to reach the Earth. 54. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639  102 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460 ft2/gal becomes 2

 460 ft 2  1 m   1 gal  2 460 ft /gal        11.3 m L.  gal   3.28 ft   3.79 L  2

(b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes

 11.3 m2  1000 L  11.3 m /L    1.13  10 4 m 1.  3  L  1 m   2

(c) The inverse of the original quantity is (460 ft2/gal)1 = 2.17  103 gal/ft2. (d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the paint thickness [it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b)]. 55. (a) The receptacle is a volume of (40 cm)(40 cm)(30 cm) = 48000 cm3 = 48 L = (48)(16)/11.356 = 67.63 standard bottles, which is a little more than 3 nebuchadnezzars (the largest bottle indicated). The remainder, 7.63 standard bottles, is just a little less than 1 methuselah. Thus, the answer to part (a) is 3 nebuchadnezzars and 1 methuselah. (b) Since 1 methuselah.= 8 standard bottles, then the extra amount is 8  7.63 = 0.37 standard bottle. (c) Using the conversion factor 16 standard bottles = 11.356 L, we have 11.356 L   0.37 standard bottle  (0.37 standard bottle)    0.26 L.  16 standard bottles 

56. The mass of the pig is 3.108 slugs, or (3.108)(14.59) = 45.346 kg. Referring now to the corn, a U.S. bushel is 35.238 liters. Thus, a value of 1 for the corn-hog ratio would be equivalent to 35.238/45.346 = 0.7766 in the indicated metric units. Therefore, a value of 5.7 for the ratio corresponds to 5.7(0.777)  4.4 in the indicated metric units. 57. Two jalapeño peppers have spiciness = 8000 SHU, and this amount multiplied by 400 (the number of people) is 3.2 106 SHU, which is roughly ten times the SHU value for a

CHAPTER 1

22 single habanero pepper. required SHU value.

More precisely, 10.7 habanero peppers will provide that total

58. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where x = 0.05 m is the increase in horizontal depth per step)  4.57  x  Nsteps x     0.05 m   1.2 m.  0.19 

However, we can approach this more carefully by noting that if there are N = 4.57/.19  24 rises then under normal circumstances we would expect N  1 = 23 runs (horizontal pieces) in that staircase. This would yield (23)(0.05 m) = 1.15 m, which - to two significant figures - agrees with our first result. 59. The volume of the filled container is 24000 cm3 = 24 liters, which (using the conversion given in the problem) is equivalent to 50.7 pints (U.S). The expected number is therefore in the range from 1317 to 1927 Atlantic oysters. Instead, the number received is in the range from 406 to 609 Pacific oysters. This represents a shortage in the range of roughly 700 to 1500 oysters (the answer to the problem). Note that the minimum value in our answer corresponds to the minimum Atlantic minus the maximum Pacific, and the maximum value corresponds to the maximum Atlantic minus the minimum Pacific. 60. (a) We reduce the stock amount to British teaspoons: 1 breakfastcup = 2  8  2  2 = 64 teaspoons 1 teacup

= 8  2  2 = 32 teaspoons

6 tablespoons = 6  2  2  24 teaspoons 1 dessertspoon = 2 teaspoons

which totals to 122 British teaspoons, or 122 U.S. teaspoons since liquid measure is being used. Now with one U.S cup equal to 48 teaspoons, upon dividing 122/48  2.54, we find this amount corresponds to 2.5 U.S. cups plus a remainder of precisely 2 teaspoons. In other words, 122 U.S. teaspoons = 2.5 U.S. cups + 2 U.S. teaspoons.

(b) For the nettle tops, one-half quart is still one-half quart. (c) For the rice, one British tablespoon is 4 British teaspoons which (since dry-goods measure is being used) corresponds to 2 U.S. teaspoons. (d) A British saltspoon is 21 British teaspoon which corresponds (since dry-goods measure is again being used) to 1 U.S. teaspoon.

Chapter 2 1. The speed (assumed constant) is v = (90 km/h)(1000 m/km)  (3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s)  13 m. 2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find 73.2 m  73.2 m vavg  73.2 m 73.2 m  1.74 m/s. 1.22 m/s  3.05 m (b) Using the fact that distance = vt while the velocity v is constant, we find vavg 

(122 . m / s)(60 s)  (3.05 m / s)(60 s)  2.14 m / s. 120 s

(c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer.

3. THINK This one-dimensional kinematics problem consists of two parts, and we are asked to solve for the average velocity and average speed of the car. EXPRESS Since the trip consists of two parts, let the displacements during first and second parts of the motion be x1 and x2, and the corresponding time intervals be t1 and t2, respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is simply x = x1 + x2, and the total time for the trip is t = t1 + t2. Using the definition of average velocity given in Eq. 2-2, we have x x1  x2 vavg   . t t1  t2

CHAPTER 2

To find the average speed, we note that during a time t if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with d  | x |  vt. ANALYZE (a) During the first part of the motion, the displacement is x1 = 40 km and the time taken is (40 km) t1   133 . h. (30 km / h) Similarly, during the second part of the trip the displacement is x2 = 40 km and the time interval is (40 km) t2   0.67 h. (60 km / h) The total displacement is x = x1 + x2 = 40 km + 40 km = 80 km, and the total time elapsed is t = t1 + t2 = 2.00 h. Consequently, the average velocity is

vavg 

x (80 km)   40 km/h. t (2.0 h)

(b) In this case, the average speed is the same as the magnitude of the average velocity: savg  40 km/h. (c) The graph of the entire trip, shown below, consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (t1, x1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (t1, x1) to (t, x) = (2.00 h, 80 km).

From the graphical point of view, the slope of the dashed line drawn from the origin to (t, x) represents the average velocity. LEARN The average velocity is a vector quantity that depends only on the net displacement (also a vector) between the starting and ending points. 4. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the

25 motion) we have speed = D/t. Thus, the average speed is Dup  Ddown t up  tdown



D D  vup vdown

which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the average speed. 5. THINK In this one-dimensional kinematics problem, we’re given the position function x(t), and asked to calculate the position and velocity of the object at a later time. EXPRESS The position function is given as x(t) = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3. The position of the object at some instant t0 is simply given by x(t0). For the time interval t1  t  t2 , the displacement is x  x(t2 )  x(t1 ) . Similarly, using Eq. 2-2, the average velocity for this time interval is x x(t2 )  x(t1 ) vavg   . t t2  t1 ANALYZE (a) Plugging in t = 1 s into x(t) yields x(1 s) = (3 m/s)(1 s) – (4 m/s2)(1 s)2 + (1 m/s3)(1 s)3 = 0. (b) With t = 2 s we get x(2 s) = (3 m/s)(2 s) – (4 m/s2) (2 s)2 + (1 m/s3)(2 s)3 = –2 m. (c) With t = 3 s we have x (3 s) = (3 m/s)(3 s) – (4 m/s2) (3 s)2 + (1 m/s3)(3 s)3 = 0 m. (d) Similarly, plugging in t = 4 s gives x(4 s) = (3 m/s)(4 s) – (4 m/s2)(4 s)2 + (1 m/s3) (4 s)3 = 12 m. (e) The position at t = 0 is x = 0. Thus, the displacement between t = 0 and t = 4 s is x  x(4 s)  x(0)  12 m  0  12 m. (f) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement: x  x(4 s)  x(2 s)  12 m  (2 m)  14 m . Thus, the average velocity is x 14 m vavg    7 m/s. t 2s (g) The position of the object for the interval 0  t  4 is plotted below. The straight line drawn from the point at (t, x) = (2 s, –2 m) to (4 s, 12 m) would represent the average velocity, answer for part (f).

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LEARN Our graphical representation illustrates once again that the average velocity for a time interval depends only on the net displacement between the starting and ending points. 6. Huber’s speed is v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m for Whittingham is  x 200 m t    5.554 s. v1 36 m/s 7. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time that elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a distance of x = vt = (60 km/h)(1.0 h) = 60 km. 8. The amount of time it takes for each person to move a distance L with speed vs is t  L / vs . With each additional person, the depth increases by one body depth d (a) The rate of increase of the layer of people is R

dv (0.25 m)(3.50 m/s) d d   s   0.50 m/s t L / vs L 1.75 m

(b) The amount of time required to reach a depth of D  5.0 m is D 5.0 m t   10 s R 0.50 m/s 9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, respectively. If the runners were equally fast, then savg1  savg 2

From this we obtain



L1 L2  . t1 t2

t   148.15  L2  L1   2  1  L1    1  L1  0.00135L1  1.4 m  147.95   t1  where we set L1  1000 m in the last step. Thus, if L1 and L2 are no different than about 1.4 m, then runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 by more than 1.4 m, then runner 2 would actually be faster. 10. Let vw be the speed of the wind and vc be the speed of the car. (a) Suppose during time interval t1 , the car moves in the same direction as the wind. Then the effective speed of the car is given by veff ,1  vc  vw , and the distance traveled is d  veff ,1t1  (vc  vw )t1 . On the other hand, for the return trip during time interval t2, the car moves in the opposite direction of the wind and the effective speed would be veff ,2  vc  vw . The distance traveled is d  veff ,2t2  (vc  vw )t2 . The two expressions can be rewritten as d d vc  vw  and vc  vw  t1 t2

1d d  Adding the two equations and dividing by two, we obtain vc     . Thus, 2  t1 t2  method 1 gives the car’s speed vc a in windless situation. (b) If method 2 is used, the result would be d 2d vc    (t1  t2 ) / 2 t1  t2

d d  vc  vw vc  vw

  v 2  vc2  vw2   vc 1   w   . vc   vc  

The fractional difference is 2

vc  vc  vw      (0.0240)2  5.76 104 . vc  vc 

11. The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is 1.25 hour, and the distance that remains is 160 km. Thus, a speed v = (160 km)/(1.25 h) = 128 km/h is needed. 12. (a) Let the fast and the slow cars be separated by a distance d at t = 0. If during the time interval t  L / vs  (12.0 m) /(5.0 m/s)  2.40 s in which the slow car has moved a distance of L  12.0 m , the fast car moves a distance of vt  d  L to join the line of slow cars, then the shock wave would remain stationary. The condition implies a separation of d  vt  L  (25 m/s)(2.4 s) 12.0 m  48.0 m. (b) Let the initial separation at t  0 be d  96.0 m. At a later time t, the slow and

CHAPTER 2

the fast cars have traveled x  vst and the fast car joins the line by moving a distance d  x . From

t

x dx  , vs v

we get x

vs 5.00 m/s d (96.0 m)  24.0 m, v  vs 25.0 m/s  5.00 m/s

which in turn gives t  (24.0 m) /(5.00 m/s)  4.80 s. Since the rear of the slow-car pack has moved a distance of x  x  L  24.0 m 12.0 m  12.0 m downstream, the speed of the rear of the slow-car pack, or equivalently, the speed of the shock wave, is vshock 

x 12.0 m   2.50 m/s. t 4.80 s

(c) Since x  L , the direction of the shock wave is downstream. 13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is savg1 

D (55 km/h)(T/2)  (90 km/h)(T / 2)   72.5 km/h T T

which should be rounded to 73 km/h. (b) Using the fact that time = distance/speed while the speed is constant, we find

savg2 

D D  D/2  68.3 km/h /2 T 55 km/h  90Dkm/h

which should be rounded to 68 km/h. (c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip 2D savg   70 km/h. D D 72.5 km/h  68.3 km/h (d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. (e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent is not to make the student go to an atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and connecting the origin to (t1, x1) = (T/2, 55T/2) and the second having a slope of 90 and connecting (t1, x1) to (T, D) where D = (55 + 90)T/2. The average velocity, from the

29 graphical point of view, is the slope of a line drawn from the origin to (T, D). The graph (not drawn to scale) is depicted below:

14. Using the general property

v

d dx

exp(bx)  b exp(bx) , we write

FG H

IJ K

FG IJ . H K

dx d (19t ) de  t   e  t  (19t )  dt dt dt

If a concern develops about the appearance of an argument of the exponential (–t) apparently having units, then an explicit factor of 1/T where T = 1 second can be inserted and carried through the computation (which does not change our answer). The result of this differentiation is v  16(1  t )e  t with t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s. Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given function x = 16te–t with x in meters. Therefore, we find x = 5.9 m. 15. We use Eq. 2-4 to solve the problem. (a) The velocity of the particle is v

dx d  (4  12t  3t 2 )  12  6t . dt dt

Thus, at t = 1 s, the velocity is v = (–12 + (6)(1)) = –6 m/s. (b) Since v  0, it is moving in the –x direction at t = 1 s. (c) At t = 1 s, the speed is |v| = 6 m/s. (d) For 0  t  2 s, |v| decreases until it vanishes. For 2  t  3 s, |v| increases from zero to the value it had in part (c). Then, |v| is larger than that value for t  3 s. (e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t  + , we have v  + ). One can check that v = 0 when t  2 s.

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30 (f) No. In fact, from v = –12 + 6t, we know that v  0 for t  2 s.

16. We use the functional notation x(t), v(t), and a(t) in this solution, where the latter two quantities are obtained by differentiation:

b g dxdtbt g   12t and abt g  dvdtbt g   12

vt  with SI units understood.

(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0. (b) We obtain x(0) = 4.0 m. (c) and (d) Requiring x(t) = 0 in the expression x(t) = 4.0 – 6.0t2 leads to t = 0.82 s for the times when the particle can be found passing through the origin. (e) We show both the asked-for graph (on the left) as well as the “shifted” graph that is relevant to part (f). In both cases, the time axis is given by –3  t  3 (SI units understood).

(f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression. (g) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = 0 point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first). 17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds. (a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 = 21.75 cm and x3 = 50.25 cm, respectively. The average velocity during the time interval 2.00  t  3.00 s is x 50.25 cm  2175 . cm vavg   t 3.00 s  2.00 s which yields vavg = 28.5 cm/s. 2 (b) The instantaneous velocity is v  dx dt  4.5t , which, at time t = 2.00 s, yields v = (4.5)(2.00)2 = 18.0 cm/s.

31 (c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s. (d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s. (e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the particle is at xm = (x2 + x3)/2 = 36 cm). Therefore, xm  9.75  15 . tm3

 tm  2.596

in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s. (f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate points.

18. (a) Taking derivatives of x(t) = 12t2 – 2t3 we obtain the velocity and the acceleration functions: v(t) = 24t – 6t2 and a(t) = 24 – 12t with length in meters and time in seconds. Plugging in the value t = 3 yields x(3)  54 m . (b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s. (c) For t = 3, a(3) = –12 m/s2. (d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity equation reveals t = 24/6 = 4 s for the time of maximum x. Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle. (e) From (d), we see that the x reaches its maximum at t = 4.0 s. (f) A maximum v requires a = 0, which occurs when t = 24/12 = 2.0 s. This, inserted into the velocity equation, gives vmax = 24 m/s. (g) From (f), we see that the maximum of v occurs at t = 24/12 = 2.0 s. (h) In part (e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be 24 – 12(4) = –24 m/s2.

CHAPTER 2

(i) The average velocity is defined by Eq. 2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)). Thus, 54  0 vavg = = 18 m/s. 3 0 19. THINK In this one-dimensional kinematics problem, we’re given the speed of a particle at two instants and asked to calculate its average acceleration. EXPRESS We represent the initial direction of motion as the +x direction. The average acceleration over a time interval t1  t  t2 is given by Eq. 2-7: aavg 

v v(t2 )  v(t1 )  . t t2  t1

ANALYZE Let v1 = +18 m/s at t1  0 and v2 = –30 m/s at t2 = 2.4 s. Using Eq. 2-7 we find v(t )  v(t1 ) (30 m/s)  (1 m/s) aavg  2    20 m/s 2 . t2  t1 2.4 s  0 LEARN The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the particle’s initial velocity. This makes sense because the velocity of the particle is decreasing over the time interval. With t1  0 , the velocity of the particle as a function of time can be written as v  v0  at  (18 m/s)  (20 m/s2 )t .

20. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating: dx t dv t vt    15t 2  20 and a t    30t t dt

with SI units understood. These expressions are used in the parts that follow. (a) From 0   15t 2  20 , we see that the only positive value of t for which the particle is (momentarily) stopped is t  20 / 15  12 . s. (b) From 0 = – 30t, we find a(0) = 0 (that is, it vanishes at t = 0). (c) It is clear that a(t) = – 30t is negative for t > 0. (d) The acceleration a(t) = – 30t is positive for t < 0. (e) The graphs are shown below. SI units are understood.

21. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during 5 min  t  10 min is taken to be the positive x direction. We also use the fact that x  vt ' when the velocity is constant during a time interval t' . (a) The entire interval considered is t = 8 – 2 = 6 min, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only t'  8  5  3min  180 s. His position at t = 2 min is x = 0 and his position at t = 8 min is x  vt   (2.2)(180)  396 m . Therefore, 396 m  0 vavg   110 . m / s. 360 s (b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 significant figures, aavg 

2.2 m / s  0  0.00611 m / s2 . 360 s

(c) Now, the entire interval considered is t = 9 – 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is t   9  5  4min  240 s ). His position at t  3 min is x = 0 and his position at t = 9 min is x  vt   (2.2)(240)  528 m . Therefore, 528 m  0 vavg   147 . m / s. 360 s (d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, aavg = 2.2/360 = 0.00611 m/s2 just as in part (b). (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for 0  t < 300 s and the linearly rising line for 300  t  600 s represents his constant-velocity motion. The lines represent the answers to part (a) and (c) in the sense that their slopes yield those results. The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one at v = 0 for 0  t < 300 s and the next at v = 2.2 m/s for 300 

CHAPTER 2

t  600 s). The indications of the average accelerations found in parts (b) and (d) would be dotted lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of aavg). 22. In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings. (a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI system. (b) Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3. (c) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by v = dx/dt = 2ct – 3bt2, v = 0 occurs for t = 0 and for 2c 2(3.0 m / s2 ) t   10 . s. 3b 3(2.0 m / s3 ) For t = 0, x = x0 = 0 and for t = 1.0 s, x = 1.0 m > x0. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1s). (d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to x(4 s)  (30 . m / s2 )(4.0 s) 2  (2.0 m / s3 )(4.0 s) 3   80 m . The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m. (e) Its displacement is x = x2 – x1, where x1 = 0 and x2 = –80 m. Thus, x  80 m . The velocity is given by v = 2ct – 3bt2 = (6.0 m/s2)t – (6.0 m/s3)t2. (f) Plugging in t = 1 s, we obtain v(1 s)  (6.0 m/s2 )(1.0 s)  (6.0 m/s3 )(1.0 s)2  0.

(g) Similarly, v(2 s)  (6.0 m/s2 )(2.0 s)  (6.0 m/s3 )(2.0 s) 2  12m/s . (h) v(3 s)  (6.0 m/s2 )(3.0 s)  (6.0 m/s3 )(3.0 s)2  36 m/s . (i) v(4 s)  (6.0 m/s2 )(4.0 s)  (6.0 m/s3 )(4.0 s) 2  72 m/s . The acceleration is given by a = dv/dt = 2c – 6b = 6.0 m/s2 – (12.0 m/s3)t. (j) Plugging in t = 1 s, we obtain a(1 s)  6.0 m/s2  (12.0 m/s3 )(1.0 s)  6.0 m/s2 . (k) a(2 s)  6.0 m/s2  (12.0 m/s3 )(2.0 s)  18 m/s 2 .

35 (l) a(3 s)  6.0 m/s2  (12.0 m/s3 )(3.0 s)  30 m/s2 . (m) a(4 s)  6.0 m/s2  (12.0 m/s3 )(4.0 s)   42 m/s 2 . 23. THINK The electron undergoes a constant acceleration. Given the final speed of the electron and the distance it has traveled, we can calculate its acceleration. EXPRESS Since the problem involves constant acceleration, the motion of the electron can be readily analyzed using the equations given in Table 2-1:

v  v0  at

(2  11)

1 x  x0  v0t  at 2 2

(2  15)

v 2  v02  2a( x  x0 )

(2  16)

The acceleration can be found by solving Eq. 2-16. ANALYZE With v0  1.50 105 m/s , v  5.70 106 m/s , x0 = 0 and x = 0.010 m, we find the average acceleration to be

a

v 2  v02 (5.7 106 m/s)2  (1.5 105 m/s) 2   1.62 1015 m/s 2 . 2x 2(0.010 m)

LEARN It is always a good idea to apply other equations in Table 2-1 not used for solving the problem as a consistency check. For example, since we now know the value of the acceleration, using Eq. 2-11, the time it takes for the electron to reach its final speed would be v  v0 5.70 106 m/s  1.5 105 m/s t   3.426 109 s a 1.62 1015 m/s 2 Substituting the value of t into Eq. 2-15, the distance the electron travels is 1 1 x  x0  v0t  at 2  0  (1.5 105 m/s)(3.426 109 s)  (1.62 1015 m/s 2 )(3.426 109 s)2 2 2  0.01 m

This is what was given in the problem statement. So we know the problem has been solved correctly. 24. In this problem we are given the initial and final speeds, and the displacement, and are asked to find the acceleration. We use the constant-acceleration equation given in Eq. 2-16, v2 = v20 + 2a(x – x0). (a) Given that v0  0 , v  1.6 m/s, and x  5.0 m, the acceleration of the spores during the launch is

CHAPTER 2

a

v 2  v02 (1.6 m/s)2   2.56 105 m/s 2  2.6 104 g 6 2x 2(5.0 10 m)

(b) During the speed-reduction stage, the acceleration is

a

v 2  v02 0  (1.6 m/s)2   1.28 103 m/s 2  1.3 102 g 2x 2(1.0 103 m)

The negative sign means that the spores are decelerating. 25. We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given v0 = 0; v = 20 m/s and a = 2.0 m/s2. In part 2, the vehicle decelerates from its highest speed to a halt; we are given v0 = 20 m/s; v = 0 and a = –1.0 m/s2 (negative because the acceleration vector points opposite to the direction of motion). (a) From Table 2-1, we find t1 (the duration of part 1) from v = v0 + at. In this way, 20  0  2.0t1 yields t1 = 10 s. We obtain the duration t2 of part 2 from the same equation. Thus, 0 = 20 + (–1.0)t2 leads to t2 = 20 s, and the total is t = t1 + t2 = 30 s. (b) For part 1, taking x0 = 0, we use the equation v2 = v20 + 2a(x – x0) from Table 2-1 and find v 2  v02 (20 m/s)2  (0) 2 x   100 m . 2a 2(2.0 m/s 2 ) This position is then the initial position for part 2, so that when the same equation is used in part 2 we obtain v 2  v02 (0)2  (20 m/s) 2 . x  100 m   2a 2(1.0 m/s 2 ) Thus, the final position is x = 300 m. That this is also the total distance traveled should be evident (the vehicle did not "backtrack" or reverse its direction of motion). 26. The constant-acceleration condition permits the use of Table 2-1. (a) Setting v = 0 and x0 = 0 in v2  v02  2a( x  x0 ) , we find 1 v02 1 (5.00  106 ) 2 x    0.100 m . 2 a 2 1.25  1014

Since the muon is slowing, the initial velocity and the acceleration must have opposite signs. (b) Below are the time plots of the position x and velocity v of the muon from the moment it enters the field to the time it stops. The computation in part (a) made no reference to t, so that other equations from Table 2-1 (such as v  v0  at and

x  v0t  12 at 2 ) are used in making these plots.

27. We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s. (a) Since we wish to calculate the velocity for a time before t = 0, we set t = –2.5 s. Thus, Eq. 2-11 gives v  (9.6 m / s)  3.2 m / s2 ( 2.5 s)  16 . m / s.

(b) Now, t = +2.5 s and we find v  (9.6 m / s)  3.2 m / s2 (2.5 s)  18 m / s. 28. We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s2. We also take x0 = 0. (a) The time to come to a halt is found using Eq. 2-11: 0  v0  at  t 

24.6 m/s  5.00 s .  4.92 m/s 2

(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 2-16 (since it does not depend on our answer to part (a)).

0  v02  2ax  x  

(24.6 m/s) 2  61.5 m . 2   4.92 m/s 2 

(c) Using these results, we plot v0t  12 at 2 (the x graph, shown next, on the left) and v0 + at (the v graph, on the right) over 0  t  5 s, with SI units understood.

CHAPTER 2

29. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s2 and a2 = –1.22 m/s2. We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s. (a) We denote x as the distance moved during t1, and use Eq. 2-16: v 2  v02  2a1x  x 

(b) Using Eq. 2-11, we have t1 

(5.08 m/s)2  10.59 m  10.6 m. 2(1.22 m/s 2 )

v  v0 5.08 m/s   4.17 s. a1 1.22 m/s 2

The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances traveled during t1 and t2 are the same so that they total to 2(10.59 m) = 21.18 m. This implies that for a distance of 190 m – 21.18 m = 168.82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is

t3 

168.82 m  33.21 s. 5.08 m / s

Therefore, the total time is 8.33 s + 33.21 s  41.5 s. 30. We choose the positive direction to be that of the initial velocity of the car (implying that a < 0 since it is slowing down). We assume the acceleration is constant and use Table 2-1. (a) Substituting v0 = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = –5.2 m/s2 into v = v0 + at, we obtain t

25 m / s  38 m / s  2.5 s . 5.2 m / s2

(b) We take the car to be at x = 0 when the brakes are applied (at time t = 0). Thus, the coordinate of the car as a function of time is given by x   38 m/s  t 

1 5.2 m/s 2  t 2  2

in SI units. This function is plotted from t = 0 to t = 2.5 s on the graph to the right. We have not shown the v-vs-t graph here; it is a descending straight line from v0 to v. 31. THINK The rocket ship undergoes a constant acceleration from rest, and we want to know the time elapsed and the distance traveled when the rocket reaches a certain speed.

39 EXPRESS Since the problem involves constant acceleration, the motion of the rocket can be readily analyzed using the equations in Table 2-1: v  v0  at (2  11)

1 x  x0  v0t  at 2 2 v 2  v02  2a( x  x0 )

(2  15) (2  16)

ANALYZE (a) Given that a  9.8 m/s2 , v0  0 and v  0.1c  3.0 107 m/s , we can solve v  v0  at for the time:

t

v  v0 3.0 107 m/s  0   3.1106 s 2 a 9.8 m/s

which is about 1.2 months. So it takes 1.2 months for the rocket to reach a speed of 0.1c starting from rest with a constant acceleration of 9.8 m/s2. (b) To calculate the distance traveled during this time interval, we evaluate x  x0  v0t  12 at 2 , with x0 = 0 and v0 0 . The result is 1 x   9.8 m/s2  (3.1106 s)2  4.6 1013 m. 2 LEARN In solving parts (a) and (b), we did not use Eq. (2-16): v2  v02  2a( x  x0 ) . This equation can be used to check our answers. The final velocity based on this equation is

v  v02  2a( x  x0 )  0  2(9.8 m/s2 )(4.6 1013 m  0)  3.0 107 m/s , which is what was given in the problem statement. So we know the problems have been solved correctly. 32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).

a

F 1000 m / kmIJ 1020 km / hg G b H 3600 s / h K v t



14 . s

 202.4 m / s2 .

In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows:  202.4 m/s 2  a g  21g . 2   9.8 m/s  33. THINK The car undergoes a constant negative acceleration to avoid impacting a barrier. Given its initial speed, we want to know the distance it has traveled and the time elapsed prior to the impact.

CHAPTER 2

EXPRESS Since the problem involves constant acceleration, the motion of the car can be readily analyzed using the equations in Table 2-1:

v  v0  at

(2  11)

1 x  x0  v0t  at 2 2

(2  15)

v 2  v02  2a( x  x0 )

(2  16)

We take x0 = 0 and v0 = 56.0 km/h = 15.55 m/s to be the initial position and speed of the car. Solving Eq. 2-15 with t = 2.00 s gives the acceleration a. Once a is known, the speed of the car upon impact can be found by using Eq. 2-11. ANALYZE (a) Using Eq. 2-15, we find the acceleration to be a

2( x  v0t ) 2 (24.0 m)  (15.55 m/s)(2.00 s)     3.56m/s 2 , 2 2 t (2.00 s)

or | a | 3.56 m/s2 . The negative sign indicates that the acceleration is opposite to the direction of motion of the car; the car is slowing down. (b) The speed of the car at the instant of impact is v  v0  at  15.55 m/s  (3.56 m/s2 )(2.00 s)  8.43 m/s

which can also be converted to 30.3 km/h. LEARN In solving parts (a) and (b), we did not use Eq. 1-16. This equation can be used as a consistency check. The final velocity based on this equation is

v  v02  2a( x  x0 )  (15.55 m/s)2  2(3.56 m/s2 )(24 m  0)  8.43 m/s , which is what was calculated in (b). This indicates that the problems have been solved correctly. 34. Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h =100/9 m/s speed (corresponding to a passing point of x2 = 76.6 m) of the red car. We have two equations (based on Eq. 2-17): 1

where t1 = x1  v1

where t2 = x2  v2

d – x1 = vo t1 + 2 a t12 d – x2 = vo t2 + 2 a t22

We simultaneously solve these equations and obtain the following results:

41 (a) The initial velocity of the green car is vo =  13.9 m/s. or roughly  50 km/h (the negative sign means that it’s along the –x direction). (b) The corresponding acceleration of the car is a =  2.0 m/s2 (the negative sign means that it’s along the –x direction). 35. The positions of the cars as a function of time are given by 1 1 xr (t )  xr 0  ar t 2  (35.0 m)  ar t 2 2 2 xg (t )  xg 0  vg t  (270 m)  (20 m/s)t

where we have substituted the velocity and not the speed for the green car. The two cars pass each other at t  12.0 s when the graphed lines cross. This implies that 1 (270 m)  (20 m/s)(12.0 s)  30 m  (35.0 m)  ar (12.0 s) 2 2

which can be solved to give ar  0.90 m/s2 . 36. (a) Equation 2-15 is used for part 1 of the trip and Eq. 2-18 is used for part 2: 1

where a1 = 2.25 m/s2 and x1 =

where a2 = 0.75 m/s2 and x2 =

x1 = vo1 t1 + 2 a1 t12 x2 = v2 t2  2 a2 t22

900 m 4 3(900) m 4

In addition, vo1 = v2 = 0. Solving these equations for the times and adding the results gives t = t1 + t2 = 56.6 s. (b) Equation 2-16 is used for part 1 of the trip:  900  2 2 v2 = (vo1)2 + 2a1x1 = 0 + 2(2.25)   = 1013 m /s  4 

which leads to v = 31.8 m/s for the maximum speed. 37. (a) From the figure, we see that x0 = –2.0 m. From Table 2-1, we can apply x – x0 = v0t +

1 2

at2

with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a:

CHAPTER 2

1 2 0.0   2.0 m   v0 1.0 s   a 1.0 s  2 1 2 6.0 m   2.0 m   v0  2.0 s   a  2.0 s  . 2 Solving these simultaneous equations yields the results v0 = 0 and a = 4.0 m/s2. (b) The fact that the answer is positive tells us that the acceleration vector points in the +x direction. 38. We assume the train accelerates from rest ( v0  0 and x0  0 ) at . m / s2 until the midway point and then decelerates at a2  134 a1  134 . m / s2 until it comes to a stop v2  0 at the next station. The velocity at the midpoint is v1, which occurs at x1 = 806/2 = 403m.

(a) Equation 2-16 leads to

v12  v02  2a1 x1  v1  2 1.34 m/s2   403 m   32.9 m/s. (b) The time t1 for the accelerating stage is (using Eq. 2-15) x1  v0t1 

2  403 m  1 2 a1t1  t1   24.53 s . 2 1.34 m/s2

Since the time interval for the decelerating stage turns out to be the same, we double this result and obtain t = 49.1 s for the travel time between stations. (c) With a “dead time” of 20 s, we have T = t + 20 = 69.1 s for the total time between start-ups. Thus, Eq. 2-2 gives 806 m . m/s .  117 vavg  69.1 s (d) The graphs for x, v and a as a function of t are shown below. The third graph, a(t), consists of three horizontal “steps” — one at 1.34 m/s2 during 0 < t < 24.53 s, and the next at –1.34 m/s2 during 24.53 s < t < 49.1 s and the last at zero during the “dead time” 49.1 s < t < 69.1 s).

39. (a) We note that vA = 12/6 = 2 m/s (with two significant figures understood). Therefore, with an initial x value of 20 m, car A will be at x = 28 m when t = 4 s. This must be the value of x for car B at that time; we use Eq. 2-15: 1

28 m = (12 m/s)t + 2 aB t2

where t = 4.0 s .

This yields aB = – 2.5 m/s2. (b) The question is: using the value obtained for aB in part (a), are there other values of t (besides t = 4 s) such that xA = xB ? The requirement is 1

20 + 2t = 12t + 2 aB t2 where aB  5 / 2. There

are

two

distinct

roots

unless

the

discriminant

10  2(20)(aB) is zero. In our case, it is zero – which means there is only one root. The cars are side by side only once at t = 4 s. 2

(d) We only care about real roots, which means 102  2(20)(aB)  0. If |aB| > 5/2 then there are no (real) solutions to the equation; the cars are never side by side. (e) Here we have 102  2(20)(aB) > 0 at two different times.

 two real roots. The cars are side by side

40. We take the direction of motion as +x, so a = –5.18 m/s2, and we use SI units, so v0 = 55(1000/3600) = 15.28 m/s.

CHAPTER 2

(a) The velocity is constant during the reaction time T, so the distance traveled during it is dr = v0T – (15.28 m/s) (0.75 s) = 11.46 m. We use Eq. 2-16 (with v = 0) to find the distance db traveled during braking:

v 2  v02  2adb  db  

(15.28 m/s)2 2  5.18 m/s 2 

which yields db = 22.53 m. Thus, the total distance is dr + db = 34.0 m, which means that the driver is able to stop in time. And if the driver were to continue at v0, the car would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s, which is (barely) enough time to enter the intersection before the light turns, which many people would consider an acceptable situation. (b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to the intersection, so the driver cannot stop without the front end of the car being a couple of meters into the intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is caught between a rock and a hard place. 41. The displacement (x) for each train is the “area” in the graph (since the displacement is the integral of the velocity). Each area is triangular, and the area of a triangle is 1/2(base) × (height). Thus, the (absolute value of the) displacement for one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) = 60 m. The initial “gap” between the trains was 200 m, and according to our displacement computations, the gap has narrowed by 160 m. Thus, the answer is 200 – 160 = 40 m. 42. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval, you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In light of the fact that the initial “gap” between cars was 25 m, this means the gap has narrowed by 10.0 m – that is, to a distance of 15.0 m between cars. (b) First, we add 0.4 s to the considerations of part (a). During a 2.4 s interval, you travel 73.33 m. The decelerating police car travels (using Eq. 2-15) 58.93 m during that time. The initial distance between cars of 25 m has therefore narrowed by 14.4 m. Thus, at the start of your braking (call it t0) the gap between the cars is 10.6 m. The speed of the police car at t0 is 30.56 – 5(2.4) = 18.56 m/s. Collision occurs at time t when xyou = xpolice (we choose coordinates such that your position is x = 0 and the police car’s position is x = 10.6 m at t0). Eq. 2-15 becomes, for each car: 1

xpolice – 10.6 = 18.56(t  t0) – 2 (5)(t  t0)2 1

xyou = 30.56(t  t0) – 2 (5)(t  t0)2

Subtracting equations, we find

10.6 = (30.56 – 18.56)(t  t0)  0.883 s = t  t0.

45 At that time your speed is 30.56 + a(t  t0) = 30.56 – 5(0.883)  26 m/s (or 94 km/h). 43. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity as vt and the locomotive’s velocity as v (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance x consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive v t . Therefore, vt  v x D  v t D     v . 2 t t t

We now use Eq. 2-11 to eliminate time from the equation. Thus,

vt  v D   v 2 v  vt / a

g which leads to F v  v  v IJ FG v  v IJ   1 bv  v g . aG H 2 K H D K 2D Hence, 1 FG 29 km  161 kmIJ  12888 km / h a 2(0.676 km) H h h K t



which we convert as follows:

F 1000 mIJ FG 1 h IJ  0.994 m / s a  c12888 km / h h G H 1 km K H 3600 sK 2

so that its magnitude is |a| = 0.994 m/s2. A graph is shown here for the case where a collision is just avoided (x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. The other case (where the collision is not quite avoided) would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet. 44. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. (a) Using y  v0t  21 gt 2 , with y = 0.544 m and t = 0.200 s, we find

CHAPTER 2

v0 

y  gt 2 / 2 0.544 m  (9.8 m/s 2 ) (0.200 s) 2 / 2   3.70 m/s . t 0.200 s

(b) The velocity at y = 0.544 m is v  v0  gt  3.70 m/s  (9.8 m/s2 ) (0.200 s)  1.74 m/s .

(c) Using v 2  v02  2 gy (with different values for y and v than before), we solve for the value of y corresponding to maximum height (where v = 0). y

v02 (3.7 m/s) 2   0.698 m. 2 g 2(9.8 m/s 2 )

Thus, the armadillo goes 0.698 – 0.544 = 0.154 m higher. 45. THINK As the ball travels vertically upward, its motion is under the influence of gravitational acceleration. The kinematics is one-dimensional. EXPRESS We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = –g = –9.8 m/s2 (we take downward to be the –y direction). We use the equations in Table 2-1 (with y replacing x) because this is a = constant motion: v  v0  gt (2  11)

1 y  y0  v0t  gt 2 2 v 2  v02  2 g ( y  y0 )

(2  15) (2  16)

We set y0 = 0. Upon reaching the maximum height y, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to y via the equation 0  v2  v02  2 gy. The time it takes for the ball to reach maximum height is given by v  v0  gt  0 , or t  v0 / g . Therefore, for the entire trip (from the time it leaves the ground until the time it returns to the ground), the total flight time is T  2t  2v0 / g. ANALYZE (a) At the highest point v = 0 and v0  2 gy . With y = 50 m, we find the initial speed of the ball to be v0  2 gy  2(9.8 m/s2 )(50 m)  31.3 m/s.

(b) Using the result from (a) for v0, the total flight time of the ball is T

2v0 2(31.3 m/s)   6.39 s g 9.8 m/s 2

47 (c) The plots of y, v and a as a function of time are shown below. The acceleration graph is a horizontal line at –9.8 m/s2. At t = 3.19 s, y = 50 m.

LEARN In calculating the total flight time of the ball, we could have used Eq. 2-15. At t  T  0 , the ball returns to its original position ( y  0 ). Therefore,

2v 1 y  v0T  gT 2  0  T  0 g 2 46. Neglect of air resistance justifies setting a = –g = –9.8 m/s2 (where down is our –y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with y replacing x). (a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have

v   v02  2 g y   0  2(9.8 m/s2 )(1700 m)  183 m/s . Its magnitude is therefore 183 m/s. (b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a raindrop to be about a gram or less, so that its mass and speed (from part (a)) would be less than that of a typical bullet, which is good news. But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the relatively healthy situation with which we are familiar. 47. THINK The wrench is in free fall with an acceleration a = –g = –9.8 m/s2. EXPRESS We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with y replacing x):

v  v0  gt

(2  11)

1 y  y0  v0t  gt 2 2

(2  15)

v 2  v02  2 g ( y  y0 )

(2  16)

Since the wrench had an initial speed v0 = 0, knowing its speed of impact allows us to apply Eq. 2-16 to calculate the height from which it was dropped.

CHAPTER 2

48 ANALYZE (a) Using v2  v02  2ay , we find the initial height to be y 

v02  v 2 0  (24 m/s) 2   29.4 m. 2a 2(9.8 m/s 2 )

So that it fell through a height of 29.4 m. (b) Solving v = v0 – gt for time, we obtain a flight time of t

v0  v 0  (24 m/s)   2.45 s. g 9.8 m/s 2

(c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. The acceleration graph is a horizontal line at –9.8 m/s2.

LEARN As the wrench falls, with a   g  0 , its speed increases but its velocity becomes more negative, as indicated by the second graph above. 48. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1 (with y replacing x). (a) Noting that y = y – y0 = –30 m, we apply Eq. 2-15 and the quadratic formula (Appendix E) to compute t:

v0  v02  2 gy 1 2 y  v0t  gt  t  2 g which (with v0 = –12 m/s since it is downward) leads, upon choosing the positive root (so that t > 0), to the result:

t

12 m/s  (12 m/s) 2  2(9.8 m/s 2 )(30 m)  1.54 s. 9.8 m/s 2

(b) Enough information is now known that any of the equations in Table 2-1 can be used to obtain v; however, the one equation that does not use our result from part (a) is Eq. 2-16: v  v02  2 gy  27.1 m / s

49 where the positive root has been chosen in order to give speed (which is the magnitude of the velocity vector). 49. THINK In this problem a package is dropped from a hot-air balloon which is ascending vertically upward. We analyze the motion of the package under the influence of gravity. EXPRESS We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. This allows us to use Table 2-1 (with y replacing x): v  v0  gt (2  11)

1 y  y0  v0t  gt 2 2

(2  15)

v 2  v02  2 g ( y  y0 )

(2  16)

We place the coordinate origin on the ground and note that the initial velocity of the package is the same as the velocity of the balloon, v0 = +12 m/s and that its initial coordinate is y0 = +80 m. The time it takes for the package to hit the ground can be found by solving Eq. 2-15 with y = 0. ANALYZE (a) We solve 0  y  y0  v0t  12 gt 2 for time using the quadratic formula (choosing the positive root to yield a positive value for t): 2 2 v0  v02  2 gy0 12 m/s  (12 m/s)  2  9.8 m/s   80 m  t   5.45 s . g 9.8 m/s2

(b) The speed of the package when it hits the ground can be calculated using Eq. 2-11. The result is

v  v0  gt  12 m/s  (9.8 m/s2 )(5.447 s)  41.38 m/s . Its final speed is 41.38 m/s. LEARN Our answers can be readily verified by using Eq. 2-16 which was not used in either (a) or (b). The equation leads to

v   v02  2 g ( y  y0 )   (12 m/s)2  2(9.8 m/s2 )(0  80 m)  41.38 m/s which agrees with that calculated in (b). 1

50. The y coordinate of Apple 1 obeys y – yo1 = – 2 g t2 where y = 0 when t = 2.0 s. This allows us to solve for yo1, and we find yo1 = 19.6 m.

The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with

CHAPTER 2

50 velocity v2) is

y – yo2 = v2(t – 1.0) – 2 g (t – 1.0)2

where yo2 = yo1 = 19.6 m and where y = 0 when t = 2.25 s. Thus, we obtain |v2| = 9.6 m/s, approximately. 51. (a) With upward chosen as the +y direction, we use Eq. 2-11 to find the initial velocity of the package: v = vo + at

 0 = vo – (9.8 m/s2)(2.0 s)

which leads to vo = 19.6 m/s. Now we use Eq. 2-15: 1

y = (19.6 m/s)(2.0 s) + 2 (–9.8 m/s2)(2.0 s)2  20 m . We note that the “2.0 s” in this second computation refers to the time interval 2 < t < 4 in the graph (whereas the “2.0 s” in the first computation referred to the 0 < t < 2 time interval shown in the graph). (b) In our computation for part (b), the time interval (“6.0 s”) refers to the 2 < t < 8 portion of the graph: 1

y = (19.6 m/s)(6.0 s) + 2 (–9.8 m/s2)(6.0 s)2  –59 m ,

or | y | 59 m .

52. The full extent of the bolt’s fall is given by 1

y – y0 = –2 g t2 where y – y0 = –90 m (if upward is chosen as the positive y direction). Thus the time for the full fall is found to be t = 4.29 s. The first 80% of its free-fall distance is given by –72 = –g 2/2, which requires time  = 3.83 s. (a) Thus, the final 20% of its fall takes t –  = 0.45 s. (b) We can find that speed using v = g. Therefore, |v| = 38 m/s, approximately. (c) Similarly, vfinal = g t 

|vfinal| = 42 m/s.

53. THINK This problem involves two objects: a key dropped from a bridge, and a boat moving at a constant speed. We look for conditions such that the key will fall into the boat. EXPRESS The speed of the boat is constant, given by vb = d/t, where d is the distance of the boat from the bridge when the key is dropped (12 m) and t is the time the key takes in falling.

51 To calculate t, we take the time to be zero at the instant the key is dropped, we compute the time t when y = 0 using y  y0  v0t  21 gt 2 , with y0  45 m. Once t is known, the speed of the boat can be readily calculated. ANALYZE Since the initial velocity of the key is zero, the coordinate of the key is given by y0  12 gt 2 . Thus, the time it takes for the key to drop into the boat is 2 y0 2(45 m)   3.03 s . g 9.8 m/s 2 12 m Therefore, the speed of the boat is vb   4.0 m/s. 3.03 s d d g LEARN From the general expression vb   , we see that d t 2 y0 2 y0 / g t

vb 1/ y0 . This agrees with our intuition that the lower the height from which the key is dropped, the greater the speed of the boat in order to catch it. 54. (a) We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Eq. 2-15 (with y replacing x) because this is constant acceleration motion. We use primed variables (except t) with the first stone, which has zero initial velocity, and unprimed variables with the second stone (with initial downward velocity –v0, so that v0 is being used for the initial speed). SI units are used throughout.

y  0  t  

1 2 gt 2

y   v0  t  1 

1 2 g  t  1 2

Since the problem indicates y’ = y = –43.9 m, we solve the first equation for t (finding t = 2.99 s) and use this result to solve the second equation for the initial speed of the second stone:

43.9 m   v0  1.99 s  

which leads to v0 = 12.3 m/s. (b) The velocity of the stones are given by

1 2 9.8 m/s2  1.99 s   2

CHAPTER 2

52 d (y)   gt , dt The plot is shown below: vy 

vy 

d (y)  v0  g (t  1) dt

55. THINK The free-falling moist-clay ball strikes the ground with a non-zero speed, and it undergoes deceleration before coming to rest. EXPRESS During contact with the ground its average acceleration is given by v , where v is the change in its velocity during contact with the ground and aavg  t t  20.0 103 s is the duration of contact. Thus, we must first find the velocity of the ball just before it hits the ground (y = 0). ANALYZE (a) Now, to find the velocity just before contact, we take t = 0 to be when it is dropped. Using Eq. 2-16 with y0  15.0 m , we obtain v   v02  2 g ( y  y0 )   0  2(9.8 m/s2 )(0  15 m)  17.15 m/s

where the negative sign is chosen since the ball is traveling downward at the moment of contact. Consequently, the average acceleration during contact with the ground is aavg 

v 0  (17.1 m/s)   857 m/s 2 . 3 t 20.0  10 s

(b) The fact that the result is positive indicates that this acceleration vector points upward. LEARN Since t is very small, it is not surprising to have a very large acceleration to stop the motion of the ball. In later chapters, we shall see that the acceleration is directly related to the magnitude and direction of the force exerted by the ground on the ball during the course of collision. 56. We use Eq. 2-16,

vB2 = vA2 + 2a(yB – yA), 1

with a = –9.8 m/s2, yB – yA = 0.40 m, and vB = 3 vA. It is then straightforward to solve: vA = 3.0 m/s, approximately.

57. The average acceleration during contact with the floor is aavg = (v2 – v1) / t,

53 where v1 is its velocity just before striking the floor, v2 is its velocity just as it leaves the floor, and t is the duration of contact with the floor (12  10–3 s). (a) Taking the y axis to be positively upward and placing the origin at the point where the ball is dropped, we first find the velocity just before striking the floor, using v12  v02  2 gy . With v0 = 0 and y = – 4.00 m, the result is v1   2 gy   2(9.8 m/s2 ) (4.00 m)  8.85 m/s

where the negative root is chosen because the ball is traveling downward. To find the velocity just after hitting the floor (as it ascends without air friction to a height of 2.00 m), we use v2  v22  2 g ( y  y0 ) with v = 0, y = –2.00 m (it ends up two meters below its initial drop height), and y0 = – 4.00 m. Therefore,

v2  2 g ( y  y0 )  2(9.8 m/s2 ) (2.00 m  4.00 m)  6.26 m/s . Consequently, the average acceleration is aavg 

v2  v1 6.26 m/s  ( 8.85 m/s)   1.26  103 m/s 2 . 3 t 12.0  10 s

(b) The positive nature of the result indicates that the acceleration vector points upward. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision. 58. We choose down as the +y direction and set the coordinate origin at the point where it was dropped (which is when we start the clock). We denote the 1.00 s duration mentioned in the problem as t – t' where t is the value of time when it lands and t' is one second prior to that. The corresponding distance is y – y' = 0.50h, where y denotes the location of the ground. In these terms, y is the same as h, so we have h –y' = 0.50h or 0.50h = y' . (a) We find t' and t from Eq. 2-15 (with v0 = 0): 1 2 y y  gt 2  t   2 g y

1 2 2y gt  t  . 2 g

Plugging in y = h and y' = 0.50h, and dividing these two equations, we obtain

2 0.50h / g t   0.50 . t 2h / g Letting t' = t – 1.00 (SI units understood) and cross-multiplying, we find

CHAPTER 2

t  100 .  t 0.50  t 

which yields t = 3.41 s.

100 . 1  0.50

(b) Plugging this result into y  12 gt 2 we find h = 57 m. (c) In our approach, we did not use the quadratic formula, but we did “choose a root” when we assumed (in the last calculation in part (a)) that 0.50 = +0.707 instead of –0.707. If we had instead let 0.50 = –0.707 then our answer for t would have been roughly 0.6 s, which would imply that t' = t – 1 would equal a negative number (indicating a time before it was dropped), which certainly does not fit with the physical situation described in the problem. 59. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y-axis. (a) The time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the floor t1 can be computed from Eq. 2-15, with v0 = 0 and y1 = –2.00 m.

1 2 y 2(2.00 m) y1   gt12  t1    0.639 s . 2 g 9.8 m/s 2 At that moment, the fourth drop begins to fall, and from the regularity of the dripping we conclude that drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves the nozzle at t = 2(0.213 s) = 0.426 s. Therefore, the time in free fall (up to the moment drop 1 lands) for drop 2 is t2 = t1 – 0.213 s = 0.426 s. Its position at the moment drop 1 strikes the floor is 1 1 y2   gt22   (9.8 m/s 2 )(0.426 s)2  0.889 m, 2 2

or about 89 cm below the nozzle. (b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 –0.426 s = 0.213 s. Its position at the moment drop 1 strikes the floor is 1 1 y3   gt32   (9.8 m/s 2 )(0.213 s)2  0.222 m, 2 2

or about 22 cm below the nozzle. 60. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero)

55 v  v0  gt  0  v0   9.8 m/s2   2.5 s 

so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the tower (taking y0 = 0 at the ground level) y  y0  v0t 

1 2 1 2 at  y  0   24.5 m/s 1.5 s    9.8 m/s 2  1.5 s  . 2 2

Thus, we obtain y = 26 m. 61. We choose down as the +y direction and place the coordinate origin at the top of the building (which has height H). During its fall, the ball passes (with velocity v1) the top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at time t2. We are told y2 – y1 = 1.20 m and t2 – t1 = 0.125 s. Using Eq. 2-15 we have

g 21 g bt  t g

y2  y1  v1 t2  t1 

which immediately yields v1 

1.20 m  12  9.8 m/s 2   0.125 s 

0.125 s

 8.99 m/s.

From this, Eq. 2-16 (with v0 = 0) reveals the value of y1: v12  2 gy1

 y1 

(8.99 m/s)2  4.12 m. 2(9.8 m/s 2 )

It reaches the ground (y3 = H) at t3. Because of the symmetry expressed in the problem (“upward flight is a reverse of the fall’’) we know that t3 – t2 = 2.00/2 = 1.00 s. And this means t3 – t1 = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces

1 y3  y1  v1 (t3  t1 )  g (t3  t1 ) 2 2 1 y3  4.12 m  (8.99 m/s) (1.125 s)  (9.8 m/s 2 ) (1.125 s) 2 2 which yields y3 = H = 20.4 m. 62. The height reached by the player is y = 0.76 m (where we have taken the origin of the y axis at the floor and +y to be upward). (a) The initial velocity v0 of the player is v0  2 gy  2(9.8 m/s2 ) (0.76 m)  3.86 m/s .

This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1

CHAPTER 2

= 0.76 m – 0.15 m = 0.61 m, his speed v1 satisfies v02  v12  2 gy1 , which yields

v1  v02  2 gy1  (3.86 m/s)2  2(9.80 m/s2 ) (0.61 m)  1.71 m/s . The time t1 that the player spends ascending in the top y1 = 0.15 m of the jump can now be found from Eq. 2-17: y1 

2  0.15 m  1  0.175 s  v1  v  t1  t1  2 1.71 m/s  0

which means that the total time spent in that top 15 cm (both ascending and descending) is 2(0.175 s) = 0.35 s = 350 ms. (b) The time t2 when the player reaches a height of 0.15 m is found from Eq. 2-15: 0.15 m  v0t2 

1 2 1 gt2  (3.86 m/s)t2  (9.8 m/s 2 )t22 , 2 2

which yields (using the quadratic formula, taking the smaller of the two positive roots) t2 = 0.041 s = 41 ms, which implies that the total time spent in that bottom 15 cm (both ascending and descending) is 2(41 ms) = 82 ms. 63. The time t the pot spends passing in front of the window of length L = 2.0 m is 0.25 s each way. We use v for its velocity as it passes the top of the window (going up). Then, with a = –g = –9.8 m/s2 (taking down to be the –y direction), Eq. 2-18 yields 1 L 1 L  vt  gt 2  v   gt . 2 t 2 The distance H the pot goes above the top of the window is therefore (using Eq. 2-16 with the final velocity being zero to indicate the highest point)

 2.00 m / 0.25 s  (9.80 m/s )(0.25 s) / 2  2.34 m. v 2  L / t  gt / 2  H   2g 2g 2(9.80 m/s2 ) 2

64. The graph shows y = 25 m to be the highest point (where the speed momentarily vanishes). The neglect of “air friction” (or whatever passes for that on the distant planet) is certainly reasonable due to the symmetry of the graph. (a) To find the acceleration due to gravity gp on that planet, we use Eq. 2-15 (with +y up) 1 1 2 y  y0  vt  g pt 2  25 m  0   0  2.5 s   g p  2.5 s  2 2 so that gp = 8.0 m/s2. (b) That same (max) point on the graph can be used to find the initial velocity.

y  y0 

1  v0  v  t 2



25 m  0 

1  v0  0  2.5 s  2

Therefore, v0 = 20 m/s. 65. The key idea here is that the speed of the head (and the torso as well) at any given time can be calculated by finding the area on the graph of the head’s acceleration versus time, as shown in Eq. 2-26:

 area between the acceleration curve  v1  v0     and the time axis, from t0 to t1  (a) From Fig. 2.15a, we see that the head begins to accelerate from rest (v0 = 0) at t0 = 110 ms and reaches a maximum value of 90 m/s2 at t1 = 160 ms. The area of this region is 1 area  (160  110) 103s   90 m/s 2   2.25 m/s 2 which is equal to v1, the speed at t1. (b) To compute the speed of the torso at t1=160 ms, we divide the area into 4 regions: From 0 to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the shape of a triangle with area 1 area B  (0.0600 s)(50.0 m/s 2 )  1.50 m/s . 2 From 100 to 120 ms, region C has the shape of a rectangle with area

area C  (0.0200 s) (50.0 m/s2 ) = 1.00 m/s. From 110 to 160 ms, region D has the shape of a trapezoid with area

1 (0.0400 s) (50.0  20.0) m/s 2  1.40 m/s. 2 Substituting these values into Eq. 2-26, with v0 = 0 then gives area D 

v1  0  0  1.50 m/s + 1.00 m/s + 1.40 m/s = 3.90 m/s,

or v1  3.90 m/s. 66. The key idea here is that the position of an object at any given time can be calculated by finding the area on the graph of the object’s velocity versus time, as shown in Eq. 2-30:  area between the velocity curve  x1  x0   .  and the time axis, from t0 to t1  (a) To compute the position of the fist at t = 50 ms, we divide the area in Fig. 2-37 into two regions. From 0 to 10 ms, region A has the shape of a triangle with area

CHAPTER 2

area A =

1 (0.010 s) (2 m/s) = 0.01 m. 2

From 10 to 50 ms, region B has the shape of a trapezoid with area area B =

1 (0.040 s) (2 + 4) m/s = 0.12 m. 2

Substituting these values into Eq. 2-30 with x0 = 0 then gives x1  0  0  0.01 m + 0.12 m = 0.13 m,

or x1  0.13 m. (b) The speed of the fist reaches a maximum at t1 = 120 ms. From 50 to 90 ms, region C has the shape of a trapezoid with area area C =

1 (0.040 s) (4 + 5) m/s = 0.18 m. 2

From 90 to 120 ms, region D has the shape of a trapezoid with area area D =

1 (0.030 s) (5 + 7.5) m/s = 0.19 m. 2

Substituting these values into Eq. 2-30, with x0 = 0 then gives x1  0  0  0.01 m + 0.12 m + 0.18 m + 0.19 m = 0.50 m,

or x1  0.50 m. 67. The problem is solved using Eq. 2-31:

 area between the acceleration curve  v1  v0     and the time axis, from t0 to t1  To compute the speed of the unhelmeted, bare head at t1 = 7.0 ms, we divide the area under the a vs. t graph into 4 regions: From 0 to 2 ms, region A has the shape of a triangle with area 1 area A = (0.0020 s) (120 m/s 2 ) = 0.12 m/s. 2 From 2 ms to 4 ms, region B has the shape of a trapezoid with area area B =

1 (0.0020 s) (120 + 140) m/s2 = 0.26 m/s. 2

59 From 4 to 6 ms, region C has the shape of a trapezoid with area

area C =

1 (0.0020 s) (140 + 200) m/s2 = 0.34 m/s. 2

From 6 to 7 ms, region D has the shape of a triangle with area 1 area D  (0.0010 s) (200 m/s 2 )  0.10 m/s. 2

Substituting these values into Eq. 2-31, with v0=0 then gives vunhelmeted  0.12 m/s  0.26 m/s  0.34 m/s  0.10 m/s  0.82 m/s.

Carrying out similar calculations for the helmeted head, we have the following results: From 0 to 3 ms, region A has the shape of a triangle with area 1 (0.0030 s) (40 m/s 2 ) = 0.060 m/s. 2 From 3 ms to 4 ms, region B has the shape of a rectangle with area area A =

area B  (0.0010 s) (40 m/s2 )  0.040 m/s.

From 4 to 6 ms, region C has the shape of a trapezoid with area area C =

1 (0.0020 s) (40 + 80) m/s 2 = 0.12 m/s. 2

From 6 to 7 ms, region D has the shape of a triangle with area 1 area D  (0.0010 s) (80 m/s 2 )  0.040 m/s. 2 Substituting these values into Eq. 2-31, with v0 = 0 then gives vhelmeted  0.060 m/s  0.040 m/s  0.12 m/s  0.040 m/s  0.26 m/s.

Thus, the difference in the speed is v  vunhelmeted  vhelmeted  0.82 m/s  0.26 m/s  0.56 m/s.

68. This problem can be solved by noting that velocity can be determined by the graphical integration of acceleration versus time. The speed of the tongue of the salamander is simply equal to the area under the acceleration curve: 1 1 1 v  area  (102 s)(100 m/s 2 )  (102 s)(100 m/s 2  400 m/s 2 )  (102 s)(400 m/s2 ) 2 2 2  5.0 m/s.

CHAPTER 2

69. Since v  dx / dt (Eq. 2-4), then x  v dt , which corresponds to the area under the v vs t graph. Dividing the total area A into rectangular (baseheight) and triangular 21 base  height areas, we have

 A0 t 2  A2 t 10  A10 t 12  A12 t 16 

FG H

IJ K

1 1 (2)(8)  (8)(8)  (2)(4)  (2)(4)  (4)(4) 2 2

with SI units understood. In this way, we obtain x = 100 m. 70. To solve this problem, we note that velocity is equal to the time derivative of a position function, as well as the time integral of an acceleration function, with the integration constant being the initial velocity. Thus, the velocity of particle 1 can be written as dx d v1  1   6.00t 2  3.00t  2.00   12.0t  3.00 . dt dt Similarly, the velocity of particle 2 is v2  v20   a2 dt  20.0   (8.00t )dt  20.0  4.00t 2 .

The condition that v1  v2 implies 12.0t  3.00  20.0  4.00t 2 

4.00t 2  12.0t  17.0  0

which can be solved to give (taking positive root) t  (3  26) / 2  1.05 s. Thus, the velocity at this time is v1  v2  12.0(1.05)  3.00  15.6 m/s. 71. (a) The derivative (with respect to time) of the given expression for x yields the “velocity” of the spot: 9 v(t) = 9 – 4 t2 with 3 significant figures understood. It is easy to see that v = 0 when t = 2.00 s. (b) At t = 2 s, x = 9(2) – ¾(2)3 = 12. Thus, the location of the spot when v = 0 is 12.0 cm from left edge of screen. 9

 9.00 cm/m2 (negative sign indicating leftward) when the spot is 12 cm from the left edge of screen. (d) Since v > 0 for times less than t = 2 s, then the spot had been moving rightward.

61 (e) As implied by our answer to part (c), it moves leftward for times immediately after t = 2 s. In fact, the expression found in part (a) guarantees that for all t > 2, v < 0 (that is, until the clock is “reset” by reaching an edge). (f) As the discussion in part (e) shows, the edge that it reaches at some t > 2 s cannot be the right edge; it is the left edge (x = 0). Solving the expression given in the problem statement (with x = 0) for positive t yields the answer: the spot reaches the left edge at t = 12 s  3.46 s. 72. We adopt the convention frequently used in the text: that "up" is the positive y direction. (a) At the highest point in the trajectory v = 0. Thus, with t = 1.60 s, the equation v = v0 – gt yields v0 = 15.7 m/s. 1

(b) One equation that is not dependent on our result from part (a) is y – y0 = vt + 2gt2;

this readily gives ymax – y0 = 12.5 m for the highest ("max") point measured relative to where it started (the top of the building). 1

(c) Now we use our result from part (a) and plug into y y0 = v0t + 2gt2 with t = 6.00 s and y = 0 (the ground level). Thus, we have

0 – y0 = (15.68 m/s)(6.00 s) – 2 (9.8 m/s2)(6.00 s)2. Therefore, y0 (the height of the building) is equal to 82.3 m. 73. We denote the required time as t, assuming the light turns green when the clock reads zero. By this time, the distances traveled by the two vehicles must be the same. (a) Denoting the acceleration of the automobile as a and the (constant) speed of the truck as v then 1 x  at 2  vt truck 2 car which leads to

FG H

t

Therefore,

IJ b g K

2v 2  9.5 m/s    8.6 s . a 2.2 m/s 2

x  vt   9.5 m/s 8.6 s   82 m .

(b) The speed of the car at that moment is vcar  at   2.2 m/s2  8.6 s   19 m/s .

74. If the plane (with velocity v) maintains its present course, and if the terrain continues its upward slope of 4.3°, then the plane will strike the ground after traveling

CHAPTER 2

x 

h 35 m   4655 . m  0.465 km. tan  tan 4.3

This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant) x 0.465 km t   0.000358 h  1.3 s. v 1300 km / h This, then, estimates the time available to the pilot to make his correction. 75. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-velocity (call it v0) type. Then the position of the car is given by x  v0tr  v0tb 

1 2 atb 2

where v0 is the initial velocity and a is the acceleration (which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating). After the brakes are applied the velocity of the car is given by v = v0 + atb. Using this equation, with v = 0, we eliminate tb from the first equation and obtain v 2 1 v02  v02 x  v0tr  0   v0tr  . a 2 a 2 a We write this equation for each of the initial velocities:

x1  v01tr 

2 1 v01 , 2 a

x2  v02tr 

2 1 v02 . 2 a

Solving these equations simultaneously for tr and a we get tr 

and a

2 2 v02 x1  v01 x2 v01v02 v02  v01

2 2 1 v02 v01  v01v02 . 2 v02 x1  v01 x2

(a) Substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m and v02 = 48.3 km/h = 13.4 m/s, we find tr 

2 2 v02 x1  v01 x2 (13.4 m/s) 2 (56.7 m)  (22.4 m/s) 2 (24.4 m)  v01v02 (v02  v01 ) (22.4 m/s)(13.4 m/s)(13.4 m/s  22.4 m/s)

 0.74 s.

(b) Similarly, substituting x1 = 56.7 m, v01 = 80.5 km/h = 22.4 m/s, x2 = 24.4 m, and

63 v02 = 48.3 km/h = 13.4 m/s gives 2 2  v01v02 1 v02v01 1 (13.4 m/s)(22.4 m/s) 2  (22.4 m/s)(13.4 m/s) 2 a  2 v02 x1  v01 x2 2 (13.4 m/s)(56.7 m)  (22.4 m/s)(24.4 m)

 6.2 m/s 2 .

The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded-off values are displayed in the above substitutions, what we have input into our calculators are the “exact” values (such as v02  161 12 m/s). 76. (a) A constant velocity is equal to the ratio of displacement to elapsed time. Thus, for the vehicle to be traveling at a constant speed v p over a distance D23 , the time delay should be t  D23 / v p . (b) The time required for the car to accelerate from rest to a cruising speed v p is t0  v p / a . During this time interval, the distance traveled is x0  at02 / 2  v2p / 2a.

The car then moves at a constant speed v p over a distance D12  x0  d to reach intersection 2, and the time elapsed is t1  ( D12  x0  d ) / v p . Thus, the time delay at intersection 2 should be set to v p D12  (v 2p / 2a)  d D12  x0  d ttotal  tr  t0  t1  tr    tr   a vp a vp vp

 tr 

1 v p D12  d  2 a vp

77. THINK The speed of the rod changes due to a nonzero acceleration. EXPRESS Since the problem involves constant acceleration, the motion of the rod can be readily analyzed using the equations given in Table 2-1. We take +x to be in the direction of motion, so 1000 m / km v  60 km / h   16.7 m / s 3600 s / h

g FGH

IJ K

and a > 0. The location where the rod starts from rest (v0 = 0) is taken to be x0 = 0. ANALYZE (a) Using Eq. 2-7, we find the average acceleration to be aavg 

v v  v0 16.7 m/s  0    3.09 m/s 2 . t t  t0 5.4 s  0

(b) Assuming constant acceleration a  aavg  3.09 m/s2 , the total distance traveled during the 5.4-s time interval is

CHAPTER 2

64 1 1 x  x0  v0t  at 2  0  0  (3.09 m/s 2 )(5.4 s)2  45 m 2 2

2  250 m  1 2 2x at  t    12.73 s 2 a 3.1 m/s 2

LEARN The displacement of the rod as a function of time can be written as 1 x(t )  (3.09 m/s 2 )t 2 . Note that we could have chosen Eq. 2-17 to solve for (b): 2 1 1 x   v0  v  t  16.7 m/s  5.4 s   45 m. 2 2 78. We take the moment of applying brakes to be t = 0. The deceleration is constant so that Table 2-1 can be used. Our primed variables (such as v0  72 km/h = 20 m/s ) refer to one train (moving in the +x direction and located at the origin when t = 0) and unprimed variables refer to the other (moving in the –x direction and located at x0 = +950 m when t = 0). We note that the acceleration vector of the unprimed train points in the positive direction, even though the train is slowing down; its initial velocity is v0 = –144 km/h = –40 m/s. Since the primed train has the lower initial speed, it should stop sooner than the other train would (were it not for the collision). Using Eq 2-16, it should stop (meaning v  0 ) at

 v   v0   0  (20 m/s)2  200 m . x  2

2a 

2 m/s 2

The speed of the other train, when it reaches that location, is

v  v02  2ax 

 40 m/s   2 1.0 m/s2   200 m  950 m  2

 10 m/s using Eq 2-16 again. Specifically, its velocity at that moment would be –10 m/s since it is still traveling in the –x direction when it crashes. If the computation of v had failed (meaning that a negative number would have been inside the square root) then we would have looked at the possibility that there was no collision and examined how far apart they finally were. A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still a good distance away from contact). 1

79. The y coordinate of Piton 1 obeys y – y01 = – 2 g t2 where y = 0 when t = 3.0 s.

This allows us to solve for yo1, and we find y01 = 44.1 m. The graph for the coordinate of Piton 2 (which is thrown apparently at t = 1.0 s with velocity v1) is

65 1

y – y02 = v1(t–1.0) – 2 g (t – 1.0)2 where y02 = y01 + 10 = 54.1 m and where (again) y = 0 when t = 3.0 s. obtain |v1| = 17 m/s, approximately.

Thus we

80. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus, v1 = +30 m/s, v2 = +50 m/s, and x2 – x1 = +160 m. (a) Using these subscripts, Eq. 2-16 leads to a

v22  v12 (50 m/s)2  (30 m/s) 2   5.0 m/s 2 . 2  x2  x1  2 160 m 

(b) We find the time interval corresponding to the displacement x2 – x1 using Eq. 2-17: t2  t1 

2  x2  x1  v1  v2



2 160 m  30 m/s  50 m/s

 4.0 s .

(c) Since the train is at rest (v0 = 0) when the clock starts, we find the value of t1 from Eq. 2-11: 30 m/s v1  v0  at1  t1   6.0 s . 5.0 m/s 2 (d) The coordinate origin is taken to be the location at which the train was initially at rest (so x0 = 0). Thus, we are asked to find the value of x1. Although any of several equations could be used, we choose Eq. 2-17: x1 

1 1  v0  v1  t1   30 m/s  6.0 s   90 m . 2 2

(e) The graphs are shown below, with SI units understood.

81. THINK The particle undergoes a non-constant acceleration along the +x-axis. An integration is required to calculate velocity. EXPRESS With a non-constant acceleration a(t )  dv / dt , the velocity of the

CHAPTER 2

66 t1

particle at time t1 is given by Eq. 2-27: v1  v0   a(t )dt , where v0 is the velocity at t0

time t0. In our situation, we have a  5.0t. In addition, we also know that v0  17 m/s at t0  2.0 s. ANALYZE Integrating (from t = 2 s to variable t = 4 s) the acceleration to get the velocity and using the values given in the problem, leads to t t 1 1 v  v0   adt  v0   (5.0t )dt  v0  (5.0)(t 2  t02 ) = 17 + 2 (5.0)(42 – 22) = 47 m/s. t0 t0 2

LEARN The velocity of the particle as a function of t is 1 1 v(t )  v0  (5.0)(t 2  t02 )  17  (5.0)(t 2  4)  7  2.5t 2 2 2 in SI units (m/s). Since the acceleration is linear in t, we expect the velocity to be quadratic in t, and the displacement to be cubic in t.

82. The velocity v at t = 6 (SI units and two significant figures understood) is 6

vgiven   adt . A quick way to implement this is to recall the area of a triangle ( 2 2

base × height). The result is v = 7 m/s + 32 m/s = 39 m/s.

83. The object, once it is dropped (v0 = 0) is in free fall (a = –g = –9.8 m/s2 if we take down as the –y direction), and we use Eq. 2-15 repeatedly. (a) The (positive) distance D from the lower dot to the mark corresponding to a certain reaction time t is given by y   D   21 gt 2 , or D = gt2/2. Thus, for t1  50.0 ms ,

c9.8 m / s hc50.0  10 sh  0.0123 m = 1.23 cm. D  2

3

 9.8 m/s   100  10 s   0.049 m = 4D . (b) For t = 100 ms, D  2

3

 9.8 m/s   150  10 s   0.11m = 9D . (c) For t = 150 ms, D  2

3

 9.8 m/s   200  10 s   0.196 m =16D . (d) For t = 200 ms, D  2

3

c9.8 m / s hc250  10 sh  0.306 m = 25D . (e) For t = 250 ms, D  2

3

67 84. We take the direction of motion as +x, take x0 = 0 and use SI units, so v = 1600(1000/3600) = 444 m/s. (a) Equation 2-11 gives 444 = a(1.8) or a = 247 m/s2. We express this as a multiple of g by setting up a ratio:  247 m/s 2  a g  25 g. 2   9.8 m/s  (b) Equation 2-17 readily yields x

1 1  v0  v  t   444 m/s 1.8 s   400 m. 2 2

85. Let D be the distance up the hill. Then average speed =

total distance traveled total time of travel

2D D D + 20 km/h 35 km/h

 25 km/h .

86. We obtain the velocity by integration of the acceleration: t

v  v0   (6.1 1.2t )dt  . 0

Lengths are in meters and times are in seconds. The student is encouraged to look at the discussion in Section 2-7 to better understand the manipulations here. (a) The result of the above calculation is v  v0  6.1 t  0.6 t 2 , where the problem states that v0 = 2.7 m/s. The maximum of this function is found by knowing when its derivative (the acceleration) is zero (a = 0 when t = 6.1/1.2 = 5.1 s) and plugging that value of t into the velocity equation above. Thus, we find v  18 m/s . (b) We integrate again to find x as a function of t: t

x  x0   v dt    (v0  6.1t   0.6 t 2 ) dt   v0t  3.05 t 2  0.2 t 3 . With x0 = 7.3 m, we obtain x = 83 m for t = 6. This is the correct answer, but one has the right to worry that it might not be; after all, the problem asks for the total distance traveled (and x  x0 is just the displacement). If the cyclist backtracked, then his total distance would be greater than his displacement. Thus, we might ask, "did he backtrack?" To do so would require that his velocity be (momentarily) zero at some point (as he reversed his direction of motion). We could solve the above quadratic equation for velocity, for a positive value of t where v = 0; if we did, we would find that at t = 10.6 s, a reversal does indeed happen. However, in the time interval we are concerned with in our problem (0 ≤ t ≤ 6 s), there is no reversal and the displacement is the same as the total distance traveled. 87. THINK In this problem we’re given two different speeds, and asked to find the difference in their travel times.

CHAPTER 2

EXPRESS The time is takes to travel a distance d with a speed v1 is t1  d / v1 . Similarly, with a speed v2 the time would be t2  d / v2 . The two speeds in this problem are 1609 m/mi v1  55 mi/h  (55 mi/h)  24.58 m/s 3600 s/h

v2  65 mi/h  (65 mi/h)

1609 m/mi  29.05 m/s 3600 s/h

ANALYZE With d  700 km  7.0 105 m , the time difference between the two is

1   1 1 1 t  t1  t2  d     (7.0 105 m)     4383 s v v 24.58 m/s 29.05 m/s    1 2   73 min or about 1.2 h. LEARN The travel time was reduced from 7.9 h to 6.9 h. Driving at higher speed (within the legal limit) reduces travel time. 88. The acceleration is constant and we may use the equations in Table 2-1. (a) Taking the first point as coordinate origin and time to be zero when the car is there, we apply Eq. 2-17: 1 1 x   v  v0  t  15.0 m/s  v0   6.00 s  . 2 2 With x = 60.0 m (which takes the direction of motion as the +x direction) we solve for the initial velocity: v0 = 5.00 m/s. (b) Substituting v = 15.0 m/s, v0 = 5.00 m/s, and t = 6.00 s into a = (v – v0)/t (Eq. 2-11), we find a = 1.67 m/s2. (c) Substituting v = 0 in v 2  v02  2ax and solving for x, we obtain

v02 (5.00 m/s)2 x      7.50m , 2a 2 1.67 m/s 2  or | x |  7.50 m . (d) The graphs require computing the time when v = 0, in which case, we use v = v0 + at' = 0. Thus, v 5.00 m/s t  0    3.0s a 1.67 m/s 2 indicates the moment the car was at rest. SI units are understood.

89. THINK In this problem we explore the connection between the maximum height an object reaches under the influence of gravity and the total amount of time it stays in air. EXPRESS Neglecting air resistance and setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion, we analyze the motion of the ball using Table 2-1 (with y replacing x). We set y0 = 0. Upon reaching the maximum height H, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed v0 to H via the equation 0  v2  v02  2 gH  v0  2 gH . The time it takes for the ball to reach maximum height is given by v  v0  gt  0 , or

t  v0 / g  2H / g . ANALYZE If we want the ball to spend twice as much time in air as before, i.e., t   2t , then the new maximum height H  it must reach is such that t   2H  / g . Solving for H  we obtain 1 1 1  H   gt 2  g (2t )2  4  gt 2   4H . 2 2 2  LEARN Since H t 2 , doubling t means that H must increase fourfold. Note also that for t   2t , the initial speed must be twice the original speed: v0  2v0 . 90. (a) Using the fact that the area of a triangle is 12 (base) (height) (and the fact that the integral corresponds to the area under the curve) we find, from t = 0 through t = 5 s, the integral of v with respect to t is 15 m. Since we are told that x0 = 0 then we conclude that x = 15 m when t = 5.0 s. (b) We see directly from the graph that v = 2.0 m/s when t = 5.0 s. (c) Since a = dv/dt = slope of the graph, we find that the acceleration during the interval 4 < t < 6 is uniformly equal to –2.0 m/s2. (d) Thinking of x(t) in terms of accumulated area (on the graph), we note that x(1) = 1 m; using this and the value found in part (a), Eq. 2-2 produces

CHAPTER 2

70 vavg 

x(5)  x(1) 15 m  1 m   3.5 m/s. 5 1 4s

(e) From Eq. 2-7 and the values v(t) we read directly from the graph, we find aavg 

v(5)  v(1) 2 m/s  2 m/s   0. 5 1 4s

91. Taking the +y direction downward and y0 = 0, we have y  v0t  21 gt 2 , which (with v0 = 0) yields t  2 y / g . (a) For this part of the motion, y1 = 50 m so that t1 

2(50 m)  3.2 s . 9.8 m/s 2

(b) For this next part of the motion, we note that the total displacement is y2 = 100 m. Therefore, the total time is 2(100 m) t2   4.5 s . 9.8 m/s 2 The difference between this and the answer to part (a) is the time required to fall through that second 50 m distance:  t  t2  t1  4.5 s – 3.2 s = 1.3 s. 92. Direction of +x is implicit in the problem statement. The initial position (when the clock starts) is x0 = 0 (where v0 = 0), the end of the speeding-up motion occurs at x1 = 1100/2 = 550 m, and the subway train comes to a halt (v2 = 0) at x2 = 1100 m. (a) Using Eq. 2-15, the subway train reaches x1 at t1 

2  550 m  2 x1   30.3 s . a1 1.2 m/s 2

The time interval t2 – t1 turns out to be the same value (most easily seen using Eq. 2-18 so the total time is t2 = 2(30.3) = 60.6 s. (b) Its maximum speed occurs at t1 and equals v1  v0  a1t1  36.3 m / s . (c) The graphs are shown below:

71 93. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the stone’s motion. We are allowed to use Table 2-1 (with x replaced by y) because the ball has constant acceleration motion (and we choose y0 = 0). (a) We apply Eq. 2-16 to both measurements, with SI units understood. 2

1  vB2  v02  2 gyB   v   2 g  y A  3  v02 2  2 2 vA  v0  2 gy A  v 2  2 gy A  v02 We equate the two expressions that each equal v02 and obtain

1 2 v  2 gy A  2 g 3  v 2  2 gy A 4



b g 43 v

2g 3 

which yields v  2 g 4  8.85 m / s. (b) An object moving upward at A with speed v = 8.85 m/s will reach a maximum height y – yA = v2/2g = 4.00 m above point A (this is again a consequence of Eq. 2-16, now with the “final” velocity set to zero to indicate the highest point). Thus, the top of its motion is 1.00 m above point B. 94. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y-axis. The total time of fall can be computed from Eq. 2-15 (using the quadratic formula).

y  v0t 

v  v02  2 gy 1 2 gt  t  0 2 g

with the positive root chosen. With y = 0, v0 = 0, and y0 = h = 60 m, we obtain t

2 gh 2h   35 . s. g g

Thus, “1.2 s earlier” means we are examining where the rock is at t = 2.3 s: 1 y  h  v0 (2.3 s)  g (2.3 s)2  y  34 m 2

where we again use the fact that h = 60 m and v0 = 0. 95. THINK This problem involves analyzing a plot describing the position of an iceboat as function of time. The boat has a nonzero acceleration due to the wind.

CHAPTER 2

EXPRESS Since we are told that the acceleration of the boat is constant, the equations of Table 2-1 can be applied. However, the challenge here is that v0, v, and a are not explicitly given. Our strategy to deduce these values is to apply the kinematic equation x  x0  v0t  12 at 2 to a variety of points on the graph and solve for the unknowns from the simultaneous equations. ANALYZE (a) From the graph, we pick two points on the curve: (t , x)  (2.0 s,16 m) and (3.0 s, 27 m) . The corresponding simultaneous equations are 1 16 m – 0 = v0(2.0 s) + 2 a(2.0 s)2 1 2 27 m – 0 = v0(3.0 s) + 2 a(3.0 s) Solving the equations lead to the values v0 = 6.0 m/s and a = 2.0 m/s2. (b) From Table 2-1, x – x0 = vt –

1 2 2at

1  27 m – 0 = v(3.0 s) – 2 (2.0 m/s2)(3.0 s)2

which leads to v = 12 m/s. 1

this interval (where v0 = 12.0 m/s from part (b)) to obtain x = (12.0 m/s)(3.0 s) +

1 (2.0 m/s2)(3.0 s)2 = 45 m. 2

LEARN By using the results obtained in (a), the position and velocity of the iceboat as a function of time can be written as 1 x(t )  (6.0 m/s)t  (2.0 m/s 2 )t 2 and v(t )  (6.0 m/s)  (2.0 m/s 2 )t. 2 One can readily verify that the same answers are obtained for (b) and (c) using the above expressions for x(t ) and v(t ) .

96. (a) Let the height of the diving board be h. We choose down as the +y direction and set the coordinate origin at the point where it was dropped (which is when we start the clock). Thus, y = h designates the location where the ball strikes the water. Let the depth of the lake be D, and the total time for the ball to descend be T. The speed of the ball as it reaches the surface of the lake is then v = 2gh (from Eq. 2-16), and the time for the ball to fall from the board to the lake surface is t1 = 2h / g (from Eq. 2-15). Now, the time it spends descending in the lake (at constant velocity v) is D D t2   . v 2 gh

Thus, T = t1 + t2 =

D , which gives 2 gh

2h + g

D  T 2 gh  2h   4.80 s 

 2 9.80 m/s2  5.20 m   2 5.20 m   38.1 m .

(b) Using Eq. 2-2, the magnitude of the average velocity is vavg 

D  h 38.1 m  5.20 m   9.02 m/s T 4.80 s

(c) In our coordinate choices, a positive sign for vavg means that the ball is going downward. If, however, upward had been chosen as the positive direction, then this answer in (b) would turn out negative-valued. (d) We find v0 from y  v0t  12 gt 2 with t = T and y = h + D. Thus,

h  D gT 5.20 m  38.1 m  9.8 m/s   4.80 s  v0     14.5 m/s T 2 4.80 s 2 2

(e) Here in our coordinate choices the negative sign means that the ball is being thrown upward. 97. We choose down as the +y direction and use the equations of Table 2-1 (replacing x with y) with a = +g, v0 = 0, and y0 = 0. We use subscript 2 for the elevator reaching the ground and 1 for the halfway point.

(a) Equation 2-16, v22  v02  2a y2  y0 , leads to

v2  2 gy2  2  9.8 m/s2  120 m   48.5 m/s . (b) The time at which it strikes the ground is (using Eq. 2-15) t2 

2 120 m  2 y2   4.95 s . g 9.8 m/s 2

v1  2 gy1  2(9.8 m/s2 )(60 m)  34.3m/s. (d) The time at which it reaches the halfway point is (using Eq. 2-15)

CHAPTER 2

t1 

2 y1 2(60 m)   3.50 s . g 9.8 m/s 2

98. Taking +y to be upward and placing the origin at the point from which the objects are dropped, then the location of diamond 1 is given by y1   12 gt 2 and the location

b g

of diamond 2 is given by y2   21 g t  1 . We are starting the clock when the first object is dropped. We want the time for which y2 – y1 = 10 m. Therefore, 1 1 2  g t  1  gt 2  10 2 2

b g

t  10 / g  0.5  15 . s.



99. With +y upward, we have y0 = 36.6 m and y = 12.2 m. Therefore, using Eq. 2-18 (the last equation in Table 2-1), we find y  y0  vt  12 gt 2  v   22.0 m/s

at t = 2.00 s. The term speed refers to the magnitude of the velocity vector, so the answer is |v| = 22.0 m/s. 100. During free fall, we ignore the air resistance and set a = –g = –9.8 m/s2 where we are choosing down to be the –y direction. The initial velocity is zero so that Eq. 2-15 becomes y   21 gt 2 where y represents the negative of the distance d she has fallen. Thus, we can write the equation as d  21 gt 2 for simplicity. (a) The time t1 during which the parachutist is in free fall is (using Eq. 2-15) given by d1  50 m =

1 2 1 gt1  9.80 m / s2 t12 2 2

which yields t1 = 3.2 s. The speed of the parachutist just before he opens the parachute is given by the positive root v12  2 gd1 , or

v1  2 gh1 

b2gc9.80 m / s hb50 mg  31 m / s. 2

If the final speed is v2, then the time interval t2 between the opening of the parachute and the arrival of the parachutist at the ground level is t2 

v1  v2 31 m / s  3.0 m / s   14 s. a 2 m / s2

This is a result of Eq. 2-11 where speeds are used instead of the (negative-valued) velocities (so that final-velocity minus initial-velocity turns out to equal initial-speed minus final-speed); we also note that the acceleration vector for this part of the motion is positive since it points upward (opposite to the direction of motion — which makes it a deceleration). The total time of flight is therefore t1 + t2 = 17 s.

75 (b) The distance through which the parachutist falls after the parachute is opened is given by

g b g  240 m. b gc h 2

31m / s  3.0 m / s v 2  v22 d 1  2a 2 2.0 m / s2

In the computation, we have used Eq. 2-16 with both sides multiplied by –1 (which changes the negative-valued y into the positive d on the left-hand side, and switches the order of v1 and v2 on the right-hand side). Thus the fall begins at a height of h = 50 + d  290 m. 101. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. The ground level is taken to correspond to y = 0. (a) With y0 = h and v0 replaced with –v0, Eq. 2-16 leads to

v  (v0 )2  2 g (y  y0 )  v02  2 gh . The positive root is taken because the problem asks for the speed (the magnitude of the velocity). (b) We use the quadratic formula to solve Eq. 2-15 for t, with v0 replaced with –v0,

y   v0t 

v  ( v0 ) 2  2 gy 1 2 gt  t  0 g 2

where the positive root is chosen to yield t > 0. With y = 0 and y0 = h, this becomes

t

v02  2 gh  v0 . g

(c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation). (d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +v0 in the equation where we had put in –v0 in part (b). The details follow:

1 y  v0t  gt 2 2

v0  v02  2 gy  t g

CHAPTER 2

with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain

v02  2 gh  v0 . g 102. We assume constant velocity motion and use Eq. 2-2 (with vavg = v > 0). Therefore, t

F GH

x  vt  303

km h

FG 1000 m / kmIJ I c100  10 sh  8.4 m. H 3600 s / h K JK 3

103. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is x  vt , where x is the horizontal distance traveled, t is the time, and v is the (horizontal) velocity. Converting v to meters per second, we have 160 km/h = 44.4 m/s. Thus x 18.4 m t    0.414 s. v 44.4 m / s The velocity-unit conversion implemented above can be figured “from basics” (1000 m = 1 km, 3600 s = 1 h) or found in Appendix D. 104. In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus, x(t) = 50t + 10t2 with SI units (meters and seconds) understood. (a) The average velocity during the first 3 s is given by

vavg 

x(3)  x(0) (50)(3)  (10)(3) 2  0   80 m / s. t 3

(b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0 s, v = 50 + (20)(3.0) = 110 m/s. (c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so the acceleration at any time is 20 m/s2. (d) and (e) The graphs that follow show the coordinate x and velocity v as functions of time, with SI units understood. The dashed line marked (a) in the first graph runs from t = 0, x = 0 to t = 3.0s, x = 240 m. Its slope is the average velocity during the first 3s of motion. The dashed line marked (b) is tangent to the x curve at t = 3.0 s. Its slope is the instantaneous velocity at t = 3.0 s.

77 105. We take +x in the direction of motion, so v0 = +30 m/s, v1 = +15 m/s and a < 0. The acceleration is found from Eq. 2-11: a = (v1 – v0)/t1 where t1 = 3.0 s. This gives a = –5.0 m/s2. The displacement (which in this situation is the same as the distance traveled) to the point it stops (v2 = 0) is, using Eq. 2-16, v22  v02  2ax  x  

(30 m/s)2  90 m. 2(5 m/s 2 )

106. The problem consists of two constant-acceleration parts: part 1 with v0 = 0, v = 6.0 m/s, x = 1.8 m, and x0 = 0 (if we take its original position to be the coordinate origin); and, part 2 with v0 = 6.0 m/s, v = 0, and a2 = –2.5 m/s2 (negative because we are taking the positive direction to be the direction of motion). (a) We can use Eq. 2-17 to find the time for the first part x – x0 =

1 2(v0 + v) t1

1  1.8 m – 0 = 2(0 + 6.0 m/s) t1

so that t1 = 0.6 s. And Eq. 2-11 is used to obtain the time for the second part v  v0  a2t2 

0 = 6.0 m/s + (–2.5 m/s2)t2

from which t2 = 2.4 s is computed. Thus, the total time is t1 + t2 = 3.0 s. (b) We already know the distance for part 1. We could find the distance for part 2 from several of the equations, but the one that makes no use of our part (a) results is Eq. 2-16 v 2  v02  2a2 x2  0 = (6.0 m/s)2 + 2(–2.5 m/s2)x2 which leads to x2 = 7.2 m. Therefore, the total distance traveled by the shuffleboard disk is (1.8 + 7.2) m = 9.0 m. 107. The time required is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). First, we convert the velocity change to SI units: v  (100 km / h)

FG 1000 m / kmIJ  27.8 m / s . H 3600 s / h K

Thus, t = v/a = 27.8/50 = 0.556 s. 108. From Table 2-1, v 2  v02  2ax is used to solve for a. Its minimum value is amin 

v2  v02 (360 km / h) 2   36000 km / h 2 2 xmax 2(180 . km)

which converts to 2.78 m/s2.

CHAPTER 2

109. (a) For the automobile v = 55 – 25 = 30 km/h, which we convert to SI units:

1000 m/ km v (30 km / h) 3600 s/ h a   0.28 m / s2 . t (0.50 min)(60 s / min)

(b) The change of velocity for the bicycle, for the same time, is identical to that of the car, so its acceleration is also 0.28 m/s2. 110. Converting to SI units, we have v = 3400(1000/3600) = 944 m/s (presumed constant) and t = 0.10 s. Thus, x = vt = 94 m. 111. This problem consists of two parts: part 1 with constant acceleration (so that the equations in Table 2-1 apply), v0 = 0, v = 11.0 m/s, x = 12.0 m, and x0 = 0 (adopting the starting line as the coordinate origin); and, part 2 with constant velocity (so that x – x0 = vt applies) with v = 11.0 m/s, x0 = 12.0, and x = 100.0 m. (a) We obtain the time for part 1 from Eq. 2-17 x  x0 

1 1 v0  v t1  12.0  0  0  110 . t1 2 2

so that t1 = 2.2 s, and we find the time for part 2 simply from 88.0 = (11.0)t2  t2 = 8.0 s. Therefore, the total time is t1 + t2 = 10.2 s. (b) Here, the total time is required to be 10.0 s, and we are to locate the point xp where the runner switches from accelerating to proceeding at constant speed. The equations for parts 1 and 2, used above, therefore become x p  0  12  0  11.0 m/s  t1 100.0 m  x p  11.0 m/s 10.0 s  t1 

where in the latter equation, we use the fact that t2 = 10.0 – t1. Solving the equations for the two unknowns, we find that t1 = 1.8 s and xp = 10.0 m. 112. The bullet starts at rest (v0 = 0) and after traveling the length of the barrel ( x 1.2 m ) emerges with the given velocity (v = 640 m/s), where the direction of motion is the positive direction. Turning to the constant acceleration equations in Table 2-1, we use x  12 (v0  v) t . Thus, we find t = 0.00375 s (or 3.75 ms). 113. There is no air resistance, which makes it quite accurate to set a = –g = –9.8 m/s2 (where downward is the –y direction) for the duration of the fall. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion; in fact, when the acceleration changes (during the process of catching the ball) we will again assume constant acceleration conditions; in this case, we have a2 = +25g = 245 m/s2. (a) The time of fall is given by Eq. 2-15 with v0 = 0 and y = 0. Thus,

t

2 y0 2(145 m)   5.44 s. g 9.8 m/s 2

(b) The final velocity for its free-fall (which becomes the initial velocity during the catching process) is found from Eq. 2-16 (other equations can be used but they would use the result from part (a)) v   v02  2 g( y  y0 )   2 gy0  53.3 m / s

where the negative root is chosen since this is a downward velocity. Thus, the speed is | v | 53.3 m/s. (c) For the catching process, the answer to part (b) plays the role of an initial velocity (v0 = –53.3 m/s) and the final velocity must become zero. Using Eq. 2-16, we find v 2  v02 (53.3 m/s)2 y2    5.80 m , 2a2 2(245 m/s 2 )

or | y2 |  5.80 m. The negative value of y2 signifies that the distance traveled while arresting its motion is downward. 114. During Tr the velocity v0 is constant (in the direction we choose as +x) and obeys v0 = Dr/Tr where we note that in SI units the velocity is v0 = 200(1000/3600) = 55.6 m/s. During Tb the acceleration is opposite to the direction of v0 (hence, for us, a < 0) until the car is stopped (v = 0). (a) Using Eq. 2-16 (with xb = 170 m) we find v 2  v02  2axb  a  

which yields |a| = 9.08 m/s2.

v02 2 xb

(b) We express this as a multiple of g by setting up a ratio:

 9.08 m/s 2  a g  0.926 g . 2   9.8 m/s  (c) We use Eq. 2-17 to obtain the braking time: 2 170 m  1 xb   v0  v  Tb  Tb   6.12 s . 2 55.6 m/s (d) We express our result for Tb as a multiple of the reaction time Tr by setting up a ratio:  6.12 s  Tb   T  15.3Tr . 3  r  400  10 s 

CHAPTER 2

(e) Since Tb > Tr, most of the full time required to stop is spent in braking. (f) We are only asked what the increase in distance D is, due to Tr = 0.100 s, so we simply have D  v0 Tr   55.6 m/s  0.100 s   5.56 m . 115. The total time elapsed is t  2 h 41 min  161 min and the center point is displaced by x  3.70 m  370 cm. Thus, the average velocity of the center point is vavg 

x 370 cm   2.30 cm/min. t 161 min

116. Using Eq. 2-11, v  v0  at , we find the initial speed to be v0  v  at  0  (3400)(9.8 m/s2 )(6.5 103 s)  216.6 m/s

117. The total number of days walked is (including the first and the last day, and leap year) N  340  365  365  366  365  365  261  2427 Thus, the average speed of the walk is savg 

d 3.06 107 m   0.146 m/s. t (2427 days)(86400 s/day)

118. (a) Let d be the distance traveled. The average speed with and without wings set as sails are vs  d / ts and vns  d / tns , respectively. Thus, the ratio of the two speeds is vs d / ts tns 25.0 s     3.52 vns d / tns ts 7.1s (b) The difference in time expressed in terms of vs is t  tns  ts 

d d d d d (2.0 m) 5.04 m     2.52  2.52  vns vs (vs / 3.52) vs vs vs vs

119. (a) Differentiating y(t )  (2.0 cm)sin( t / 4) with respect to t, we obtain v y (t ) 

dy      cm/s  cos( t / 4) dt  2 

The average velocity between t = 0 and t = 2.0 s is

81 2 1 1   2  t  v dt cm/s  y   0 cos   dt  0 (2.0 s) (2.0 s)  2   4  /2 1   2 cm  0 cos x dx  1.0 cm/s (2.0 s)

vavg 

(b) The instantaneous velocities of the particle at t = 0, 1.0 s, and 2.0 s are, respectively,    v y (0)   cm/s  cos(0)  cm/s 2 2   2   v y (1.0 s)   cm/s  cos( / 4)  cm/s 4 2    v y (2.0 s)   cm/s  cos( / 2)  0 2  (c) Differentiating v y (t ) with respect to t, we obtain the following expression for acceleration: dv y   2  a y (t )    cm/s 2  sin( t / 4) dt  8  The average acceleration between t = 0 and t = 2.0 s is 2  2  t  1 1  2 a dt   cm/s 2   sin   dt  y  (2.0 s) 0 (2.0 s)  8  4   0  / 2 1   1      2    cm/s  0 sin x dx    cm/s    cm/s (2.0 s)  2 (2.0 s)  2 4  

aavg 

(d) The instantaneous accelerations of the particle at t = 0, 1.0 s, and 2.0 s are, respectively,  2  a y (0)    cm/s 2  sin(0)  0  8  2    2 2 a y (1.0 s)    cm/s 2  sin( / 4)   cm/s 2 8 16   2   2 2 a y (2.0 s)    cm/s  sin( / 2)   cm/s 2 8  8 

Chapter 3 1. THINK In this problem we’re given the magnitude and direction of a vector in two dimensions, and asked to calculate its x- and y-components.  EXPRESS The x- and the y- components of a vector a lying in the xy plane are given by

ax  a cos  ,

ay  a sin 

 where a  | a |  ax2  a y2 is the magnitude and   tan 1 (ay / ax ) is the angle between a

and the positive x axis. Given that   250 , we see that the vector is in the third  quadrant, and we expect both the x- and the y-components of a to be negative.  ANALYZE (a) The x component of a is

ax  a cos  (7.3 m) cos 250    2.50 m ,

(b) and the y component is ay  a sin   (7.3 m)sin 250   6.86 m  6.9 m. The results are depicted in the figure below:

LEARN In considering the variety of ways to compute these, we note that the vector is 70° below the – x axis, so the components could also have been found from

ax  (7.3 m) cos 70   2.50 m, ay  (7.3 m)sin 70   6.86 m. Similarly, we note that the vector is 20° to the left from the – y axis, so one could also achieve the same results by using

ax  (7.3 m)sin 20   2.50 m, ay  (7.3 m) cos 20   6.86 m.

83 As a consistency check, we note that

ax2  ay2  ( 2.50 m)2  ( 6.86 m)2  7.3 m

and tan 1  ay / ax   tan 1[( 6.86 m) /(2.50 m)]  250  , which are indeed the values given in the problem statement.  2. (a) With r = 15 m and  = 30°, the x component of r is given by

rx = rcos = (15 m) cos 30° = 13 m. (b) Similarly, the y component is given by ry = r sin = (15 m) sin 30° = 7.5 m. 3. THINK In this problem we’re given the x- and y-components a vector A in two dimensions, and asked to calculate its magnitude and direction. EXPRESS A vector A can be represented in the magnitude-angle notation (A, ), where

A  Ax2  Ay2 is the magnitude and

 Ay    Ax 

  tan  1 

is the angle A makes with the positive x axis. Given that Ax = 25.0 m and Ay = 40.0 m, the above formulas can be readily used to calculate A and . ANALYZE (a) The magnitude of the vector A is

A  Ax2  Ay2  ( 25.0 m)2  (40.0 m) 2  47.2 m (b) Recalling that tan = tan ( + 180°), tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°. Noting that the vector is in the second quadrant (by the signs of its x and y components) we see that 122° is the correct answer. The results are depicted in the figure to the right.

LEARN We can check our answers by noting that the x- and the y- components of A can be written as Ax  A cos  , Ay  A sin  .

CHAPTER 3

Substituting the results calculated above, we obtain Ax  (47.2 m) cos122   25.0 m, Ay  (47.2 m)sin122   40.0 m

which indeed are the values given in the problem statement. 4. The angle described by a full circle is 360° = 2 rad, which is the basis of our conversion factor.

2 rad  0.349 rad . 360 2 rad (b) 50.0   50.0   0.873 rad . 360 2 rad (c) 100  100   1.75 rad . 360 360 (d) 0.330 rad =  0.330 rad   18.9 . 2 rad 360 (e) 2.10 rad =  2.10 rad   120 . 2 rad 360 (f) 7.70 rad =  7.70 rad   441 . 2 rad (a) 20.0   20.0 

  5. The vector sum of the displacements d storm and d new must give the same result as its originally intended displacement d  (120 km)jˆ where east is i , north is j . Thus, we

write

dstorm  (100 km) ˆi , dnew  A ˆi  B ˆj.

(a) The equation dstorm  dnew  do readily yields A = –100 km and B = 120 km. The  magnitude of d new is therefore equal to | dnew |  A2  B 2  156 km . (b) The direction is

tan–1 (B/A) = –50.2° or 180° + ( –50.2°) = 129.8°.

We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west. 6. (a) The height is h = d sin, where d = 12.5 m and  = 20.0°. Therefore, h = 4.28 m. (b) The horizontal distance is d cos = 11.7 m.

7. (a) The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), (b) anti-parallel (in opposite directions) to achieve a resultant 1 m long (primed case shown), (c) and perpendicular to achieve a resultant 32  42  5 m long (the double-primed case shown). In each sketch, the vectors are shown in a “head-to-tail” sketch but the resultant is not shown. The resultant would be a straight line drawn from beginning to end; the beginning is indicated by A (with or without primes, as the case may be) and the end is indicated by B.

   8. We label the displacement vectors A , B , and C (and denote the result of their vector  sum as r ). We choose east as the î direction (+x direction) and north as the ˆj direction (+y direction). All distances are understood to be in kilometers. (a) The vector diagram representing the motion is shown next: A  ( 3.1 km) ˆj B  (2.4 km) ˆi C  (  5.2 km) ˆj

(b) The final point is represented by r  A  B  C  (2.4 km)iˆ  (2.1 km)ˆj

whose magnitude is

r 

 2.4 km    2.1 km   3.2 km . 2

CHAPTER 3

 2.1 km    41, or 221 .  2.4 km 

  tan 1 

 We choose the latter possibility since r is in the third quadrant. It should be noted that many graphical calculators have polar  rectangular “shortcuts” that automatically produce the correct answer for angle (measured counterclockwise from the +x axis). We may phrase the angle, then, as 221° counterclockwise from East (a phrasing that sounds  peculiar, at best) or as 41° south from west or 49° west from south. The resultant r is not shown in our sketch; it would be an arrow directed from the “tail” of A to the “head” of C .

9. All distances in this solution are understood to be in meters. ˆ m. (a) a  b  [4.0  (1.0)] ˆi  [(3.0) 1.0] ˆj  (1.0  4.0)kˆ  (3.0iˆ  2.0jˆ  5.0 k) ˆ m. (b) a  b  [4.0  (1.0)]iˆ  [(3.0) 1.0]jˆ  (1.0  4.0)kˆ  (5.0 ˆi  4.0 ˆj  3.0 k)

      (c) The requirement a  b  c  0 leads to c  b  a , which we note is the opposite of ˆ m. what we found in part (b). Thus, c  (5.0 ˆi  4.0 ˆj  3.0 k)

10. The x, y, and z components of r  c  d are, respectively, (a) rx  cx  d x  7.4 m  4.4 m  12 m , (b) ry  cy  d y   3.8 m  2.0 m   5.8 m , and (c) rz  cz  d z   6.1 m  3.3 m   2.8 m. 11. THINK This problem involves the addition of two vectors a and b . We want to find the magnitude and direction of the resulting vector. EXPRESS In two dimensions, a vector a can be written as, in unit vector notation,

a  ax ˆi  a y ˆj . Similarly, a second vector b can be expressed as b  bx ˆi  by ˆj . Adding the two vectors gives

r  a  b  (ax  bx )iˆ  (ay  by )ˆj  rx ˆi  ry ˆj

ANALYZE (a) Given that a  (4.0 m)iˆ  (3.0 m)ˆj and b  (13.0 m)iˆ  (7.0 m)ˆj , we  find the x and the y components of r to be

rx = ax + bx = (4.0 m) + (–13 m) = –9.0 m ry = ay + by = (3.0 m) + (7.0 m) = 10.0 m. Thus r  (9.0m) ˆi  (10m) ˆj .  (b) The magnitude of r is r | r | rx2  ry2  (9.0 m)2  (10 m)2  13 m.

r   rx 

 10.0 m     48  or 132 .   9.0 m 

  tan 1  y   tan 1 

Since the x component of the resultant is negative and the y component is positive, characteristic of the second quadrant, we find the angle is 132° (measured counterclockwise from +x axis). LEARN The addition of the two vectors is depicted in the figure below (not to scale).  Indeed, since rx  0 and ry  0 , we expect r to be in the second quadrant.

   12. We label the displacement vectors A , B , and C (and denote the result of their vector  sum as r ). We choose east as the î direction (+x direction) and north as the ˆj direction  (+y direction). We note that the angle between C and the x axis is 60°. Thus,

A  (50 km) ˆi B  (30 km) ˆj C  (25 km) cos  60  ˆi + (25 km )sin  60  ˆj

CHAPTER 3

(a) The total displacement of the car from its initial position is represented by

r  A  B  C  (62.5 km) ˆi  (51.7 km) ˆj which means that its magnitude is

r  (62.5km)2  (51.7 km)2  81 km. (b) The angle (counterclockwise from +x axis) is tan–1 (51.7 km/62.5 km)  40°, which is  to say that it points 40° north of east. Although the resultant r is shown in our sketch, it  would be a direct line from the “tail” of A to the “head” of C . 13. We find the components and then add them (as scalars, not vectors). With d = 3.40 km and  = 35.0° we find d cos  + d sin  = 4.74 km. 14. (a) Summing the x components, we have which gives bx   80 m.

20 m + bx – 20 m – 60 m = 140 m,

(b) Summing the y components, we have which implies cy =110 m.

60 m – 70 m + cy – 70 m = 30 m,

(d) The angle is given by tan 1 (30 /( 140))  12  , (which would be 12 measured clockwise from the –x axis, or 168 measured counterclockwise from the +x axis). 15. THINK This problem involves the addition of two vectors a and b in two dimensions. We’re asked to find the components, magnitude and direction of the resulting vector. EXPRESS In two dimensions, a vector a can be written as, in unit vector notation,

a  ax ˆi  ay ˆj  (a cos  )iˆ  (a sin  )ˆj .

89 Similarly, a second vector b can be expressed as b  bx ˆi  by ˆj  (b cos  )iˆ  (b sin  )ˆj . From the figure, we have,   1 and   1  2 (since the angles are measured from the +x-axis) and the resulting vector is

r  a  b  [a cos 1  b cos(1  2 )]iˆ  [a sin 1  b sin(1  2 )]jˆ  rx ˆi  ry ˆj  ANALYZE (a) Given that a  b  10 m, 1  30 and 2  105, the x component of r is

rx  a cos1  b cos(1  2 )  (10 m) cos30  (10 m) cos(30  105)  1.59 m  (b) Similarly, the y component of r is ry  a sin 1  b sin(1  2 )  (10 m)sin 30  (10 m)sin(30  105)  12.1 m.  (c) The magnitude of r is r  | r |  (1.59 m)2  (12.1 m)2  12.2 m.  (d) The angle between r and the +x-axis is

r   rx 

 12.1 m    82.5  .  1.59 m 

  tan 1  y   tan 1 

LEARN As depicted in the figure, the resultant r lies in the first quadrant. This is what we expect. Note that the magnitude of r can also be calculated by using law of cosine  ( a , b and r form an isosceles triangle): r  a 2  b2  2ab cos(180   2 )  (10 m)2  (10 m)2  2(10 m)(10 m) cos 75  12.2 m.

16. (a) a  b  (3.0iˆ  4.0 ˆj) m  (5.0iˆ  2.0 ˆj) m  (8.0 m) ˆi  (2.0 m) ˆj.

  (b) The magnitude of a  b is | a  b | (8.0 m)2  (2.0 m)2  8.2 m.

(c) The angle between this vector and the +x axis is tan–1[(2.0 m)/(8.0 m)] = 14°. (d) b  a  (5.0iˆ  2.0 ˆj) m  (3.0iˆ  4.0 ˆj) m  (2.0 m) ˆi  (6.0 m)ˆj .

CHAPTER 3

90 (e) The magnitude of the difference vector b  a is | b  a | (2.0 m)2  (6.0 m)2  6.3 m.

(f) The angle between this vector and the +x axis is tan-1[( –6.0 m)/(2.0 m)] = –72°. The vector is 72° clockwise from the axis defined by î . 17. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular  polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood. (a) Using unit-vector notation, a  (50 m) cos(30)iˆ  (50 m) sin(30) ˆj b  (50 m) cos (195) î  (50 m) sin (195) ˆj c  (50 m) cos (315) î  (50 m) sin (315) ˆj a  b  c  (30.4 m) î  (23.3 m) ˆj.

The magnitude of this result is

(30.4 m)2  (23.3 m)2  38 m .

(b) The two possibilities presented by a simple calculation for the angle between the vector described in part (a) and the +x direction are tan–1[(–23.2 m)/(30.4 m)] = –37.5°, and 180° + ( –37.5°) = 142.5°. The former possibility is the correct answer since the vector is in the fourth quadrant (indicated by the signs of its components). Thus, the angle is –37.5°, which is to say that it is 37.5° clockwise from the +x axis. This is equivalent to 322.5° counterclockwise from +x. (c) We find a  b  c  [43.3  (48.3)  35.4] ˆi  [25  (  12.9)  (  35.4)] ˆj  (127 ˆi  2.60 ˆj) m

in unit-vector notation. The magnitude of this result is | a  b  c |  (127 m)2  (2.6 m)2  1.30 102 m.

(d) The angle between the vector described in part (c) and the +x axis is tan 1 (2.6 m/127 m)  1.2 .

91  (e) Using unit-vector notation, d is given by d  a  b  c  ( 40.4 ˆi  47.4 ˆj) m ,

which has a magnitude of

(40.4 m)2  (47.4 m)2  62 m.

(f) The two possibilities presented by a simple calculation for the angle between the vector described in part (e) and the +x axis are tan 1 (47.4 /(40.4))  50.0  , and 180  (50.0)  130 . We choose the latter possibility as the correct one since it  indicates that d is in the second quadrant (indicated by the signs of its components). 18. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle between C and the +x axis is 180° + 20.0° = 200°.

 (a) The x and y components of B are given by Bx = Cx – Ax = (15.0 m) cos 200° – (12.0 m) cos 40° = –23.3 m, By =Cy – Ay = (15.0 m) sin 200° – (12.0 m) sin 40° = –12.8 m. Consequently, its magnitude is | B |  (23.3 m)2  (12.8 m)2  26.6 m .

 (b) The two possibilities presented by a simple calculation for the angle between B and the +x axis are tan–1[( –12.8 m)/( –23.3 m)] = 28.9°, and 180° +28.9° = 209°. We choose the latter possibility as the correct one since it indicates that B is in the third quadrant (indicated by the signs of its components). We note, too, that the answer can be equivalently stated as  151  . 19. (a) With ^i directed forward and ^j directed leftward, the resultant is (5.00 ^i + 2.00 ^j) m . The magnitude is given by the Pythagorean theorem:  5.39 m.

(5.00 m)2  (2.00 m)2 = 5.385 m

(b) The angle is tan1(2.00/5.00)  21.8º (left of forward). 

20. The desired result is the displacement vector, in units of km, A = (5.6 km), 90º (measured counterclockwise from the +x axis), or A  (5.6 km)jˆ , where ˆj is the unit vector along the positive y axis (north). This consists of the sum of two displacements: during the whiteout, B  (7.8 km), 50 , or

B  (7.8 km)(cos50 ˆi  sin50 ˆj)  (5.01 km)iˆ  (5.98 km)ˆj and the unknown C . Thus, A  B  C .

CHAPTER 3

(a) The desired displacement is given by C  A  B  ( 5.01 km) ˆi  (0.38 km) ˆj . The magnitude is

( 5.01 km)2  ( 0.38 km)2  5.0 km.

(b) The angle is tan 1[( 0.38 km) /( 5.01 km)]  4.3 , south of due west. 21. Reading carefully, we see that the (x, y) specifications for each “dart” are to be interpreted as ( x, y) descriptions of the corresponding displacement vectors. We combine the different parts of this problem into a single exposition. (a) Along the x axis, we have (with the centimeter unit understood)

which gives bx = –70.0 cm.

30.0  bx  20.0  80.0  140,

(b) Along the y axis we have 40.0  70.0  cy  70.0   20.0

which yields cy = 80.0 cm. (c) The magnitude of the final location (–140 , –20.0) is

(140)2  (20.0)2  141 cm.

(d) Since the displacement is in the third quadrant, the angle of the overall displacement is given by  + tan 1[( 20.0) /(140)] or 188° counterclockwise from the +x axis (or  172  counterclockwise from the +x axis). 22. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to write the vectors in unit-vector notation before adding them. However, a very differentlooking approach using the special capabilities of most graphical calculators can be imagined. Wherever the length unit is not displayed in the solution below, the unit meter should be understood. (a) Allowing for the different angle units used in the problem statement, we arrive at

 E  3.73 i  4.70 j  F  1.29 i  4.83 j  G  1.45 i  3.73 j  H  5.20 i  3.00 j     E  F  G  H  1.28 i  6.60 j. (b) The magnitude of the vector sum found in part (a) is

(1.28 m)2  (6.60 m)2  6.72 m .

(c) Its angle measured counterclockwise from the +x axis is tan–1(6.60/1.28) = 79.0°. (d) Using the conversion factor  rad = 180 , 79.0° = 1.38 rad. 

23. The resultant (along the y axis, with the same magnitude as C ) forms (along with 



C ) a side of an isosceles triangle (with B forming the base). If the angle between C and the y axis is   tan 1 (3/ 4)  36.87  , then it should be clear that (referring to the magnitudes of the vectors) B  2C sin( / 2) . Thus (since C = 5.0) we find B = 3.2. 24. As a vector addition problem, we express the situation (described in the problem 



statement) as A + B = (3A) j , where A = A i and B = 7.0 m. Since i  j we may use the Pythagorean theorem to express B in terms of the magnitudes of the other two vectors: B = (3A)2 + A2



1 B = 2.2 m . 10 

25. The strategy is to find where the camel is ( C ) by adding the two consecutive displacements described in the problem, and then finding the difference between that 

location and the oasis ( B ). Using the magnitude-angle notation C = (24   15) + (8.0  90) = (23.25  4.41)

so B  C  (25  0)  (23.25  4.41)  (2.5   45)

which is efficiently implemented using a vector-capable calculator in polar mode. The distance is therefore 2.6 km.

       26. The vector equation is R  A  B  C  D . Expressing B and D in unit-vector ˆ m and (2.87iˆ  4.10j) ˆ m , respectively. Where the notation, we have (1.69iˆ  3.63j) length unit is not displayed in the solution below, the unit meter should be understood. (a) Adding corresponding components, we obtain R  (3.18 m)iˆ  ( 4.72 m) ˆj . (b) Using Eq. 3-6, the magnitude is | R | (3.18 m)2  (4.72 m)2  5.69 m.

 4.72 m    56.0 (with  x axis).  3.18 m 

  tan 1 

CHAPTER 3

If measured counterclockwise from +x-axis, the angle is then 180  56.0  124 . Thus, converting the result to polar coordinates, we obtain . , 4.72g  b5.69  124g b318

27. Solving the simultaneous equations yields the answers: 



(a) d1 = 4 d3 = 8 ^i + 16 ^j , and 



(b) d2 = d3 = 2 ^i + 4 ^j. 



28. Let A represent the first part of Beetle 1’s trip (0.50 m east or 0.5 ˆi ) and C represent the first part of Beetle 2’s trip intended voyage (1.6 m at 50º north of east). For 



their respective second parts: B is 0.80 m at 30º north of east and D is the unknown. The final position of Beetle 1 is

A  B  (0.5 m)iˆ  (0.8 m)(cos30 ˆi  sin30 ˆj)  (1.19 m) ˆi  (0.40 m) ˆj. The equation relating these is A  B  C  D , where C  (1.60 m)(cos50.0 iˆ  sin50.0ˆj)  (1.03 m)iˆ  (1.23 m)jˆ

(a) We find D  A  B  C  (0.16 m)iˆ  ( 0.83 m)ˆj , and the magnitude is D = 0.84 m. (b) The angle is tan 1 ( 0.83/ 0.16)  79  , which is interpreted to mean 79º south of east (or 11º east of south). 29. Let l0  2.0 cm be the length of each segment. The nest is located at the endpoint of segment w. (a) Using unit-vector notation, the displacement vector for point A is

 

d A  w  v  i  h  l0 (cos 60ˆi  sin60 ˆj)  l0 ˆj  l0 (cos120ˆi  sin120 ˆj)  l0 ˆj  (2  3)l0 ˆj.

Therefore, the magnitude of d A is | d A |  (2  3)(2.0 cm)  7.5 cm . (b) The angle of d A is   tan 1 (d A, y / d A, x )  tan 1 ()  90 . (c) Similarly, the displacement for point B is

dB  w  v  j  p  o

 

 l0 (cos 60î  sin 60 ˆj)  l0 ˆj  l0 (cos 60î  sin60 ˆj)  l0 (cos 30î  sin30 ˆj)  l0 î  (2  3 / 2)l0 î  (3 / 2  3)l0 ˆj.

Therefore, the magnitude of d B is | d B |  l0 (2  3 / 2)2  (3/ 2  3)2  (2.0 cm)(4.3)  8.6 cm .

(d) The direction of d B is  d B, y    1 3/ 2  3 1   tan    tan (1.13)  48 . d  2 3/2  B, x 

 B  tan 1 

30. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular  polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eq. 3-6).  (a) The magnitude of a is a  (4.0 m)2  ( 3.0 m)2  5.0 m.  (b) The angle between a and the +x axis is tan–1 [(–3.0 m)/(4.0 m)] = –37°. The vector is 37° clockwise from the axis defined by i .

 (c) The magnitude of b is b  (6.0 m)2  (8.0 m) 2  10 m.  (d) The angle between b and the +x axis is tan–1[(8.0 m)/(6.0 m)] = 53°.

(e) a  b  (4.0 m  6.0 m) ˆi  [(  3.0 m)  8.0 m]jˆ  (10 m)iˆ  (5.0 m)ˆj. The magnitude of this vector is | a  b | (10 m)2  (5.0 m)2  11 m; we round to two significant figures in our results. (f) The angle between the vector described in part (e) and the +x axis is tan–1[(5.0 m)/(10 m)] = 27°. (g) b  a  (6.0 m  4.0 m) ˆi  [8.0 m  (3.0 m)] ˆj  (2.0 m) iˆ  (11 m) ˆj. The magnitude of this vector is | b  a | (2.0 m)2  (11 m)2  11 m, which is, interestingly, the same result as in part (e) (exactly, not just to 2 significant figures) (this curious coincidence is   made possible by the fact that a  b ).

CHAPTER 3

(h) The angle between the vector described in part (g) and the +x axis is tan–1[(11 m)/(2.0 m)] = 80°. (i) a  b  (4.0 m  6.0 m) ˆi  [(  3.0 m)  8.0 m] ˆj  (2.0 m) ˆi  (11 m) ˆj. The magnitude of this vector is | a  b | ( 2.0 m)2  ( 11 m)2  11 m .

(j) The two possibilities presented by a simple calculation for the angle between the vector described in part (i) and the +x direction are tan–1 [(–11 m)/(–2.0 m)] = 80°, and 180° + 80° = 260°. The latter possibility is the correct answer (see part (k) for a further observation related to this result).

    (k) Since a  b  ( 1)(b  a ) , they point in opposite (anti-parallel) directions; the angle between them is 180°. 31. (a) With a = 17.0 m and  = 56.0° we find ax = a cos  = 9.51 m. (b) Similarly, ay = a sin  = 14.1 m. (c) The angle relative to the new coordinate system is ´ = (56.0° – 18.0°) = 38.0°. Thus, ax  a cos    13.4 m. (d) Similarly, ay = a sin ´ = 10.5 m. 32. (a) As can be seen from Figure 3-30, the point diametrically opposite the origin (0,0,0) has position vector a i  a j  a k and this is the vector along the “body diagonal.” (b) From the point (a, 0, 0), which corresponds to the position vector a î, the diametrically opposite point is (0, a, a) with the position vector a j  a k . Thus, the vector along the line is the difference  a ˆi  aˆj  a kˆ .

(c) If the starting point is (0, a, 0) with the corresponding position vector a ˆj , the diametrically opposite point is (a, 0, a) with the position vector a ˆi  a kˆ . Thus, the vector along the line is the difference a ˆi  a ˆj  a kˆ .

(d) If the starting point is (a, a, 0) with the corresponding position vector a î  a ˆj , the diametrically opposite point is (0, 0, a) with the position vector a k̂ . Thus, the vector along the line is the difference a î  a ˆj  a kˆ . (e) Consider the vector from the back lower left corner to the front upper right corner. It is ˆ We may think of it as the sum of the vector a i parallel to the x axis and a î  a ˆj  a k. the vector a j  a k perpendicular to the x axis. The tangent of the angle between the vector and the x axis is the perpendicular component divided by the parallel component. Since the magnitude of the perpendicular component is





a 2  a 2  a 2 and the

magnitude of the parallel component is a, tan   a 2 / a  2 . Thus   54.7 . The angle between the vector and each of the other two adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonal vectors and any of the cube sides adjacent to them.

a 2  a 2  a 2  a 3.

(f) The length of any of the diagonals is given by 



33. Examining the figure, we see that a + b + c = 0, where a  b .





(a)  a  b = (3.0)(4.0) = 12 since the angle between them is 90º. (b) Using the Right-Hand Rule, the vector a  b points in the ˆi  ˆj  kˆ , or the +z direction. 



(c) | a  c | = | a  ( a  b )| = |  a  b )| =  (d) The vector a  b points in the ˆi  ˆj  kˆ , or the z direction. 



(e) | b  c | = | b  ( a  b )| = |  b  a ) | = |  a  b ) | = 12. (f) The vector points in the +z direction, as in part (a). 34. We apply Eq. 3-23 and Eq. 3-27.

 (a) a  b = (axby  a y bx ) kˆ since all other terms vanish, due to the fact that neither a nor  b have any z components. Consequently, we obtain [(3.0)(4.0)  (5.0)(2.0)]kˆ  2.0 kˆ . (b) a  b  axbx  a yby yields (3.0)(2.0) + (5.0)(4.0) = 26. (c) a  b  (3.0  2.0) ˆi  (5.0  4.0) ˆj  (a + b )  b = (5.0) (2.0) + (9.0) (4.0) = 46 .

CHAPTER 3

(d) Several approaches are available. In this solution, we will construct a b unit-vector  and “dot” it (take the scalar product of it) with a . In this case, we make the desired unitvector by b 2.0 ˆi  4.0 ˆj  bˆ  . |b | (2.0)2  (4.0)2 We therefore obtain

(3.0)(2.0)  (5.0)(4.0) ab  a  bˆ   5.8. (2.0)2  (4.0) 2

35. (a) The scalar or dot product is (4.50)(7.30)cos(320º – 85.0º) = – 18.8 . ^

(b) The vector or cross product is in the k direction (by the right-hand rule) with magnitude |(4.50)(7.30) sin(320º – 85.0º)| = 26.9 . 



36. First, we rewrite the given expression as 4( dplane · dcross ) where dplane = d1 +   d2 and in the plane of d1 and d 2 , and dcross  d1  d2 . Noting that dcross is perpendicular 

to the plane of d1 and d 2 , we see that the answer must be 0 (the scalar or dot product of perpendicular vectors is zero). 37. We apply Eq. 3-23 and Eq.3-27. If a vector-capable calculator is used, this makes a good exercise for getting familiar with those features. Here we briefly sketch the method. (a) We note that b  c   8.0 ˆi  5.0 ˆj  6.0kˆ . Thus, a  (b  c ) = (3.0) (  8.0)  (3.0)(5.0)  (  2.0) (6.0) =  21. ˆ Thus, (b) We note that b + c = 1.0 ˆi  2.0 ˆj + 3.0 k.

a  (b  c )  (3.0) (1.0)  (3.0) (  2.0)  (  2.0) (3.0)  9.0. (c) Finally, a  (b + c )  [(3.0)(3.0)  (  2.0)(  2.0)] ˆi  [(  2.0)(1.0)  (3.0)(3.0)] ˆj . [(3.0)(  2.0)  (3.0)(1.0)] kˆ  5iˆ  11jˆ  9kˆ

38. Using the fact that we obtain

î  ˆj  k, ˆ ˆj  kˆ  î, kˆ  î  ˆj





2 A  B  2 2.00iˆ  3.00jˆ  4.00kˆ  3.00iˆ  4.00jˆ  2.00kˆ



ˆ  44.0iˆ  16.0jˆ  34.0k.

Next, making use of

we have

î  î = ˆj  ˆj = kˆ  kˆ = 1 î  ˆj = ˆj  kˆ = kˆ  î = 0

 





3C  2 A  B  3 7.00 ˆi  8.00ˆj  44.0 iˆ  16.0 ˆj  34.0 kˆ



 3[(7.00) (44.0)+(  8.00) (16.0)  (0) (34.0)]  540.

39. From the definition of the dot product between A and B , A  B  AB cos , we have

cos 

A B AB

With A  6.00 , B  7.00 and A  B  14.0 , cos  0.333 , or   70.5  . 40. The displacement vectors can be written as (in meters)

ˆ  (2.04 m) ˆj  (4.01 m) kˆ d1  (4.50 m)(cos 63 ˆj  sin 63 k) ˆ  (1.21 m) ˆi  (0.70 m) kˆ . d  (1.40 m)(cos 30 ˆi  sin 30 k) 2

(a) The dot product of d1 and d 2 is

ˆ  (1.21iˆ  0.70k) ˆ = (4.01k) ˆ  (0.70k) ˆ = 2.81 m2 . d1  d2  (2.04 ˆj  4.01k) (b) The cross product of d1 and d 2 is

ˆ  (1.21iˆ  0.70 k) ˆ d1  d 2  (2.04 ˆj  4.01k) ˆ + (2.04)(0.70)iˆ  (4.01)(1.21)ˆj  (2.04)(1.21)(k) ˆ m2 .  (1.43 ˆi  4.86 ˆj  2.48 k) (c) The magnitudes of d1 and d 2 are

d1  (2.04 m) 2  (4.01 m) 2  4.50 m d 2  (1.21 m)2  (0.70 m) 2  1.40 m. Thus, the angle between the two vectors is

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100

 d1  d 2   2.81 m2 1    cos    cos    63.5.  (4.50 m)(1.40 m)   d1d 2  1

41. THINK The angle between two vectors can be calculated using the definition of scalar product. EXPRESS Since the scalar product of two vectors a and b is a  b  ab cos   axbx  a yby  az bz ,

the angle between them is given by

cos  

axbx  a y by  az bz ab

 axbx  a yby  az bz     cos 1  . ab  

Once the magnitudes and components of the vectors are known, the angle  can be readily calculated. ANALYZE Given that a  (3.0)iˆ  (3.0)ˆj  (3.0)kˆ and b  (2.0)iˆ  (1.0)ˆj  (3.0)kˆ , the magnitudes of the vectors are a  | a |  ax2  a y2  az2  (3.0) 2  (3.0) 2  (3.0) 2  5.20 b  | b |  bx2  by2  bz2  (2.0) 2  (1.0) 2  (3.0) 2  3.74.

The angle between them is found to be cos  

(3.0) (2.0)  (3.0) (1.0)  (3.0) (3.0)  0.926, (5.20) (3.74)

or = 22°. LEARN As the name implies, the scalar product (or dot product) between two vectors is a scalar quantity. It can be regarded as the product between the magnitude of one of the vectors and the scalar component of the second vector along the direction of the first one, as illustrated below (see also in Fig. 3-18 of the text):

a  b  ab cos   (a)(b cos  )

101 42. The two vectors are written as, in unit of meters,

d1  4.0 ˆi+5.0 ˆj  d1x ˆi  d1 y ˆj,

d2   3.0 ˆi+4.0 ˆj  d2 x ˆi  d2 y ˆj

(a) The vector (cross) product gives

ˆ d1  d2  (d1x d2 y  d1 y d2 x )kˆ  [(4.0)(4.0)  (5.0)(3.0)]k=31 kˆ (b) The scalar (dot) product gives

d1  d2  d1x d2 x  d1 y d2 y  (4.0)( 3.0)  (5.0)(4.0)  8.0. (c)

(d1  d2 )  d2  d1  d2  d22  8.0  ( 3.0)2  (4.0)2  33. (d) Note that the magnitude of the d1 vector is 16+25 = 6.4. Now, the dot product is (6.4)(5.0)cos = 8. Dividing both sides by 32 and taking the inverse cosine yields  = 75.5. Therefore the component of the d1 vector along the direction of the d2 vector is 6.4cos 1.6. 43. THINK In this problem we are given three vectors a , b and c on the xy-plane, and asked to calculate their components.    EXPRESS From the figure, we note that c  b , which implies that the angle between c and the +x axis is  + 90°. In unit-vector notation, the three vectors can be written as a  ax î b  b ˆi  b ˆj  (b cos  )iˆ  (b sin  )ˆj x

ˆ c  cx ˆi  c y ˆj  [c cos(  90)]iˆ  [c sin(  90)]j.

The above expressions allow us to evaluate the components of the vectors. ANALYZE (a) The x-component of a is ax = a cos 0° = a = 3.00 m. (b) Similarly, the y-componnet of a is ay = a sin 0° = 0. (c) The x-component of b is bx = b cos 30° = (4.00 m) cos 30° = 3.46 m, (d) and the y-component is by = b sin 30° = (4.00 m) sin 30° = 2.00 m. (e) The x-component of c is cx = c cos 120° = (10.0 m) cos 120° = –5.00 m,

CHAPTER 3

102 (f) and the y-component is cy = c sin 30° = (10.0 m) sin 120° = 8.66 m. (g) The fact that c  pa  qb implies

c  cx î  cy ˆj  p(ax î)  q(bx î  by ˆj)  ( pax  qbx )iˆ  qby ˆj or

cx  pax  qbx ,

cy  qby .

Substituting the values found above, we have 5.00 m  p (3.00 m)  q (3.46 m) 8.66 m  q (2.00 m).

Solving these equations, we find p = –6.67. (h) Similarly, q = 4.33 (note that it’s easiest to solve for q first). The numbers p and q have no units. LEARN This exercise shows that given two (non-parallel) vectors in two dimensions, the third vector can always be written as a linear combination of the first two.

   44. Applying Eq. 3-23, F  qv  B (where q is a scalar) becomes Fx ˆi  Fy ˆj  Fz kˆ  q  vy Bz  vz By  ˆi  q  vz Bx  vx Bz  ˆj  q  vx By  vy Bx  kˆ

which — plugging in values — leads to three equalities:

4.0  2 (4.0 Bz  6.0 By ) 20  2 (6.0 Bx  2.0 Bz ) 12  2 (2.0 By  4.0 Bx ) Since we are told that Bx = By, the third equation leads to By = –3.0. Inserting this value into the first equation, we find Bz = –4.0. Thus, our answer is ˆ B  3.0 ˆi  3.0 ˆj  4.0 k.

45. The two vectors are given by

A  8.00(cos130 î  sin130 ˆj)  5.14 î  6.13 ˆj B  B î  B ˆj  7.72 î  9.20 ˆj. x

103 (a) The dot product of 5A  B is 5 A  B  5(5.14 ˆi  6.13 ˆj)  (7.72 ˆi  9.20 ˆj)  5[(5.14)(7.72)  (6.13)(9.20)]  83.4.

(b) In unit vector notation ˆ  1.14 103 kˆ 4 A  3B  12 A  B  12(5.14iˆ  6.13 ˆj)  (7.72iˆ  9.20 ˆj)  12(94.6k)

(c) We note that the azimuthal angle is undefined for a vector along the z axis. Thus, our result is “1.14103,  not defined, and  = 0.” 

(d) Since A is in the xy plane, and A  B is perpendicular to that plane, then the answer is 90. 

(e) Clearly, A + 3.00 k = –5.14 ^i + 6.13 ^j + 3.00 k^ . (f) The Pythagorean theorem yields magnitude A  (5.14)2  (6.13)2  (3.00)2  8.54 . 

The azimuthal angle is  = 130, just as it was in the problem statement ( A is the projection onto the xy plane of the new vector created in part (e)). The angle measured from the +z axis is  = cos1(3.00/8.54) = 69.4. 46. The vectors are shown on the diagram. The x axis runs from west to east and the y axis runs from south to north. Then ax = 5.0 m, ay = 0, bx = –(4.0 m) sin 35° = –2.29 m, by = (4.0 m) cos 35° = 3.28 m.

   (a) Let c  a  b . Then cx  ax  bx = 5.00 m  2.29 m = 2.71 m and cy  ay  by = 0 + 3.28 m = 3.28 m . The magnitude of c is

c  cx2  cy2 

 2.71m    3.28m   4.2 m. 2

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104    (b) The angle  that c  a  b makes with the +x axis is

c   cx 

 3.28    50.5  50.  2.71 

  tan 1  y   tan 1 

The second possibility ( = 50.4° + 180° = 230.4°) is rejected because it would point in a  direction opposite to c .

  (c) The vector b  a is found by adding a to b . The result is shown on the diagram to the right. Let c  b  a. The components are cx  bx  ax  2.29 m  5.00 m  7.29 m cy  by  ay  3.28 m.

The magnitude of c is c  cx2  c 2y  8.0 m .

(d) The tangent of the angle  that c makes with the +x axis (east) is tan  

cy cx



3.28 m  4.50. 7.29 m

There are two solutions: –24.2° and 155.8°. As the diagram shows, the second solution is    correct. The vector c  a  b is 24° north of west. 47. Noting that the given 130 is measured counterclockwise from the +x axis, the two vectors can be written as

A  8.00(cos130 î  sin130 ˆj)  5.14 î  6.13 ˆj B  B î  B ˆj  7.72 î  9.20 ˆj. x

(a) The angle between the negative direction of the y axis ( ˆj ) and the direction of A is

105

   A  (ˆj)  6.13  6.13  1   cos1    cos    140. 2 2   A 8.00   (  5.14)  (6.13)    

  cos1 

Alternatively, one may say that the y direction corresponds to an angle of 270, and the answer is simply given by 270130 = 140. (b) Since the y axis is in the xy plane, and A  B is perpendicular to that plane, then the answer is 90.0. (c) The vector can be simplified as ˆ  (5.14 î  6.13 ˆj)  (7.72 î  9.20 ˆj  3.00 k) ˆ A  ( B  3.00 k)  18.39 î  15.42 ˆj  94.61kˆ

ˆ |  97.6. The angle between the negative direction of the Its magnitude is | A  ( B  3.00k) y axis ( ˆj ) and the direction of the above vector is  15.42    99.1.  97.6 

  cos1 

48. Where the length unit is not displayed, the unit meter is understood. (a) We first note that the magnitudes of the vectors are a  | a |  (3.2)2  (1.6)2  3.58 and b  | b |  (0.50)2  (4.5)2  4.53 . Now,

a  b  axbx  a y by  ab cos  (3.2) (0.50)  (1.6) (4.5)  (3.58) (4.53) cos  which leads to  = 57° (the inverse cosine is double-valued as is the inverse tangent, but we know this is the right solution since both vectors are in the same quadrant).  (b) Since the angle (measured from +x) for a is tan–1(1.6/3.2) = 26.6°, we know the angle  for c is 26.6° –90° = –63.4° (the other possibility, 26.6° + 90° would lead to a cx < 0). Therefore, cx = c cos (–63.4° )= (5.0)(0.45) = 2.2 m.

(c) Also, cy = c sin (–63.4°) = (5.0)( –0.89) = – 4.5 m.  (d) And we know the angle for d to be 26.6° + 90° = 116.6°, which leads to

CHAPTER 3

106

dx = d cos(116.6°) = (5.0)( –0.45) = –2.2 m. (e) Finally, dy = d sin 116.6° = (5.0)(0.89) = 4.5 m. 49. THINK This problem deals with the displacement of a sailboat. We want to find the displacement vector between two locations. EXPRESS The situation is depicted in the figure below. Let a represent the first part of his actual voyage (50.0 km east) and c represent the intended voyage (90.0 km north). We look for a vector b such that c  a  b .

ANALYZE (a) Using the Pythagorean theorem, the distance traveled by the sailboat is b  (50.0 km)2  (90.0 km)2  103 km.

(b) The direction is

 50.0 km    29.1   90.0 km 

  tan 1 

west of north (which is equivalent to 60.9 north of due west). LEARN This problem could also be solved by first expressing the vectors in unit-vector ˆ c  (90.0 km)ˆj . This gives notation: a  (50.0 km)i,

b  c  a  (50.0 km)iˆ  (90.0 km)ˆj . The angle between b and the +x-axis is  90.0 km    119.1  .   50.0 km 

  tan 1 

The angle  is related to  by   90    .

107

50. The two vectors d1 and d 2 are given by d1  d1 ˆj and d2  d2 î. (a) The vector d2 / 4  (d2 / 4) î points in the +x direction. The ¼ factor does not affect the result. (b) The vector d1 /(4)  (d1 / 4)ˆj points in the +y direction. The minus sign (with the “4”) does affect the direction: (–y) = + y. (c) d1  d2  0 since î  ˆj = 0. The two vectors are perpendicular to each other. (d) d1  (d2 / 4)  (d1  d2 ) / 4  0 , as in part (c). (e) d1  d2  d1d2 (ˆj  î) = d1d2 kˆ , in the +z-direction. (f) d2  d1  d2 d1 (iˆ  ˆj) = d1d 2 kˆ , in the z-direction. (g) The magnitude of the vector in (e) is d1d 2 . (h) The magnitude of the vector in (f) is d1d 2 . (i) Since d1  (d2 / 4)  (d1d2 / 4)kˆ , the magnitude is d1d2 / 4. (j) The direction of d1  (d2 / 4)  (d1d2 / 4)kˆ is in the +z-direction. 51. Although we think of this as a three-dimensional movement, it is rendered effectively two-dimensional by referring measurements to its well-defined plane of the fault. (a) The magnitude of the net displacement is 

| AB |  | AD |2  | AC |2  (17.0 m) 2  (22.0 m) 2  27.8m. 

(b) The magnitude of the vertical component of AB is |AD| sin 52.0° = 13.4 m. 52. The three vectors are

d1  4.0 î  5.0ˆj  6.0 kˆ d 2   1.0 î  2.0ˆj+3.0 kˆ d3  4.0 î  3.0ˆj+2.0 kˆ

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108 (a) r  d1  d2  d3  (9.0 m)iˆ  (6.0 m)ˆj  ( 7.0 m)kˆ . 

(b) The magnitude of r is | r | (9.0 m)2  (6.0 m) 2  ( 7.0 m) 2  12.9 m. The angle between r and the z-axis is given by cos  

which implies   123  .

r  k̂  7.0 m    0.543 | r | 12.9 m

 is the angle between d1 and d 2 , and û is the unit vector in the direction of d 2 . Using the properties of the scalar (dot) product, we have  d d  d d (4.0)(  1.0)  (5.0)(2.0)  ( 6.0)(3.0)  12 d  d1  1 2  = 1 2     3.2 m. d2 14 ( 1.0)2  (2.0) 2  (3.0) 2  d1d 2 

(d) Now we are looking for d  such that d12  (4.0)2  (5.0)2  ( 6.0)2  77  d 2  d2 . From (c), we have d  77 m2  ( 3.2 m)2  8.2 m.

This gives the magnitude of the perpendicular component (and is consistent with what one would get using Eq. 3-24), but if more information (such as the direction, or a full specification in terms of unit vectors) is sought then more computation is needed. 53. THINK This problem involves finding scalar and vector products between two vectors a and b . EXPRESS We apply Eqs. 3-20 and 3-24 to calculate the scalar and vector products between two vectors: a  b  ab cos  | a  b |  ab sin .

ANALYZE (a) Given that a  | a |  10 , b  | b |  6.0 and   60  , the scalar (dot) product of a and b is a  b  ab cos   (10) (6.0) cos 60  30.

(b) Similarly, the magnitude of the vector (cross) product of the two vectors is

109 | a  b |  ab sin   (10) (6.0) sin 60  52.

LEARN When two vectors a and b are parallel (   0 ), their scalar and vector products are a  b  ab cos   ab and | a  b |  ab sin   0 , respectively. However, when they are perpendicular (   90  ), we have a  b  ab cos   0 and | a  b |  ab sin   ab . 



54. From the figure, it is clear that a + b + c = 0, where a  b . 



(a) a · b = 0 since the angle between them is 90º. 



(b) a · c = a · ( a  b ) =  a  =  



(c) Similarly, b · c = 9.0 . 55. We choose +x east and +y north and measure all angles in the “standard” way  (positive ones are counterclockwise from +x). Thus, vector d1 has magnitude d1 = 4.00 m  (with the unit meter) and direction 1 = 225°. Also, d 2 has magnitude d2 = 5.00 m and  direction 2 = 0°, and vector d 3 has magnitude d3 = 6.00 m and direction 3 = 60°.  (a) The x-component of d1 is d1x = d1 cos 1 = –2.83 m.  (b) The y-component of d1 is d1y = d1 sin 1 = –2.83 m.

 d 2 is d2x = d2 cos 2 = 5.00 m.

 (d) The y-component of d 2 is d2y = d2 sin 2 = 0.  (e) The x-component of d 3 is d3x = d3 cos 3 = 3.00 m.  (f) The y-component of d 3 is d3y = d3 sin 3 = 5.20 m.

(g) The sum of x-components is dx = d1x + d2x + d3x = –2.83 m + 5.00 m + 3.00 m = 5.17 m. (h) The sum of y-components is dy = d1y + d2y + d3y = –2.83 m + 0 + 5.20 m = 2.37 m. (i) The magnitude of the resultant displacement is

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110

d  d x2  d y2  (5.17 m)2  (2.37 m)2  5.69 m. (j) And its angle is

 = tan–1 (2.37/5.17) = 24.6°,

which (recalling our coordinate choices) means it points at about 25° north of east. (k) and (l) This new displacement (the direct line home) when vectorially added to the previous (net) displacement must give zero. Thus, the new displacement is the negative, or opposite, of the previous (net) displacement. That is, it has the same magnitude (5.69 m) but points in the opposite direction (25° south of west). 56. If we wish to use Eq. 3-5 directly, we should note that the angles for Q, R, and S are 100°, 250°, and 310°, respectively, if they are measured counterclockwise from the +x axis. (a) Using unit-vector notation, with the unit meter understood, we have P  10.0 cos  25.0  î  10.0sin  25.0  ˆj Q  12.0 cos 100  î  12.0sin 100  ˆj R  8.00 cos  250  î  8.00sin  250  ˆj S  9.00 cos  310  î  9.00sin  310  ˆj P  Q  R  S  (10.0 m )iˆ  (1.63 m)ˆj

(b) The magnitude of the vector sum is

(10.0 m)2  (1.63 m)2  10.2 m .

(c) The angle is tan–1 (1.63 m/10.0 m)  9.24° measured counterclockwise from the +x axis. 57. THINK This problem deals with addition and subtraction of two vectors. EXPRESS From the problem statement, we have

A  B  (6.0)iˆ  (1.0)ˆj,

A  B  (4.0)iˆ  (7.0)ˆj

Solving the simultaneous equations gives A and B .

111 ANALYZE Adding the above equations and dividing by 2 leads to A  (1.0)iˆ  (4.0)ˆj . The magnitude of A is

A  | A |  Ax2  Ay2  (1.0)2  (4.0)2  4.1 LEARN The vector B is B  (5.0)iˆ  ( 3.0)ˆj , and its magnitude is

B  | B |  Bx2  By2  (5.0)2  (3.0)2  5.8 . The results are summarized in the figure to the right.

58. The vector can be written as d  (2.5 m)jˆ , where we have taken ˆj to be the unit vector pointing north. (a) The magnitude of the vector a  4.0 d is (4.0)(2.5 m) = 10 m.    (b) The direction of the vector a = 4.0d is the same as the direction of d (north).

(c) The magnitude of the vector c =  3.0d is (3.0)(2.5 m) = 7.5 m.  (d) The direction of the vector c =  3.0d is the opposite of the direction of d . Thus, the  direction of c is south.

59. Reference to Figure 3-18 (and the accompanying material in that section) is helpful. If we  convert B to the magnitude-angle notation (as A already is) we have B  14.4  33.7 (appropriate notation especially if we are using a vector capable calculator in polar mode). Where the length unit is not displayed in the solution, the unit meter should be understood. In the magnitude-angle notation, rotating the axis by +20° amounts to subtracting that angle from the angles previously specified. Thus,   A  12.0  40.0  and B  (14.4  13.7 )  , where the ‘prime’ notation indicates that

the description is in terms of the new coordinates. Converting these results to (x, y) representations, we obtain (a) A  (9.19 m) ˆi  (7.71 m) ˆj.

CHAPTER 3

112 (b) Similarly, B  (14.0 m) ˆi  (3.41 m) ˆj . 60. The two vectors can be found be solving the simultaneous equations.

  (a) If we add the equations, we obtain 2a  6c , which leads to a  3c  9 ˆi  12 ˆj .

  (b) Plugging this result back in, we find b  c  3i  4j .

61. The three vectors given are

a  5.0 î  4.0 ˆj  6.0 kˆ b   2.0 î  2.0 ˆj  3.0 kˆ c  4.0 î  3.0 ˆj  2.0 kˆ

(a) The vector equation r  a  b  c is r  [5.0  ( 2.0)  4.0]iˆ  (4.0  2.0  3.0)ˆj  (  6.0  3.0  2.0)kˆ ˆ ˆ  7.0k. ˆ =11i+5.0j   Noting (b) We find the angle from +z by “dotting” (taking the scalar product) r with k. that r = |r | = (11.0)2 + (5.0)2 + (  7.0)2 = 14,

Eq. 3-20 with Eq. 3-23 leads to

r  k   7.0  14 1 cos     120 . (c) To find the component of a vector in a certain direction, it is efficient to “dot” it (take the scalar product of it) with a unit-vector in that direction. In this case, we make the desired unit-vector by ˆ ˆ +3.0kˆ b 2.0i+2.0j bˆ   . 2 2 2 |b |  2.0  (2.0)  (3.0)   We therefore obtain

 5.0  2.0    4.0  2.0    6.0 3.0    4.9 . ab  a  bˆ  2  2.0  (2.0)2  (3.0)2 (d) One approach (if all we require is the magnitude) is to use the vector cross product, as the problem suggests; another (which supplies more information) is to subtract the result  in part (c) (multiplied by b ) from a . We briefly illustrate both methods. We note that if

113   a cos  (where  is the angle between a and b ) gives ab (the component along b ) then we expect a sin  to yield the orthogonal component:

a sin  

  a b b

 7.3

(alternatively, one might compute  form part (c) and proceed more directly). The second method proceeds as follows:

g c

gh cb g b

 a  ab b  5.0  2.35 i  4.0  2.35 j + 6.0  353 . k = 2.65i  6.35j  2.47 k  This describes the perpendicular part of a completely. To find the magnitude of this part, we compute

(2.65)2  (6.35)2  ( 2.47)2  7.3

which agrees with the first method. 62. We choose +x east and +y north and measure all angles in the “standard” way  (positive ones counterclockwise from +x, negative ones clockwise). Thus, vector d1 has magnitude d1 = 3.66 (with the unit meter and three significant figures assumed) and  direction 1 = 90°. Also, d 2 has magnitude d2 = 1.83 and direction 2 = –45°, and vector  d 3 has magnitude d3 = 0.91 and direction 3 = –135°. We add the x and y components, respectively: x : d1 cos 1  d 2 cos  2  d3 cos 3  0.65 m

y : d1 sin 1  d 2 sin  2  d3 sin 3  1.7 m.    (a) The magnitude of the direct displacement (the vector sum d1 + d2 + d3 ) is (0.65 m)2  (1.7 m)2  1.8 m.

(b) The angle (understood in the sense described above) is tan–1 (1.7/0.65) = 69°. That is, the first putt must aim in the direction 69° north of east. 63. The three vectors are

d1  3.0 ˆi  3.0 ˆj  2.0 kˆ d  2.0 ˆi  4.0 ˆj  2.0 kˆ 2

ˆ d3  2.0 ˆi  3.0 ˆj  1.0 k.

CHAPTER 3

114 (a) Since d2  d3  0 î  1.0 ˆj  3.0 kˆ , we have ˆ  (0 î  1.0 ˆj  3.0 k) ˆ d1  (d 2  d3 )  (3.0 î  3.0 ˆj  2.0 k)  0  3.0 + 6.0  3.0 m 2 .

ˆ Thus, (b) Using Eq. 3-27, we obtain d2  d3  10 î  6.0 ˆj  2.0 k. ˆ  (10 î  6.0 ˆj  2.0 k) ˆ d1  (d 2  d3 )  (3.0 î  3.0 ˆj  2.0 k)  30  18  4.0  52 m3. 



(c) We found d2 + d3 in part (a). Use of Eq. 3-27 then leads to ˆ  (0 ˆi  1.0 ˆj  3.0 k) ˆ d1  (d 2  d3 )  (3.0 ˆi  3.0 ˆj  2.0 k) = (11iˆ + 9.0 ˆj + 3.0 kˆ ) m 2

64. THINK This problem deals with the displacement and distance traveled by a fly from one corner of a room to the diagonally opposite corner. The displacement vector is threedimensional. EXPRESS The displacement of the fly is illustrated in the figure below:

A coordinate system such as the one shown (above right) allows us to express the displacement as a three-dimensional vector. ANALYZE (a) The magnitude of the displacement from one corner to the diagonally opposite corner is

d  | d |  w2  l 2  h2 Substituting the values given, we obtain d  | d |  w2  l 2  h2  (3.70 m)2  (4.30 m)2  (3.00 m)2  6.42 m.

115 (b) The displacement vector is along the straight line from the beginning to the end point of the trip. Since a straight line is the shortest distance between two points, the length of the path cannot be less than d, the magnitude of the displacement. (c) The length of the path of the fly can be greater than d, however. The fly might, for example, crawl along the edges of the room. Its displacement would be the same but the path length would be  w  h 11.0 m. (d) The path length is the same as the magnitude of the displacement if the fly flies along the displacement vector. (e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to be upward (as shown in the figure above). Then the x component of the displacement is w = 3.70 m, the y component of the displacement is 4.30 m, and the z component is 3.00 m . Thus, the displacement vector can be written as ˆ d  (3.70 m) ˆi  ( 4.30 m) ˆj  (3.00 m)k.

(f) Suppose the path of the fly is as shown by the dotted lines on the diagram (below left). Pretend there is a hinge where the front wall of the room joins the floor and lay the wall down as shown (above right).

The shortest walking distance between the lower left back of the room and the upper right front corner is the dotted straight line shown on the diagram. Its length is

smin 

 w  h   l 2  3.70 m  3.00 m   (4.30 m)2  7.96 m. 2

LEARN To show that the shortest path is indeed given by smin , we write the length of the path as s  y 2  w2  (l  y)2  h2 .

The condition for minimum is given by

CHAPTER 3

116 ds  dy

y y 2  w2

ly



(l  y )2  h2

 0.

A little algebra shows that the condition is satisfied when y  lw /(w  h) , which gives

   l2 l2 2 smin  w2 1   h 1   ( w  h) 2  l 2 .  2  2  ( w  h ) ( w  h )     Any other path would be longer than 7.96 m. 65. (a) This is one example of an answer: (40 î – 20 ^j + 25 k^ ) m, with î directed antiparallel to the first path, ^j directed anti-parallel to the second path, and k^ directed upward (in order to have a right-handed coordinate system). Other examples include (40 î + 20 ^j + 25 k^ ) m and (40i^ – 20 ^j – 25 k^ ) m (with slightly different interpretations for the unit vectors). Note that the product of the components is positive in each example. (b) Using the Pythagorean theorem, we have

(40 m)2  (20 m)2 = 44.7 m  45 m.

66. The vectors can be written as a  a ˆi and b  bˆj where a, b  0. (a) We are asked to consider

 b b   j d d

FG IJ H K

in the case d > 0. Since the coefficient of j is positive, then the vector points in the +y direction. (b) If, however, d < 0, then the coefficient is negative and the vector points in the –y direction. (c) Since cos 90° = 0, then a  b  0 , using Eq. 3-20.  (d) Since b / d is along the y axis, then (by the same reasoning as in the previous part) a  (b / d )  0 .

  (e) By the right-hand rule, a  b points in the +z-direction.   (f) By the same rule, b  a points in the –z-direction. We note that b  a  a  b is true in this case and quite generally.

117  (g) Since sin 90° = 1, Eq. 3-24 gives | a  b |  ab where a is the magnitude of a .

(h) Also, | a  b |  | b  a |  ab . (i) With d > 0, we find that a  (b / d ) has magnitude ab/d. (j) The vector a  (b / d ) points in the +z direction. 67. We note that the set of choices for unit vector directions has correct orientation (for a right-handed coordinate system). Students sometimes confuse “north” with “up”, so it might be necessary to emphasize that these are being treated as the mutually perpendicular directions of our real world, not just some “on the paper” or “on the blackboard” representation of it. Once the terminology is clear, these questions are basic to the definitions of the scalar (dot) and vector (cross) products.

ˆ (a) ˆi  k=0 since ˆi  kˆ ˆ  (  ˆj)=0 since kˆ  ˆj . (b) (k)

68. A sketch of the displacements is shown. The resultant (not shown) would be a straight line from start (Bank) to finish (Walpole). With a careful drawing, one should find that the resultant vector has length 29.5 km at 35° west of south.

69. The point P is displaced vertically by 2R, where R is the radius of the wheel. It is displaced horizontally by half the circumference of the wheel, or R. Since R = 0.450 m,

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118

the horizontal component of the displacement is 1.414 m and the vertical component of the displacement is 0.900 m. If the x axis is horizontal and the y axis is vertical, the vector displacement (in meters) is r  1.414 ˆi + 0.900 ˆj . The displacement has a magnitude of



r  and an angle of



 R    2R   R 2  4  1.68m 2

 2R  1  2  tan 1    tan    32.5 R   

above the floor. In physics there are no “exact” measurements, yet that angle computation seemed to yield something exact. However, there has to be some uncertainty in the observation that the wheel rolled half of a revolution, which introduces some indefiniteness in our result. 70. The diagram of her walk,  shows the displacement vectors for the two segments  labeled A and B , and the total (“final”) displacement vector, labeled r . We take east to  be the +x direction and north to be the +y direction. We observe that the angle between A and the x axis is 60°. Where the units are not explicitly shown, the distances are  understood to be in meters. Thus, the components of A are Ax = 250 cos60° = 125 and Ay  = 250 sin60° = 216.5. The components of B are Bx = 175 and By = 0. The components of the total displacement are rx = Ax + Bx = 125 + 175 = 300 ry = Ay + By = 216.5 + 0 = 216.5.

(a) The magnitude of the resultant displacement is

| r |  rx2  ry2  (300 m)2  (216.5 m)2  370m. (b) The angle the resultant displacement makes with the +x axis is

 ry   216.5 m  tan 1    tan 1    36.  300 m   rx 

119 The direction is 36 north of due east. (c) The total distance walked is d = 250 m + 175 m = 425 m. (d) The total distance walked than the magnitude of the resultant displacement.  is greater  The diagram shows why: A and B are not collinear. 71. The vector d (measured in meters) can be represented as d  (3.0 m)(ˆj) , where ˆj is the unit vector pointing south. Therefore, 5.0d  5.0(3.0 m ˆj)  (15 m) ˆj. (a) The positive scalar factor (5.0) affects the magnitude but not the direction. The magnitude of 5.0d is 15 m. (b) The new direction of 5d is the same as the old: south. The vector 2.0d can be written as 2.0d  (6.0 m) ˆj. (c) The absolute value of the scalar factor (|2.0| = 2.0) affects the magnitude. The new magnitude is 6.0 m. (d) The minus sign carried by this scalar factor reverses the direction, so the new direction is  ĵ , or north. 72. The ant’s trip consists of three displacements:

d1  (0.40 m)(cos 225 î  sin 225 ˆj)  (0.28 m) î  (0.28 m) ˆj d  (0.50 m) î 2

d3  (0.60 m)(cos 60 ˆi  sin 60 ˆj)  (0.30 m) ˆi  (0.52 m )ˆj, where the angle is measured with respect to the positive x axis. We have taken the positive x and y directions to correspond to east and north, respectively. (a) The x component of d1 is d1x  (0.40 m) cos 225  0.28 m . (b) The y component of d1 is d1 y  (0.40 m)sin 225  0.28 m . (c) The x component of d 2 is d2 x  0.50 m . (d) The y component of d 2 is d2 y  0 m .

CHAPTER 3

120 (e) The x component of d 3 is d3 x  (0.60 m) cos 60  0.30 m . (f) The y component of d 3 is d3 y  (0.60 m)sin 60  0.52 m . (g) The x component of the net displacement d net is

dnet, x  d1x  d2 x  d3 x  (0.28 m)  (0.50 m)  (0.30 m)  0.52 m.

(h) The y component of the net displacement d net is dnet, y  d1 y  d2 y  d3 y  (0.28 m)  (0 m)  (0.52 m)  0.24 m.

(i) The magnitude of the net displacement is 2 2 dnet  dnet, (0.52 m)2  (0.24 m)2  0.57 m. x  d net, y 

(j) The direction of the net displacement is  d net, y  1  0.24 m    tan    25 (north of east)  0.52 m   d net, x 

  tan 1 

If the ant has to return directly to the starting point, the displacement would be d net . (k) The distance the ant has to travel is | dnet |  0.57 m. (l) The direction the ant has to travel is 25 (south of west) . 73. We apply Eq. 3-23 and Eq. 3-27.  (a) a  b  (axby  a ybx ) kˆ since all other terms vanish, due to the fact that neither a nor  b have any z components. Consequently, we obtain ((3.0)(4.0)  (5.0)(2.0))kˆ  2.0kˆ .

(b) a  b  axbx + a yby yields (3.0)(2.0) + (5.0)(4.0) = 26. (c) a  b = (3.0  2.0) ˆi  (5.0  4.0) ˆj  (a + b )  b = (5.0) (2.0) + (9.0) (4.0) = 46 . (d) Several approaches are available. In this solution, we will construct a b unit-vector  and “dot” it (take the scalar product of it) with a . In this case, we make the desired unitvector by

121 b  bˆ  |b |

We therefore obtain

2.0 ˆi  4.0 ˆj (2.0)2  (4.0)2

(3.0)(2.0)  (5.0)(4.0) ab  a  bˆ   5.81. (2.0)2  (4.0)2

74. The two vectors a and b are given by ˆ  1.45 ˆj  2.85 kˆ a  3.20(cos 63 ˆj  sin 63 k) ˆ  0.937iˆ  1.04 kˆ b  1.40(cos 48 ˆi  sin 48 k)  The components of a are ax = 0, ay = 3.20 cos 63° = 1.45, and az = 3.20 sin 63° = 2.85.  The components of b are bx = 1.40 cos 48° = 0.937, by = 0, and bz = 1.40 sin 48° = 1.04.

(a) The scalar (dot) product is therefore

  a b  axbx  a yby  azbz  0 0.937  145 . 0  2.85 104 .  2.97.

b gb

g b gb g b gb g

(b) The vector (cross) product is a  b   a y bz  az by  ˆi +  a z bx  axbz  ˆj +  axby  a y bx  kˆ



  1.451.04   0   ˆi +   2.85  0.937   0  ˆj   0  1.45  0.937   kˆ  1.51iˆ + 2.67 ˆj  1.36kˆ .   (c) The angle  between a and b is given by

  a b 2.97 1   48.5 .   cos   ab    3.20 1.40  

  cos 1 

75. We orient i eastward, j northward, and k upward, and use the following fundamental products: î  ˆj   ˆj  î  kˆ ˆj  kˆ   kˆ  ˆj  î kˆ  î   î  kˆ  ˆj (a) “north cross west” = ˆj (î)  kˆ = “up.”

CHAPTER 3

122 ˆ  ( ˆj)  0 . (b) “down dot south” = (k)

ˆ   ˆj = “south.” (c) “east cross up” = î  (k) (d) “west dot west” = ( î)  ( î )  1 . (e) “south cross south” = ( ˆj)  ( ˆj)  0 .

 76. Let A denote the magnitude of ; similarly for the other vectors. The vector equation A  is A  B = C where B = 8.0  m and C = 2A. We are also told that the angle (measured in the ‘standard’ sense) for A is 0° and the angle for C is 90°, which makes this a right triangle (when drawn in a “head-to-tail” fashion) where B is the size of the hypotenuse. Using the Pythagorean theorem, B

A2  C 2  8.0 

A 2  4 A2

which leads to A = 8 / 5 = 3.6 m. 77. We orient i eastward, j northward, and k upward. (a) The displacement is d  (1300 m)iˆ  ( 2200 m)ˆj  (  410 m)kˆ . (b) The displacement for the return portion is d   (1300 m)iˆ  ( 2200 m)ˆj and the magnitude is d   (1300 m)2  (  2200 m)2  2.56 103 m . The net displacement is zero since his final position matches his initial position.

     78. Let c  b  a . Then the magnitude of c is c = ab sin . Since c is perpendicular    to a the magnitude of a  c is ac. The magnitude of a  (b  a ) is consequently | a  (b  a ) |  ac  a 2b sin  . Substituting the values given, we obtain

| a  (b  a ) |  a 2b sin   (3.90)2 (2.70)sin 63.0  36.6 . 79. The area of a triangle is half the product of its base and altitude. The base is the side  formed by vector a. Then the altitude is b sin  and the area is A  12 ab sin   12 | a  b | . Substituting the values given, we have A

1 1 ab sin   (4.3)(5.4)sin 46  8.4 . 2 2

Chapter 4  1. (a) The magnitude of r is

| r | (5.0 m)2  (  3.0 m)2  (2.0 m)2  6.2 m.

(b) A sketch is shown. The coordinate values are in meters. 2. (a) The position vector, according to Eq. 4-1, is r = (  5.0 m) ˆi + (8.0 m)jˆ . (b) The magnitude is |r | x2 + y 2 + z 2  (5.0 m)2  (8.0 m)2  (0 m) 2  9.4 m. (c) Many calculators have polar  rectangular conversion capabilities that make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 3-6, we obtain:  8.0 m    58 or 122  5.0 m 

  tan 1 

where the latter possibility (122° measured counterclockwise from the +x direction) is chosen since the signs of the components imply the vector is in the second quadrant. (d) The sketch is shown to the right. The vector is 122° counterclockwise from the +x direction.  (e) The displacement is r  r  r where r is given in part (a) and ˆ Therefore, r  (8.0 m)iˆ  (8.0 m)jˆ . r   (3.0 m)i.

(f) The magnitude of the displacement is | r | (8.0 m)2  (  8.0 m)2  11 m.

(g) The angle for the displacement, using Eq. 3-6, is  8.0 m  tan 1   =  45 or 135  8.0 m 

123

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124

where we choose the former possibility (45°, or 45° measured clockwise from +x) since the signs of the components imply the vector is in the fourth quadrant. A sketch of r is shown on the right.     3. The initial position vector ro satisfies r  ro  r , which results in

ˆ  (2.0iˆ  3.0jˆ  6.0k)m ˆ  (2.0 m) ˆi  (6.0 m) ˆj  (10 m) kˆ . ro  r  r  (3.0jˆ  4.0k)m

4. We choose a coordinate system with origin at the clock center and +x rightward (toward the “3:00” position) and +y upward (toward “12:00”). (a) In unit-vector notation, we have r1  (10 cm)iˆ and r2  (10 cm)ˆj. Thus, Eq. 4-2 gives r  r2  r1  (10 cm)iˆ  (10 cm)ˆj.

The magnitude is given by |  r | ( 10 cm)2  ( 10 cm)2  14 cm. (b) Using Eq. 3-6, the angle is   10 cm    45  or  135  .   10 cm 

  tan 1 

We choose  135  since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write  r  r2  r1  ( 10 cm)iˆ  ( 10 cm)ˆj  (14 cm   135 ).

(c) In this case, we have r1  (10 cm)ˆj and r2  (10 cm)ˆj, and r  (20 cm)ˆj. Thus, |  r | 20 cm. (d) Using Eq. 3-6, the angle is given by

 20 cm    90  .  0 cm 

  tan 1 

(e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a full-hour sweep is also zero.

125 5. THINK This problem deals with the motion of a train in two dimensions. The entire trip consists of three parts, and we’re interested in the overall average velocity. EXPRESS The average velocity of the entire trip is given by Eq. 4-8, vavg  r / t , where the total displacement  r   r1   r2   r3 is the sum of three displacements (each result of a constant velocity during a given time), and  t   t1   t2   t3 is the total amount of time for the trip. We use a coordinate system with +x for East and +y for North. ANALYZE (a) In unit-vector notation, the first displacement is given by km   40.0 min  ˆ  ˆ r1 =  60.0   i = (40.0 km)i. h   60 min/h   20.0 min The second displacement has a magnitude of (60.0 km h )  ( 60 min/h )  20.0 km, and its direction is 40° north of east. Therefore,

r2  (20.0 km) cos(40.0) ˆi  (20.0 km) sin(40.0) ˆj  (15.3 km) ˆi  (12.9 km) ˆj.

Similarly, the third displacement is km   50.0 min  ˆ  ˆ r3    60.0   i = (  50.0 km) i. h 60 min/h    Thus, the total displacement is

r  r1  r2  r3  (40.0 km)iˆ  (15.3 km) î  (12.9 km) ˆj  (50.0 km) î  (5.30 km) î  (12.9 km) ˆj. The time for the trip is  t  (40.0 + 20.0 + 50.0) min = 110 min, which is equivalent to 1.83 h. Eq. 4-8 then yields vavg 

(5.30 km) ˆi  (12.9 km) ˆj  (2.90 km/h) ˆi  (7.01 km/h) ˆj. 1.83 h

The magnitude of vavg is | vavg |  (2.90 km/h)2  (7.01 km/h)2  7.59 km/h. (b) The angle is given by v  v   avg, x 

 7.01 km/h    67.5  (north of east),  2.90 km/h 

  tan 1  avg, y   tan 1  or 22.5 east of due north.

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126

LEARN The displacement of the train is depicted in the figure below:

Note that the net displacement  r is found by adding  r1 ,  r2 and  r3 vectorially. 6. To emphasize the fact that the velocity is a function of time, we adopt the notation v(t) for dx / dt. (a) Equation 4-10 leads to v(t ) 

d ˆ  (3.00 m/s)iˆ  (8.00 m/s2 )t ˆj (3.00tˆi  4.00t 2ˆj + 2.00k) dt

ˆ m/s. (b) Evaluating this result at t = 2.00 s produces v = (3.00iˆ  16.0j)

 16.0 m/s  tan 1    79.4 or 101  3.00 m/s 

where we choose the first possibility (79.4° measured clockwise from the +x direction, or 281° counterclockwise from +x) since the signs of the components imply the vector is in the fourth quadrant. 7. Using Eq. 4-3 and Eq. 4-8, we have vavg 

ˆ m  (5.0iˆ  6.0jˆ + 2.0k) ˆ m (  2.0iˆ + 8.0jˆ  2.0k) ˆ m/s.  (0.70iˆ +1.40jˆ  0.40k) 10 s

8. Our coordinate system has i pointed east and j pointed north. The first displacement is r  (483 km)iˆ and the second is r  (966 km)ˆj. AB

127 (a) The net displacement is rAC  rAB  rBC  (483 km)iˆ  (966 km)jˆ

which yields | rAC |  (483 km) 2  (  966 km) 2  1.08 103 km. (b) The angle is given by

 966 km    63.4.  483 km 

  tan 1 

We observe that the angle can be alternatively expressed as 63.4° south of east, or 26.6° east of south.  (c) Dividing the magnitude of rAC by the total time (2.25 h) gives

vavg 

(483 km)iˆ  (966 km)jˆ  (215 km/h)iˆ  (429 km/h)ˆj 2.25 h

with a magnitude | vavg |  (215 km/h) 2  ( 429 km/h) 2  480 km/h. (d) The direction of vavg is 26.6° east of south, same as in part (b). In magnitude-angle notation, we would have vavg  (480 km/h   63.4 ).  (e) Assuming the AB trip was a straight one, and similarly for the BC trip, then | rAB | is the  distance traveled during the AB trip, and | rBC | is the distance traveled during the BC trip. Since the average speed is the total distance divided by the total time, it equals

483 km  966 km  644 km/h. 2.25 h

9. The (x,y) coordinates (in meters) of the points are A = (15, 15), B = (30, 45), C = (20, 15), and D = (45, 45). The respective times are tA = 0, tB = 300 s, tC = 600 s, and tD =  900 s. Average velocity is defined by Eq. 4-8. Each displacement r is understood to originate at point A. (a) The average velocity having the least magnitude (5.0 m/600 s) is for the displacement ending at point C: | vavg |  0.0083 m/s. (b) The direction of vavg is 0 (measured counterclockwise from the +x axis).

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128

(c) The average velocity having the greatest magnitude ( (15 m)2  (30 m)2 / 300 s ) is for the displacement ending at point B: | vavg | 0.11 m/s. (d) The direction of vavg is 297 (counterclockwise from +x) or 63

(which is

equivalent to measuring 63 clockwise from the +x axis). 10. We differentiate r  5.00t ˆi  (et  ft 2 )ˆj . ^

(a) The particle’s motion is indicated by the derivative of r : v = 5.00 i + (e + 2ft) j . The angle of its direction of motion is consequently

 = tan1(vy /vx ) = tan1[(e + 2ft)/5.00]. The graph indicates o = 35.0, which determines the parameter e: e = (5.00 m/s) tan(35.0) = 3.50 m/s. (b) We note (from the graph) that  = 0 when t = 14.0 s. Thus, e + 2ft = 0 at that time. This determines the parameter f : f 

e 3.5 m/s   0.125 m/s2 . 2t 2(14.0 s)

11. In parts (b) and (c), we use Eq. 4-10 and Eq. 4-16. For part (d), we find the direction of the velocity computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the given expression, we obtain

r t 2.00  [2.00(8)  5.00(2)]iˆ + [6.00  7.00(16)] ˆj  (6.00 ˆi  106 ˆj) m (b) Taking the derivative of the given expression produces v (t ) = (6.00t 2  5.00) ˆi  28.0t 3 ˆj

where we have written v(t) to emphasize its dependence on time. This becomes, at t = 2.00 s, v = (19.0 î  224 ˆj) m/s.  (c) Differentiating the v (t ) found above, with respect to t produces 12.0t î  84.0t 2 ˆj, which yields a =(24.0 î  336 ˆj) m/s2 at t = 2.00 s.

 (d) The angle of v , measured from +x, is either

129  224 m/s  tan 1     85.2 or 94.8  19.0 m/s 

where we settle on the first choice (–85.2°, which is equivalent to 275° measured counterclockwise from the +x axis) since the signs of its components imply that it is in the fourth quadrant. 12. We adopt a coordinate system with i pointed east and j pointed north; the coordinate origin is the flagpole. We “translate” the given information into unit-vector notation as follows: ro  (40.0 m)iˆ and vo = (10.0 m/s)jˆ ˆ r  (40.0 m)ˆj and v  (10.0 m/s)i.  (a) Using Eq. 4-2, the displacement r is

 r  r  ro  ( 40.0 m)iˆ  (40.0 m)ˆj with a magnitude |  r | ( 40.0 m)2  (40.0 m)2  56.6 m.  (b) The direction of r is

 y   1  40.0 m    tan     45.0  or 135.  x    40.0 m 

  tan 1 

Since the desired angle is in the second quadrant, we pick 135 ( 45 north of due west). Note that the displacement can be written as  r  r  ro   56.6 135   in terms of the magnitude-angle notation.  (c) The magnitude of vavg is simply the magnitude of the displacement divided by the

time (t = 30.0 s). Thus, the average velocity has magnitude (56.6 m)/(30.0 s) = 1.89 m/s.   (d) Equation 4-8 shows that vavg points in the same direction as r , that is, 135 ( 45 north of due west).

(e) Using Eq. 4-15, we have

The

magnitude

aavg 

v  vo ˆ  (0.333 m/s2 )iˆ  (0.333 m/s2 )j. t

the

average

of 2 2

2 2

acceleration 2

| aavg |  (0.333 m/s )  (0.333 m/s )  0.471 m/s .

vector

therefore

equal

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130

(f) The direction of aavg is

 0.333 m/s 2   45  or  135 . 2   0.333 m/s 

  tan 1 

Since the desired angle is now in the first quadrant, we choose 45 , and aavg points north of due east. 13. THINK Knowing the position of a particle as function of time allows us to calculate its corresponding velocity and acceleration by taking time derivatives. EXPRESS From the position vector r (t ) , the velocity and acceleration of the particle can be found by differentiating r (t ) with respect to time: v

dr , dt

a

dv d 2 r  . dt dt 2

ANALYZE (a) Taking the derivative of the position vector r (t )  ˆi  (4t 2 )jˆ  t kˆ with respect to time, we have, in SI units (m/s), d ˆ ˆ  8t ˆj  k. ˆ (i  4t 2 ˆj  t k) dt (b) Taking another derivative with respect to time leads to, in SI units (m/s2), v

a

d ˆ  8 ˆj. (8t ˆj  k) dt

LEARN The particle undergoes constant acceleration in the +y-direction. This can be seen by noting that the y component of r (t ) is 4t2, which is quadratic in t.   14. We use Eq. 4-15 with v1 designating the initial velocity and v2 designating the later one.

(a) The average acceleration during the t = 4 s interval is aavg 

ˆ m/s  (4.0 î  22 ˆj+3.0 k) ˆ m/s (  2.0 î  2.0 ˆj+5.0 k) ˆ  (  1.5 m/s 2 ) î  (0.5m/s 2 ) k. 4s

 (b) The magnitude of aavg is

(1.5 m/s2 )2  (0.5 m/s2 )2  1.6 m/s2 .

131

 0.5 m/s 2  tan 1   18 or 162 2   1.5 m/s   where we settle on the second choice since the signs of its components imply that it is in the second quadrant. 15. THINK Given the initial velocity and acceleration of a particle, we’re interested in finding its velocity and position at a later time. EXPRESS Since the acceleration, a  ax ˆi  ay ˆj  ( 1.0 m/s 2 )iˆ  ( 0.50 m/s 2 )ˆj , is constant in both x and y directions, we may use Table 2-1 for the motion along each direction. This can be handled individually (for x and y) or together with the unit-vector notation (for  r ). Since the particle started at the origin, the coordinates of the particle at any time t are       given by r  v0 t  21 at 2 . The velocity of the particle at any time t is given by v  v0  at ,   where v0 is the initial velocity and a is the (constant) acceleration. Along the x-direction, we have 1 x(t )  v0 xt  axt 2 , vx (t )  v0 x  axt 2 Similarly, along the y-direction, we get

1 y(t )  v0 y t  a y t 2 , 2

v y (t )  v0 y  a yt .

Known: v0 x  3.0 m/s, v0 y  0, ax  1.0 m/s 2 , ay  0.5 m/s 2 . ANALYZE (a) Substituting the values given, the components of the velocity are vx (t )  v0 x  axt  (3.0 m/s)  (1.0 m/s 2 )t v y (t )  v0 y  a y t  (0.50 m/s 2 )t

When the particle reaches its maximum x coordinate at t = tm, we must have vx = 0. Therefore, 3.0 – 1.0tm = 0 or tm = 3.0 s. The y component of the velocity at this time is

vy (t  3.0 s)  (0.50 m/s 2 )(3.0)  1.5 m/s Thus, vm  ( 1.5 m/s)jˆ . (b) At t = 3.0 s , the components of the position are

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1 1 x(t  3.0 s)  v0 xt  axt 2  (3.0 m/s)(3.0 s)  (  1.0 m/s 2 )(3.0 s) 2  4.5 m 2 2 1 2 1 2 y (t  3.0 s)  v0 y t  a y t  0  ( 0.5 m/s )(3.0 s) 2   2.25 m 2 2 Using unit-vector notation, the results can be written as rm  (4.50 m) ˆi  (2.25 m) ˆj. LEARN The motion of the particle in this problem is two-dimensional, and the kinematics in the x- and y-directions can be analyzed separately. 16. We make use of Eq. 4-16. (a) The acceleration as a function of time is a

dv d  dt dt

 6.0t  4.0t  ˆi + 8.0 ˆj   6.0  8.0t  ˆi 2

in SI units. Specifically, we find the acceleration vector at t  3.0 s to be  6.0  8.0(3.0)  ˆi  (18 m/s2 )i.ˆ

 (b) The equation is a  6.0  8.0t i = 0 ; we find t = 0.75 s. (c) Since the y component of the velocity, vy = 8.0 m/s, is never zero, the velocity cannot vanish. (d) Since speed is the magnitude of the velocity, we have v |v | 

 6.0t  4.0t   8.0  10 2 2

in SI units (m/s). To solve for t, we first square both sides of the above equation, followed by some rearrangement:

 6.0t  4.0t   64  100   6.0t  4.0t   36 2 2

2 2

Taking the square root of the new expression and making further simplification lead to 6.0t  4.0t 2  6.0  4.0t 2  6.0t  6.0  0

Finally, using the quadratic formula, we obtain

133

t

6.0  36  4  4.0  6.0  2  8.0 

where the requirement of a real positive result leads to the unique answer: t = 2.2 s. 17. We find t by applying Eq. 2-11 to motion along the y axis (with vy = 0 characterizing y = ymax ): 0 = (12 m/s) + (2.0 m/s2)t  t = 6.0 s. Then, Eq. 2-11 applies to motion along the x axis to determine the answer: vx = (8.0 m/s) + (4.0 m/s2)(6.0 s) = 32 m/s. Therefore, the velocity of the cart, when it reaches y = ymax , is (32 m/s)i^. 1 18. We find t by solving  x  x0  v0 xt  axt 2 : 2

1 12.0 m  0  (4.00 m/s)t  (5.00 m/s 2 )t 2 2

where we have used x = 12.0 m, vx = 4.00 m/s, and ax = 5.00 m/s2 . We use the quadratic formula and find t = 1.53 s. Then, Eq. 2-11 (actually, its analog in two dimensions) applies with this value of t. Therefore, its velocity (when x = 12.00 m) is

v  v0  at  (4.00 m/s)iˆ  (5.00 m/s 2 )(1.53 s)iˆ  (7.00 m/s2 )(1.53 s)jˆ  (11.7 m/s) ˆi  (10.7 m/s) ˆj. Thus, the magnitude of v is | v |  (11.7 m/s)2  (10.7 m/s)2  15.8 m/s.  (b) The angle of v , measured from +x, is

 10.7 m/s  tan 1    42.6.  11.7 m/s 

19. We make use of Eq. 4-16 and Eq. 4-10. Using a  3t ˆi  4tˆj , we have (in m/s) t ˆ  t (3 t ˆi  4tˆj) dt  5.00  3t 2 / 2  iˆ   2.00  2t 2  ˆj v (t )  v0   a dt  (5.00iˆ  2.00j)  0

Integrating using Eq. 4-10 then yields (in meters)

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134

t t ˆ dt r (t )  r0   vdt  (20.0iˆ  40.0ˆj)   [(5.00  3t 2 / 2)iˆ  (2.00  2t 2 )j] 0

 (20.0iˆ  40.0ˆj)  (5.00t  t 3 / 2)iˆ  (2.00t  2t 3 /3)jˆ  (20.0  5.00t  t 3 / 2)iˆ  (40.0  2.00t  2t 3 /3)jˆ

(a) At t  4.00 s , we have r (t  4.00 s)  (72.0 m)iˆ  (90.7 m)ˆj. (b) v (t  4.00 s)  (29.0 m/s)iˆ  (34.0 m/s)ˆj . Thus, the angle between the direction of travel and +x, measured counterclockwise, is   tan 1[(34.0 m/s) /(29.0 m/s)]  49.5 . 20. The acceleration is constant so that use of Table 2-1 (for both the x and y motions) is permitted. Where units are not shown, SI units are to be understood. Collision between particles A and B requires two things. First, the y motion of B must satisfy (using Eq. 2-15 and noting that  is measured from the y axis) y

1 2 1 a y t  30 m  (0.40 m/s 2 ) cos   t 2 . 2 2

Second, the x motions of A and B must coincide: vt 

1 2 1 axt  (3.0 m/s)t  (0.40 m/s 2 ) sin   t 2 . 2 2

We eliminate a factor of t in the last relationship and formally solve for time: t

2v 2(3.0 m/s)  . ax (0.40 m/s 2 ) sin 

This is then plugged into the previous equation to produce   1 2(3.0 m/s) 30 m  (0.40 m/s 2 ) cos     2 2  (0.40 m/s ) sin  

which, with the use of sin2  = 1 – cos2 , simplifies to

30 

9.0 cos  9.0  1  cos 2   cos  . 2 0.20 1  cos   0.20  30 

We use the quadratic formula (choosing the positive root) to solve for cos : cos  

1.5  1.52  4 1.0  1.0  2



1 2

135

which yields   cos1

FG 1IJ  60 . H 2K

21. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The initial velocity is horizontal so that v0 y  0 and

v0 x  v0  10 m s. (a) With the origin at the initial point (where the dart leaves the thrower’s hand), the y coordinate of the dart is given by y   21 gt 2 , so that with y = –PQ we have

PQ  12  9.8 m/s2   0.19 s   0.18 m. 2

(b) From x = v0t we obtain x = (10 m/s)(0.19 s) = 1.9 m. 22. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. (a) With the origin at the initial point (edge of table), the y coordinate of the ball is given by y   21 gt 2 . If t is the time of flight and y = –1.20 m indicates the level at which the ball hits the floor, then 2  1.20 m 

t

9.80 m/s 2

 0.495s.

 (b) The initial (horizontal) velocity of the ball is v  v0 i . Since x = 1.52 m is the horizontal position of its impact point with the floor, we have x = v0t. Thus,

v0 

x 1.52 m   3.07 m/s. t 0.495 s

23. (a) From Eq. 4-22 (with 0 = 0), the time of flight is t

2h 2(45.0 m)   3.03 s. g 9.80 m/s 2

(b) The horizontal distance traveled is given by Eq. 4-21: x  v0t  (250 m/s)(3.03 s)  758 m.

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136

24. We use Eq. 4-26  v2  v 2  9.50m/s  Rmax   0 sin 20   0   9.209 m  9.21m 9.80m/s2  g max g 2

to compare with Powell’s long jump; the difference from Rmax is only R =(9.21m – 8.95m) = 0.259 m. 25. Using Eq. (4-26), the take-off speed of the jumper is

v0 

gR (9.80 m/s 2 )(77.0 m)   43.1 m/s sin 20 sin 2(12.0)

26. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is the throwing point (the stone’s initial position). The x component of its initial velocity is given by v0 x  v0 cos 0 and the y component is given by v0 y  v0 sin  0 , where v0 = 20 m/s is the initial speed and 0 = 40.0° is the launch angle. (a) At t = 1.10 s, its x coordinate is

x  v0t cos  0  20.0 m / s 110 . s cos 40.0  16.9 m

(b) Its y coordinate at that instant is y  v0t sin 0 

1 2 1 2 gt   20.0m/s 1.10s  sin 40.0   9.80m/s 2  1.10s   8.21 m. 2 2

(c) At t' = 1.80 s, its x coordinate is x  20.0 m / s 180 . s cos 40.0  27.6 m. (d) Its y coordinate at t' is y   20.0m/s 1.80s  sin 40.0 

1 9.80m/s 2  1.80s 2   7.26m.  2

(e) The stone hits the ground earlier than t = 5.0 s. To find the time when it hits the ground solve y  v0t sin  0  21 gt 2  0 for t. We find t

2 20.0 m / s 2v0 sin  0  sin 40  2.62 s. g 9.8 m / s2

137 Its x coordinate on landing is x  v0t cos 0   20.0 m/s  2.62 s  cos 40  40.2 m.

(f) Assuming it stays where it lands, its vertical component at t = 5.00 s is y = 0. 27. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below the release point. We write 0 = –30.0° since the angle shown in the figure is measured clockwise from horizontal. We note that the initial speed of the decoy is the plane’s speed at the moment of release: v0 = 290 km/h, which we convert to SI units: (290)(1000/3600) = 80.6 m/s. (a) We use Eq. 4-12 to solve for the time: x  (v0 cos 0 ) t

 t

700 m  10.0 s. (80.6 m/s) cos (30.0)

(b) And we use Eq. 4-22 to solve for the initial height y0: 1 1 y  y0  (v0 sin 0 ) t  gt 2  0  y0  (40.3 m/s)(10.0 s)  (9.80 m/s 2 )(10.0 s)2 2 2

which yields y0 = 897 m. 28. (a) Using the same coordinate system assumed in Eq. 4-22, we solve for y = h: h  y0  v0 sin  0t 

1 2 gt 2

which yields h = 51.8 m for y0 = 0, v0 = 42.0 m/s, 0 = 60.0°, and t = 5.50 s. (b) The horizontal motion is steady, so vx = v0x = v0 cos 0, but the vertical component of velocity varies according to Eq. 4-23. Thus, the speed at impact is

v

 v0 cos0    v0 sin 0  gt   27.4 m/s. 2

bv sin g  67.5 m. H 2

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138

29. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at its initial position (where it is launched). At maximum height, we observe vy = 0 and denote vx = v (which is also equal to v0x). In this notation, we have v0  5v. Next, we observe v0 cos 0 = v0x = v, so that we arrive at an equation (where v  0 cancels) which can be solved for 0: 1 (5v) cos 0  v  0  cos 1    78.5. 5

30. Although we could use Eq. 4-26 to find where it lands, we choose instead to work with Eq. 4-21 and Eq. 4-22 (for the soccer ball) since these will give information about where and when and these are also considered more fundamental than Eq. 4-26. With y = 0, we have 1 (19.5 m/s)sin 45.0 y  (v0 sin 0 ) t  gt 2  t   2.81 s. 2 (9.80 m/s 2 ) / 2 Then Eq. 4-21 yields x = (v0 cos 0)t = 38.7 m. Thus, using Eq. 4-8, the player must have an average velocity of vavg 

r (38.7 m) ˆi  (55 m)iˆ   (5.8 m/s) ˆi t 2.81s

which means his average speed (assuming he ran in only one direction) is 5.8 m/s. 31. We first find the time it takes for the volleyball to hit the ground. Using Eq. 4-22, we have 1 1 y  y0  (v0 sin 0 ) t  gt 2  0  2.30 m  (20.0 m/s)sin(18.0)t  (9.80 m/s2 )t 2 2 2 which gives t  0.30 s . Thus, the range of the volleyball is

R   v0 cos0  t  (20.0 m/s) cos18.0(0.30 s)  5.71 m On the other hand, when the angle is changed to 0  8.00 , using the same procedure as shown above, we find 1 1 y  y0  (v0 sin 0 ) t   gt 2  0  2.30 m  (20.0 m/s)sin(8.00)t   (9.80 m/s2 )t 2 2 2

which yields t   0.46 s , and the range is

R   v0 cos0  t   (20.0 m/s) cos18.0(0.46 s)  9.06 m

139 Thus, the ball travels an extra distance of R  R  R  9.06 m  5.71 m  3.35 m

32. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4-5), and we let 0 be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is vx = v0 cos 40.0°, the time it takes for the ball to hit the wall is t

x 22.0 m   1.15 s. vx (25.0 m/s) cos 40.0

(a) The vertical distance is

1 1 y  (v0 sin 0 )t  gt 2  (25.0 m/s)sin 40.0(1.15 s)  (9.80 m/s 2 )(1.15 s)2  12.0 m. 2 2 (b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: vx = v0 cos 40.0° = 19.2 m/s. (c) The vertical component becomes (using Eq. 4-23)

vy  v0 sin 0  gt  (25.0 m/s) sin 40.0  (9.80 m/s2 )(1.15 s)  4.80 m/s. (d) Since vy > 0 when the ball hits the wall, it has not reached the highest point yet. 33. THINK This problem deals with projectile motion. We’re interested in the horizontal displacement and velocity of the projectile before it strikes the ground. EXPRESS We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below the release point. We write 0 = –37.0° for the angle measured from +x, since the angle 0  53.0  given in the problem is measured from the –y direction. The initial setup of the problem is shown in the figure to the right (not to scale). ANALYZE (a) The initial speed of the projectile is the plane’s speed at the moment of release. Given that y0  730 m and y  0 at t  5.00 s , we use Eq. 4-22 to find v0:

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140

1 1 y  y0  (v0 sin 0 ) t  gt 2  0  730 m  v0 sin(37.0)(5.00 s)  (9.80 m/s 2 )(5.00 s)2 2 2

which yields v0 = 202 m/s. (b) The horizontal distance traveled is R  vxt  (v0 cos 0 )t  [(202 m/s)cos(37.0 )](5.00 s)  806 m .

(d) The y component of the velocity (just before impact) is

vy  v0 sin 0  gt  (202 m/s)sin(37.0 )  (9.80 m/s 2 )(5.00 s)  171 m/s . LEARN In this projectile problem we analyzed the kinematics in the vertical and horizontal directions separately since they do not affect each other. The x-component of the velocity, vx  v0 cos 0 , remains unchanged throughout since there’s no horizontal acceleration. 34. (a) Since the y-component of the velocity of the stone at the top of its path is zero, its speed is

v  vx2  vy2  vx  v0 cos 0  (28.0 m/s) cos 40.0  21.4 m/s . (b) Using the fact that v y  0 at the maximum height ymax , the amount of time it takes for the stone to reach ymax is given by Eq. 4-23:

0  v y  v0 sin 0  gt  t 

v0 sin 0 . g

Substituting the above expression into Eq. 4-22, we find the maximum height to be 2

 v0 sin 0  1  v0 sin 0  v02 sin 2 0 1 2 ymax  (v0 sin 0 ) t  gt  v0 sin 0  .  g   2 g  2  g  2g 

To find the time the stone descends to y  ymax / 2 , we solve the quadratic equation given in Eq. 4-22: v 2 sin 2 0 (2  2)v0 sin 0 1 1 y  ymax  0  (v0 sin 0 ) t  gt 2  t  . 2 4g 2 2g

141

Choosing t  t (for descending), we have

vx  v0 cos 0  (28.0 m/s) cos 40.0  21.4 m/s v y  v0 sin 0  g

(2  2)v0 sin 0 2 2 v0 sin 0   (28.0 m/s)sin 40.0  12.7 m/s  2g 2 2

Thus, the speed of the stone when y  ymax / 2 is

v  vx2  vy2  (21.4 m/s)2  (12.7 m/s)2  24.9 m/s . (c) The percentage difference is 24.9 m/s  21.4 m/s  0.163  16.3% . 21.4 m/s

35. THINK This problem deals with projectile motion of a bullet. We’re interested in the firing angle that allows the bullet to strike a target at some distance away. EXPRESS We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of § 4-5), and we let 0 be the firing angle. If the target is a distance d away, then its coordinates are x = d, y = 0.

The projectile motion equations lead to d  (v0 cos 0 )t ,

1 0  v0t sin 0  gt 2 2

where  0 is the firing angle. The setup of the problem is shown in the figure above (scale exaggerated). ANALYZE The time at which the bullet strikes the target is given by t  d /(v0 cos 0 ) .

b g

Eliminating t leads to 2v02 sin  0 cos 0  gd  0 . Using sin  0 cos 0  12 sin 2 0 , we obtain

CHAPTER 4

142 v02 sin (20 )  gd  sin(20 ) 

gd (9.80 m/s 2 )(45.7 m)  v02 (460 m/s)2

which yields sin(20 )  2.11103 , or 0 = 0.0606°. If the gun is aimed at a point a distance  above the target, then tan  0   d so that  d tan 0  (45.7 m) tan(0.0606)  0.0484 m  4.84 cm.

LEARN Due to the downward gravitational acceleration, in order for the bullet to strike the target, the gun must be aimed at a point slightly above the target. 36. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below the point where the ball was hit by the racquet. (a) We want to know how high the ball is above the court when it is at x = 12.0 m. First, Eq. 4-21 tells us the time it is over the fence:

t

x 12.0 m   0.508 s. v0 cos 0  23.6 m/s  cos 0

At this moment, the ball is at a height (above the court) of y  y0   v0 sin  0  t 

1 2 gt  1.10m 2

which implies it does indeed clear the 0.90-m-high fence. (b) At t = 0.508 s, the center of the ball is (1.10 m – 0.90 m) = 0.20 m above the net. (c) Repeating the computation in part (a) with 0 = –5.0° results in t = 0.510 s and y  0.040 m , which clearly indicates that it cannot clear the net. (d) In the situation discussed in part (c), the distance between the top of the net and the center of the ball at t = 0.510 s is 0.90 m – 0.040 m = 0.86 m. 37. THINK The trajectory of the diver is a projectile motion. We are interested in the displacement of the diver at a later time. EXPRESS The initial velocity has no vertical component ( 0  0 ), but only an x component. Eqs. 4-21 and 4-22 can be simplified to

143 x  x0  v0 xt

1 1 y  y0  v0 y t  gt 2   gt 2 . 2 2

where x0  0 , v0 x  v0   2.0 m/s and y0 = +10.0 m (taking the water surface to be at y  0 ). The setup of the problem is shown in the figure below.

ANALYZE (a) At t  0.80 s , the horizontal distance of the diver from the edge is x  x0  v0 xt  0  (2.0 m/s)(0.80 s)  1.60 m.

(b) Similarly, using the second equation for the vertical motion, we obtain 1 1 y  y0  gt 2  10.0 m  (9.80 m/s 2 )(0.80 s) 2  6.86 m. 2 2

(c) At the instant the diver strikes the water surface, y = 0. Solving for t using the equation y  y0  12 gt 2  0 leads to

t

2 y0 2(10.0 m)   1.43 s. g 9.80 m/s 2

During this time, the x-displacement of the diver is R = x = (2.00 m/s)(1.43 s) = 2.86 m. LEARN Using Eq. 4-25 with 0  0 , the trajectory of the diver can also be written as

gx 2 . 2v02 Part (c) can also be solved by using this equation: y  y0 

2v02 y0 gx 2 2(2.0 m/s) 2 (10.0 m) y  y0  2  0  x  R    2.86 m . 2v0 g 9.8 m/s 2

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144

38. In this projectile motion problem, we have v0 = vx = constant, and what is plotted is

v  vx2  v y2 . We infer from the plot that at t = 2.5 s, the ball reaches its maximum height, where vy = 0. Therefore, we infer from the graph that vx = 19 m/s. (a) During t = 5 s, the horizontal motion is x – x0 = vxt = 95 m. (b) Since

(19 m/s)2  v02y  31 m/s (the first point on the graph), we find v0 y  24.5 m/s.

Thus, with t = 2.5 s, we can use ymax  y0  v0 yt  12 gt 2 or vy2  0  v0 2y  2g ymax  y0 , or





ymax  y0  12 vy  v0 y t to solve. Here we will use the latter: 1 1 ymax  y0  (v y  v0 y ) t  ymax  (0  24.5m/s)(2.5 s)  31 m 2 2

where we have taken y0 = 0 as the ground level. 39. Following the hint, we have the time-reversed problem with the ball thrown from the ground, toward the right, at 60° measured counterclockwise from a rightward axis. We see in this time-reversed situation that it is convenient to use the familiar coordinate system with +x as rightward and with positive angles measured counterclockwise. (a) The x-equation (with x0 = 0 and x = 25.0 m) leads to 25.0 m = (v0 cos 60.0°)(1.50 s), so that v0 = 33.3 m/s. And with y0 = 0, and y = h > 0 at t = 1.50 s, we have y  y0  v0 y t  21 gt 2 where v0y = v0 sin 60.0°. This leads to h = 32.3 m. (b) We have

vx = v0x = (33.3 m/s)cos 60.0° = 16.7 m/s vy = v0y – gt = (33.3 m/s)sin 60.0° – (9.80 m/s2)(1.50 s) = 14.2 m/s.

The magnitude of v is given by

| v | vx2  vy2  (16.7 m/s) 2  (14.2 m/s) 2  21.9 m/s. (c) The angle is

v   vx 

 14.2 m/s    40.4  . 16.7 m/s  

  tan 1  y   tan 1 

(d) We interpret this result (“undoing” the time reversal) as an initial velocity (from the edge of the building) of magnitude 21.9 m/s with angle (down from leftward) of 40.4°.

145 40. (a) Solving the quadratic equation Eq. 4-22: 1 1 y  y0  (v0 sin 0 ) t  gt 2  0  2.160 m  (15.00 m/s)sin(45.00)t  (9.800 m/s2 )t 2 2 2

the total travel time of the shot in the air is found to be t  2.352 s . Therefore, the horizontal distance traveled is

R   v0 cos0  t  (15.00 m/s) cos 45.00(2.352 s)  24.95 m . (b) Using the procedure outlined in (a) but for 0  42.00 , we have 1 1 y  y0  (v0 sin 0 ) t  gt 2  0  2.160 m  (15.00 m/s)sin(42.00)t  (9.800 m/s2 )t 2 2 2

and the total travel time is t  2.245 s . This gives

R   v0 cos0  t  (15.00 m/s) cos 42.00(2.245 s)  25.02 m . 41. With the Archer fish set to be at the origin, the position of the insect is given by (x, y) where x  R / 2  v02 sin 20 / 2 g , and y corresponds to the maximum height of the parabolic trajectory: y  ymax  v02 sin 2 0 / 2 g . From the figure, we have tan  

y v02 sin 2 0 / 2 g 1   tan 0 x v02 sin 20 / 2 g 2

Given that   36.0  , we find the launch angle to be

0  tan 1  2 tan    tan 1  2 tan 36.0   tan 1 1.453  55.46   55.5  . Note that  0 depends only on  and is independent of d. 42. (a) Using the fact that the person (as the projectile) reaches the maximum height over the middle wheel located at x  23 m  (23/ 2) m  34.5 m , we can deduce the initial launch speed from Eq. 4-26:

x

R v02 sin 20  2 2g

 v0 

2 gx 2(9.8 m/s 2 )(34.5 m)   26.5 m/s . sin 20 sin(2  53)

Upon substituting the value to Eq. 4-25, we obtain

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146 y  y0  x tan 0 

gx 2 (9.8 m/s 2 )(23 m)2  3.0 m  (23 m) tan 53    23.3 m. 2v02 cos2 0 2(26.5 m/s)2 (cos 53)2

Since the height of the wheel is hw  18 m, the clearance over the first wheel is y  y  hw  23.3 m  18 m  5.3 m . (b) The height of the person when he is directly above the second wheel can be found by solving Eq. 4-24. With the second wheel located at x  23 m  (23/ 2) m  34.5 m, we have gx 2 (9.8 m/s 2 )(34.5 m)2 y  y0  x tan 0  2  3.0 m  (34.5 m) tan 53  2v0 cos 2 0 2(26.52 m/s)2 (cos 53)2  25.9 m.

Therefore, the clearance over the second wheel is y  y  hw  25.9 m  18 m  7.9 m . (c) The location of the center of the net is given by 0  y  y0  x tan 0 

v02 sin 20 (26.52 m/s)2 sin(2  53) gx 2  x    69 m. 2v02 cos 2 0 g 9.8 m/s 2

43. We designate the given velocity v  (7.6 m/s)iˆ  (6.1 m/s) ˆj as v1 , as opposed to the  velocity when it reaches the max height v2 or the velocity when it returns to the ground  v3 , and take v0 as the launch velocity, as usual. The origin is at its launch point on the ground. (a) Different approaches are available, but since it will be useful (for the rest of the problem) to first find the initial y velocity, that is how we will proceed. Using Eq. 2-16, we have v12y  v02y  2 g y  (6.1 m/s)2  v02y  2(9.8 m/s2 )(9.1 m) which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use Eq. 2-16 again but now with y = h for the maximum height: v22 y  v02 y  2 gh 

0  (14.7 m/s)2  2(9.8 m/s 2 )h

which yields h = 11 m. (b) Recalling the derivation of Eq. 4-26, but using v0 y for v0 sin 0 and v0x for v0 cos 0, we have 1 0  v0 y t  gt 2 , R  v0 xt 2

147

which leads to R  2v0 x v0 y / g. Noting that v0x = v1x = 7.6 m/s, we plug in values and obtain R = 2(7.6 m/s)(14.7 m/s)/(9.8 m/s2) = 23 m. (c) Since v3x = v1x = 7.6 m/s and v3y = – v0 y = –14.7 m/s, we have v3  v32 x  v32 y  (7.6 m/s)2  (14.7 m/s)2  17 m/s.  (d) The angle (measured from horizontal) for v3 is one of these possibilities:

 14.7 m  tan 1    63 or 117  7.6 m 

where we settle on the first choice (–63°, which is equivalent to 297°) since the signs of its components imply that it is in the fourth quadrant. 44. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The initial velocity is horizontal so that v0 y  0 and v0 x  v0  161 km h . Converting to SI units, this is v0 = 44.7 m/s.

(a) With the origin at the initial point (where the ball leaves the pitcher’s hand), the y coordinate of the ball is given by y   21 gt 2 , and the x coordinate is given by x = v0t. From the latter equation, we have a simple proportionality between horizontal distance and time, which means the time to travel half the total distance is half the total time. Specifically, if x = 18.3/2 m, then t = (18.3/2 m)/(44.7 m/s) = 0.205 s. (b) And the time to travel the next 18.3/2 m must also be 0.205 s. It can be useful to write the horizontal equation as x = v0t in order that this result can be seen more clearly. (c) Using the equation y   12 gt 2 , we see that the ball has reached the height of |  12  9.80 m/s2   0.205 s  |  0.205 m at the moment the ball is halfway to the batter. 2

(d) The ball’s height when it reaches the batter is  12  9.80 m/s2   0.409 s   0.820m , 2

which, when subtracted from the previous result, implies it has fallen another 0.615 m. Since the value of y is not simply proportional to t, we do not expect equal time-intervals to correspond to equal height-changes; in a physical sense, this is due to the fact that the initial y-velocity for the first half of the motion is not the same as the “initial” y-velocity for the second half of the motion.

CHAPTER 4

148 d

45. (a) Let m = d2 = 0.600 be the slope of the ramp, so y = mx there. We choose our 1

coordinate origin at the point of launch and use Eq. 4-25. Thus,

(9.80 m/s2 ) x 2 y  tan(50.0) x   0.600 x 2(10.0 m/s)2 (cos50.0)2 which yields x = 4.99 m. This is less than d1 so the ball does land on the ramp. (b) Using the value of x found in part (a), we obtain y = mx = 2.99 m. Thus, the Pythagorean theorem yields a displacement magnitude of x2 + y2 = 5.82 m. (c) The angle is, of course, the angle of the ramp: tan1(m) = 31.0º. 46. Using the fact that v y  0 when the player is at the maximum height ymax , the amount of time it takes to reach ymax can be solved by using Eq. 4-23:

0  v y  v0 sin 0  gt  tmax 

v0 sin 0 . g

Substituting the above expression into Eq. 4-22, we find the maximum height to be 2

 v sin 0  1  v0 sin 0  v02 sin 2 0 1 2 ymax  (v0 sin 0 ) tmax  gtmax  v0 sin 0  0 .  g   2 g  2  g  2g 

To find the time when the player is at y  ymax / 2 , we solve the quadratic equation given in Eq. 4-22: v 2 sin 2 0 (2  2)v0 sin 0 1 1 y  ymax  0  (v0 sin 0 ) t  gt 2  t  . 2 4g 2 2g With t  t (for ascending), the amount of time the player spends at a height y  ymax / 2 is v sin 0 (2  2)v0 sin 0 v0 sin 0 tmax t 1 t  tmax  t  0       0.707 . 2g g tmax 2g 2 2 Therefore, the player spends about 70.7% of the time in the upper half of the jump. Note that the ratio t / tmax is independent of v0 and  0 , even though t and tmax depend on these quantities. 47. THINK The baseball undergoes projectile motion after being hit by the batter. We’d like to know if the ball clears a high fence at some distance away.

149 EXPRESS We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below impact point between bat and ball. In the absence of a fence, with 0  45  , the horizontal range (same launch level) is R  107 m . We want to know how high the ball is from the ground when it is at x  97.5 m , which requires knowing the initial velocity. The trajectory of the baseball can be described by Eq. 4-25: y  y0  (tan 0 ) x 

gx 2 . 2(v0 cos 0 ) 2

The setup of the problem is shown in the figure below (not to scale).

ANALYZE (a) We first solve for the initial speed v0. Using the range information ( y  y0 when x  R ) and 0 = 45°, Eq. 4-25 gives

gR v0   sin 20

9.8 m/s  107 m   32.4 m/s. 2

sin(2  45)

Thus, the time at which the ball flies over the fence is:

x  (v0 cos 0 )t  

t 

x 97.5 m   4.26 s. v0 cos 0  32.4 m/s  cos 45

At this moment, the ball is at a height (above the ground) of y  y0   v0 sin 0  t  

1 2 gt  2

1  1.22 m  [(32.4 m/s)sin 45 ](4.26 s)  (9.8 m/s 2 )(4. 26 s) 2 2  9.88 m

which implies it does indeed clear the 7.32 m high fence. (b) At t   4.26 s , the center of the ball is 9.88 m – 7.32 m = 2.56 m above the fence.

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150

LEARN Using the trajectory equation above, one can show that the minimum initial velocity required to clear the fence is given by gx2   , y  y0  (tan 0 ) x  2(v0 cos 0 ) 2 or about 31.9 m/s. 48. Following the hint, we have the time-reversed problem with the ball thrown from the roof, toward the left, at 60° measured clockwise from a leftward axis. We see in this time-reversed situation that it is convenient to take +x as leftward with positive angles measured clockwise. Lengths are in meters and time is in seconds. (a) With y0 = 20.0 m, and y = 0 at t = 4.00 s, we have y  y0  v0 y t  21 gt 2 where v0 y  v0 sin 60 . This leads to v0 = 16.9 m/s. This plugs into the x-equation x  x0  v0xt

(with x0 = 0 and x = d) to produce d = (16.9 m/s)cos 60°(4.00 s) = 33.7 m.

(b) We have vx  v0 x  (16.9 m/s) cos 60.0  8.43 m/s vy  v0 y  gt  (16.9 m/s)sin 60.0  (9.80m/s 2 )(4.00 s)  24.6 m/s. The magnitude of v is | v | vx2  vy2  (8.43 m/s)2  ( 24.6 m/s) 2  26.0 m/s. (c) The angle relative to horizontal is v    24.6 m/s    tan 1  y   tan 1     71.1  .  8.43 m/s   vx  We may convert the result from rectangular components to magnitude-angle representation: v  (8.43,  24.6)  (26.0   71.1) and we now interpret our result (“undoing” the time reversal) as an initial velocity of magnitude 26.0 m/s with angle (up from rightward) of 71.1°. 49. THINK In this problem a football is given an initial speed and it undergoes projectile motion. We’d like to know the smallest and greatest angles at which a field goal can be scored. EXPRESS We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the point where the ball is kicked. We use x and y to denote the coordinates of the ball at the goalpost, and try to find the kicking angle(s) 0 so that y = 3.44 m when x = 50 m. Writing the kinematic equations for projectile motion:

151 x  v0 cos 0 , y  v0t sin 0  12 gt 2 ,

we see the first equation gives t = x/v0 cos 0, and when this is substituted into the second the result is gx 2 y  x tan  0  2 . 2v0 cos2  0 ANALYZE One may solve the above equation by trial and error: systematically trying values of 0 until you find the two that satisfy the equation. A little manipulation, however, will give an algebraic solution: Using the trigonometric identity 1 / cos2 0 = 1 + tan2 0,

we obtain

1 gx 2 1 gx 2 2 tan  0  x tan  0  y  0 2 v02 2 v02

which is a second-order equation for tan 0. To simplify writing the solution, we denote c

1 2 2 1 2 2 gx / v0   9.80 m/s2   50 m  /  25 m/s   19.6 m. 2 2

Then the second-order equation becomes c tan2 0 – x tan 0 + y + c = 0. Using the quadratic formula, we obtain its solution(s).

tan 0 

x  x2  4  y  c  c 2c



50 m  (50 m)2  4  3.44 m  19.6 m 19.6 m  2 19.6 m 

The two solutions are given by tan0 = 1.95 and tan0 = 0.605. The corresponding (firstquadrant) angles are 0 = 63° and 0 = 31°. Thus, (a) The smallest elevation angle is 0 = 31°, and (b) The greatest elevation angle is 0 = 63°. LEARN If kicked at any angle between 31° and 63°, the ball will travel above the cross bar on the goalposts. 50. We apply Eq. 4-21, Eq. 4-22, and Eq. 4-23. (a) From x  v0 x t , we find v0 x  40 m / 2 s  20 m/s. (b) From y  v0 y t  21 gt 2 , we find v0 y   53 m  12 (9.8 m/s2 )(2 s)2  / 2  36 m/s.

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152

(c) From v y  v0 y  gt  with vy = 0 as the condition for maximum height, we obtain t   (36 m/s) /(9.8 m/s2 )  3.7 s. During that time the x-motion is constant, so x  x0  (20 m/s)(3.7 s)  74 m.

51. (a) The skier jumps up at an angle of 0  11.3 up from the horizontal and thus returns to the launch level with his velocity vector 11.3 below the horizontal. With the snow surface making an angle of   9.0 (downward) with the horizontal, the angle between the slope and the velocity vector is   0    11.3  9.0  2.3 . (b) Suppose the skier lands at a distance d down the slope. Using Eq. 4-25 with x  d cos  and y  d sin  (the edge of the track being the origin), we have d sin   d cos  tan 0 

Solving for d, we obtain

g (d cos  ) 2 . 2v02 cos 2 0

2v02 cos 2 0 2v02 cos 0 d  cos  tan 0  sin     cos  sin 0  cos0 sin   g cos 2  g cos 2  

2v02 cos 0 sin(0   ). g cos 2 

Substituting the values given, we find d

which gives

2(10 m/s)2 cos(11.3) sin(11.3  9.0)  7.117 m. (9.8 m/s 2 ) cos 2 (9.0)

y  d sin   (7.117 m)sin(9.0)  1.11 m.

Therefore, at landing the skier is approximately 1.1 m below the launch level. (c) The time it takes for the skier to land is t

x d cos  (7.117 m) cos(9.0)    0.72 s . vx v0 cos 0 (10 m/s) cos(11.3)

Using Eq. 4-23, the x-and y-components of the velocity at landing are vx  v0 cos 0  (10 m/s) cos(11.3 )  9.81 m/s v y  v0 sin 0  gt  (10 m/s)sin(11.3 )  (9.8 m/s 2 )(0.72 s)  5.07 m/ s

153

Thus, the direction of travel at landing is

v   vx 

  5.07 m/s     27.3  .  9.81 m/s 

  tan 1  y   tan 1 

or 27.3 below the horizontal. The result implies that the angle between the skier’s path and the slope is   27.3  9.0  18.3 , or approximately 18 to two significant figures. 52. From Eq. 4-21, we find t  x / v0 x . Then Eq. 4-23 leads to v y  v0 y  gt  v0 y 

gx . v0 x

Since the slope of the graph is 0.500, we conclude g 1  vox = 19.6 m/s.  v0 x 2

And from the “y intercept” of the graph, we find voy = 5.00 m/s. Consequently,

o = tan1(voy  vox) = 14.3  14  . 53. Let y0 = h0 = 1.00 m at x0 = 0 when the ball is hit. Let y1 = h (the height of the wall) and x1 describe the point where it first rises above the wall one second after being hit; similarly, y2 = h and x2 describe the point where it passes back down behind the wall four seconds later. And yf = 1.00 m at xf = R is where it is caught. Lengths are in meters and time is in seconds. (a) Keeping in mind that vx is constant, we have x2 – x1 = 50.0 m = v1x (4.00 s), which leads to v1x = 12.5 m/s. Thus, applied to the full six seconds of motion: xf – x0 = R = vx(6.00 s) = 75.0 m. (b) We apply y  y0  v0 y t  12 gt 2 to the motion above the wall, y2  y1  0  v1 y  4.00 s  

1 2 g  4.00 s  2

and obtain v1y = 19.6 m/s. One second earlier, using v1y = v0y – g(1.00 s), we find v0 y  29.4 m/s . Therefore, the velocity of the ball just after being hit is

v  v0 x ˆi  v0 y ˆj  (12.5 m/s) ˆi  (29.4 m/s) ˆj

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Its magnitude is | v | (12.5 m/s)2 +(29.4 m/s) 2  31.9 m/s. (c) The angle is

v   vx 

 29.4 m/s    67.0  .  12.5 m/s 

  tan 1  y   tan 1 

We interpret this result as a velocity of magnitude 31.9 m/s, with angle (up from rightward) of 67.0°. (d) During the first 1.00 s of motion, y  y0  v0 y t  21 gt 2 yields





h  1.0 m   29.4 m/s 1.00 s   12 9.8 m/s 2 1.00 s   25.5 m. 2

54. For y = 0, Eq. 4-22 leads to t = 2vosino/g, which immediately implies tmax = 2vo/g (which occurs for the “straight up” case: o = 90). Thus, 1 t = vo/g 2 max

 2 = sino.

Therefore, the half-maximum-time flight is at angle o = 30.0. Since the least speed occurs at the top of the trajectory, which is where the velocity is simply the x-component of the initial velocity (vocoso = vocos30 for the half-maximum-time flight), then we need to refer to the graph in order to find vo – in order that we may complete the solution. In the graph, we note that the range is 240 m when o = 45.0. Equation 4-26 then leads to vo = 48.5 m/s. The answer is thus (48.5 m/s)cos30.0 = 42.0 m/s. 55. THINK In this problem a ball rolls off the top of a stairway with an initial speed, and we’d like to know on which step it lands first. EXPRESS We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh and travel horizontally a distance between (n – 1)w and nw. We take the origin of a coordinate system to be at the point where the ball leaves the top of the stairway, and we choose the y axis to be positive in the upward direction, as shown in the figure. The coordinates of the ball at time t are given by x = v0xt and y   21 gt 2 (since v0y = 0). ANALYZE We equate y to –nh and solve for the time to reach the level of step n:

155 2nh . g

t

The x coordinate then is x  v0 x

2nh 2n(0.203 m)  (1.52 m/s)  (0.309 m) n. g 9.8 m/s 2

The method is to try values of n until we find one for which x/w is less than n but greater than n – 1. For n = 1, x = 0.309 m and x/w = 1.52, which is greater than n. For n = 2, x = 0.437 m and x/w = 2.15, which is also greater than n. For n = 3, x = 0.535 m and x/w = 2.64. Now, this is less than n and greater than n – 1, so the ball hits the third step. LEARN To check the consistency of our calculation, we can substitute n = 3 into the above equations. The results are t = 0.353 s, y = 0.609 m and x = 0.535 m. This indeed corresponds to the third step. 56. We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find acceleration a. (a) Since the radius of Earth is 6.37  106 m, the radius of the satellite orbit is r = (6.37  106 + 640  103 ) m = 7.01  106 m. Therefore, the speed of the satellite is

2 7.01  106 m 2r v   7.49  103 m / s. T 98.0 min 60 s / min

(b) The magnitude of the acceleration is

h  8.00 m / s .

7.49  10 m / s v2 a  r 7.01  106 m

57. The magnitude of centripetal acceleration (a = v2/r) and its direction (toward the center of the circle) form the basis of this problem.

 (a) If a passenger at this location experiences a  183 . m / s2 east, then the center of the 2 circle is east of this location. The distance is r = v /a = (3.66 m/s)2/(1.83 m/s2) = 7.32 m. (b) Thus, relative to the center, the passenger at that moment is located 7.32 m toward the west.

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 (c) If the direction of a experienced by the passenger is now south—indicating that the center of the merry-go-round is south of him, then relative to the center, the passenger at that moment is located 7.32 m toward the north.

58. (a) The circumference is c = 2r = 2(0.15 m) = 0.94 m. (b) With T = (60 s)/1200 = 0.050 s, the speed is v = c/T = (0.94 m)/(0.050 s) = 19 m/s. This is equivalent to using Eq. 4-35. (c) The magnitude of the acceleration is a = v2/r = (19 m/s)2/(0.15 m) = 2.4  103 m/s2. (d) The period of revolution is (1200 rev/min)–1 = 8.3  10–4 min, which becomes, in SI units, T = 0.050 s = 50 ms. 59. (a) Since the wheel completes 5 turns each minute, its period is one-fifth of a minute, or 12 s. (b) The magnitude of the centripetal acceleration is given by a = v2/R, where R is the radius of the wheel, and v is the speed of the passenger. Since the passenger goes a distance 2R for each revolution, his speed is v

b g

2 15 m  7.85 m / s 12 s

b7.85 m / sg  4.1 m / s . and his centripetal acceleration is a  2

15 m

(c) When the passenger is at the highest point, his centripetal acceleration is downward, toward the center of the orbit. (d) At the lowest point, the centripetal acceleration is a  4.1 m/s2 , same as part (b). (e) The direction is up, toward the center of the orbit. 60. (a) During constant-speed circular motion, the velocity vector is perpendicular to the   acceleration vector at every instant. Thus, v · a = 0. (b) The acceleration in this vector, at every instant, points toward the center of the circle, whereas the position vector points from the center of the circle to the object in motion.     Thus, the angle between r and a is 180º at every instant, so r  a = 0. 61. We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find centripetal acceleration a. (a) v = 2r/T = 2(20 km)/1.0 s = 126 km/s = 1.3  105 m/s.

157

(b) The magnitude of the acceleration is

g  7.9  10 m / s .

126 km / s v2 a  r 20 km

g  4.0 m / s .

10 m / s v2 a  r 25 m



63. We first note that a1 (the acceleration at t1 = 2.00 s) is perpendicular to a2 (the acceleration at t2=5.00 s), by taking their scalar (dot) product:

ˆ ˆ [(4.00 m/s 2 )i+( ˆ  6.00 m/s2 )j]=0. ˆ a1  a2  [(6.00 m/s2 )i+(4.00 m/s 2 )j] Since the acceleration vectors are in the (negative) radial directions, then the two positions (at t1 and t2) are a quarter-circle apart (or three-quarters of a circle, depending on whether one measures clockwise or counterclockwise). A quick sketch leads to the conclusion that if the particle is moving counterclockwise (as the problem states) then it travels three-quarters of a circumference in moving from the position at time t1 to the position at time t2 . Letting T stand for the period, then t2 – t1 = 3.00 s = 3T/4. This gives T = 4.00 s. The magnitude of the acceleration is

a  ax2  a y2  (6.00 m/s 2 )2  (4.00 m/s) 2  7.21 m/s 2. Using Eqs. 4-34 and 4-35, we have a  4 2 r / T 2 , which yields

r

aT 2 (7.21 m/s 2 )(4.00 s) 2   2.92 m. 4 2 4 2

64. When traveling in circular motion with constant speed, the instantaneous acceleration vector necessarily points toward the center. Thus, the center is “straight up” from the cited point. (a) Since the center is “straight up” from (4.00 m, 4.00 m), the x coordinate of the center is 4.00 m. (b) To find out “how far up” we need to know the radius. Using Eq. 4-34 we find

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v 2  5.00 m/s  r   2.00 m. a 12.5 m/s 2 2

Thus, the y coordinate of the center is 2.00 m + 4.00 m = 6.00 m. Thus, the center may be written as (x, y) = (4.00 m, 6.00 m). 65. Since the period of a uniform circular motion is T  2 r / v , where r is the radius and v is the speed, the centripetal acceleration can be written as 2

a

v 2 1  2 r  4 2 r    .  r r T  T2

Based on this expression, we compare the (magnitudes) of the wallet and purse accelerations, and find their ratio is the ratio of r values. Therefore, awallet = 1.50 apurse . Thus, the wallet acceleration vector is

ˆ a  1.50[(2.00 m/s 2 )iˆ +(4.00 m/s 2 )j]=(3.00 m/s 2 )iˆ +(6.00 m/s 2 )jˆ . 66. The fact that the velocity is in the +y direction and the acceleration is in the +x direction at t1 = 4.00 s implies that the motion is clockwise. The position corresponds to the “9:00 position.” On the other hand, the position at t2 = 10.0 s is in the “6:00 position” since the velocity points in the x direction and the acceleration is in the +y direction. The time interval t  10.0 s  4.00 s  6.00 s is equal to 3/4 of a period:

Equation 4-35 then yields r

3 6.00 s  T  T  8.00 s. 4 vT (3.00 m/s)(8.00 s)   3.82 m. 2 2

(a) The x coordinate of the center of the circular path is x  5.00 m  3.82 m  8.82 m. (b) The y coordinate of the center of the circular path is y  6.00 m. In other words, the center of the circle is at (x,y) = (8.82 m, 6.00 m). 67. THINK In this problem we have a stone whirled in a horizontal circle. After the string breaks, the stone undergoes projectile motion. EXPRESS The stone moves in a circular path (top view shown below left) initially, but undergoes projectile motion after the string breaks (side view shown below right). Since a  v2 / R , to calculate the centripetal acceleration of the stone, we need to know its

159 speed during its circular motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in §4-6) to find that speed.

(top view)

(side view)

Taking the +y direction to be upward and placing the origin at the point where the stone leaves its circular orbit, then the coordinates of the stone during its motion as a projectile are given by x = v0t and y   21 gt 2 (since v0y = 0). It hits the ground at x = 10 m and y   2.0 m . ANALYZE Formally solving the y-component equation for the time, we obtain t  2 y / g , which we substitute into the first equation: v0  x 

g  10 m 2y

b g  29b.82m.0/ ms g  15.7 m / s.

Therefore, the magnitude of the centripetal acceleration is

v 2 15.7 m/s  a 0   160 m/s 2 . R 1.5 m 2

gx 2 . The equation implies  2 yR that the greater the centripetal acceleration, the greater the initial speed of the projectile, and the greater the distance traveled by the stone. This is precisely what we expect.

LEARN The above equations can be combined to give a 

68. We note that after three seconds have elapsed (t2 – t1 = 3.00 s) the velocity (for this object in circular motion of period T ) is reversed; we infer that it takes three seconds to reach the opposite side of the circle. Thus, T = 2(3.00 s) = 6.00 s. (a) Using Eq. 4-35, r = vT/2, where v  (3.00 m/s)2  (4.00 m/s)2  5.00 m/s , we obtain r  4.77 m . The magnitude of the object’s centripetal acceleration is therefore a = v2/r = 5.24 m/s2.

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160 (b) The average acceleration is given by Eq. 4-15: aavg 

ˆ m/s  (3.00iˆ  4.00j) ˆ m/s v2  v1 ( 3.00iˆ  4.00j) ˆ  2.67 m/s 2 )ˆj   ( 2.00 m/s 2 )i+( t2  t1 5.00 s  2.00 s

which implies | aavg | ( 2.00 m/s 2 )2  ( 2.67 m/s 2 ) 2  3.33 m/s 2. 69. We use Eq. 4-15 first using velocities relative to the truck (subscript t) and then using velocities relative to the ground (subscript g). We work with SI units, so 20 km / h  5.6 m / s , 30 km / h  8.3 m / s , and 45 km / h  12.5 m / s . We choose east as the  i direction. (a) The velocity of the cheetah (subscript c) at the end of the 2.0 s interval is (from Eq. 4-44) vc t  vc g  vt g  (12.5 m/s) î  (5.6 m/s) î  (18.1 m/s) î relative to the truck. Since the velocity of the cheetah relative to the truck at the beginning of the 2.0 s interval is (8.3 m/s)iˆ , the (average) acceleration vector relative to the cameraman (in the truck) is (18.1 m/s)iˆ  (8.3 m/s)iˆ ˆ aavg   (13 m/s 2 )i, 2.0 s or | aavg | 13 m/s 2 . (b) The direction of aavg is +iˆ , or eastward. (c) The velocity of the cheetah at the start of the 2.0 s interval is (from Eq. 4-44) v0cg  v0ct  v0tg  (8.3 m/s)iˆ  (  5.6 m/s)iˆ  (13.9 m/s)iˆ

relative to the ground. The (average) acceleration vector relative to the crew member (on the ground) is (12.5 m/s)iˆ  (13.9 m/s)iˆ ˆ |a |  13 m/s 2 aavg   (13 m/s 2 )i, avg 2.0 s identical to the result of part (a). (d) The direction of aavg is +iˆ , or eastward. 70. We use Eq. 4-44, noting that the upstream corresponds to the +iˆ direction. (a) The subscript b is for the boat, w is for the water, and g is for the ground.

161

vbg  vbw  vwg  (14 km/h) î  (9 km/h) î  (5 km/h) î. Thus, the magnitude is | vbg | 5 km/h. (b) The direction of vbg is +x, or upstream. (c) We use the subscript c for the child, and obtain

   vc g  vc b  vb g  ( 6 km / h) i  (5 km / h) i  ( 1 km / h) i . The magnitude is | vcg | 1 km/h. (d) The direction of vcg is x, or downstream. 71. While moving in the same direction as the sidewalk’s motion (covering a distance d relative to the ground in time t1 = 2.50 s), Eq. 4-44 leads to d vsidewalk + vman running = t . 1 While he runs back (taking time t2 = 10.0 s) we have

d vsidewalk  vman running =  t . 2

Dividing these equations and solving for the desired ratio, we get

12.5 5 = 3 = 1.67. 7.5

72. We denote the velocity of the player with vPF and the relative velocity between the player and the ball be vBP . Then the velocity vBF of the ball relative to the field is given by vBF  vPF  vBP . The smallest angle

min corresponds to the case when vBF  vPF . Hence,  | vPF |  1  4.0 m/s    180  cos    130.  6.0 m/s   | vBP | 

min  180  cos1 

73. We denote the police and the motorist with subscripts p and m, respectively. The coordinate system is indicated in Fig. 4-46. (a) The velocity of the motorist with respect to the police car is

vm p  vm  v p  ( 60 km/h)ˆj  ( 80 km/h)iˆ  (80 km/h)iˆ  (60 km/h)ˆj.

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 (b) vm p does happen to be along the line of sight. Referring to Fig. 4-46, we find the vector pointing from one car to another is r  (800 m)iˆ  (600 m) ˆj (from M to P). Since   the ratio of components in r is the same as in vm p , they must point the same direction. (c) No, they remain unchanged. 74. Velocities are taken to be constant; thus, the velocity of the plane relative to the ground is vPG  (55 km)/(1/4 hour) ˆj= (220 km/h)jˆ . In addition,

ˆ vAG  (42 km/h)(cos 20 iˆ  sin 20 ˆj)  (39 km/h)iˆ  (14 km/ h)j. Using vPG  vPA  vAG , we have

ˆ vPA  vPG  vAG  (39 km/h)iˆ  (234 km/h)j. which implies | vPA | 237 km/h , or 240 km/h (to two significant figures.) 75. THINK This problem deals with relative motion in two dimensions. Raindrops appear to fall vertically by an observer on a moving train. EXPRESS Since the raindrops fall vertically relative to the train, the horizontal component of the velocity of a raindrop, vh = 30 m/s, must be the same as the speed of the train, i.e., vh  vtrain (see figure). On the other hand, if vv is the vertical component of the velocity and  is the angle between the direction of motion and the vertical, then tan  = vh/vv. Knowing vv and vh allows us to determine the speed of the raindrops. ANALYZE With   70  , we find the vertical component of the velocity to be vv = vh/tan  = (30 m/s)/tan 70° = 10.9 m/s. Therefore, the speed of a raindrop is v  vh2  vv2  (30 m / s) 2  (10.9 m / s) 2  32 m / s .

163 LEARN As long as the horizontal component of the velocity of the raindrops coincides with the speed of the train, the passenger on board will see the rain falling perfectly vertically. 

76. The destination is D = 800 km j where we orient axes so that +y points north and +x points east. This takes two hours, so the (constant) velocity of the plane (relative to the ^ ground) is vpg = (400 km/h) j . This must be the vector sum of the plane’s velocity with respect to the air which has (x,y) components (500cos70º, 500sin70º), and the velocity of the air (wind) relative to the ground vag . Thus, ^

(400 km/h) j = (500 km/h) cos70º i + (500 km/h) sin70º j + vag which yields

vag =( –171 km/h)i –( 70.0 km/h)j .

(a) The magnitude of vag is | vag | ( 171 km/h) 2  ( 70.0 km/h) 2  185 km/h. (b) The direction of vag is   70.0 km/h    22.3  (south of west).   171 km/h 

  tan 1 

77. THINK This problem deals with relative motion in two dimensions. Snowflakes falling vertically downward are seen to fall at an angle by a moving observer. EXPRESS Relative to the car the velocity of the snowflakes has a vertical component of vv  8.0 m/s and a horizontal component of vh  50 km/h  13.9 m/s . ANALYZE The angle  from the vertical is found from tan  

which yields  = 60°.

vh 13.9 m/s   1.74 vv 8.0 m/s

LEARN The problem can also be solved by expressing the velocity relation in vector notation: vrel  vcar  vsnow , as shown in the figure. 78. We make use of Eq. 4-44 and Eq. 4-45. The velocity of Jeep P relative to A at the instant is

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vPA  (40.0 m/s)(cos 60ˆi  sin 60ˆj)  (20.0 m/s)iˆ  (34.6 m/s)ˆj. Similarly, the velocity of Jeep B relative to A at the instant is

vBA  (20.0 m/s)(cos30ˆi  sin 30ˆj)  (17.3 m/s)iˆ  (10.0 m/s)ˆj. Thus, the velocity of P relative to B is

ˆ m/s  (17.3iˆ  10.0j) ˆ m/s  (2.68 m/s)iˆ  (24.6 m/s)j. ˆ vPB  vPA  vBA  (20.0iˆ  34.6j) (a) The magnitude of vPB is | vPB | (2.68 m/s)2  (24.6 m/s) 2  24.8 m/s. (b) The direction of vPB is   tan 1[(24.6 m/s) /(2.68 m/s)]  83.8 north of east (or 6.2º east of north). (c) The acceleration of P is

aPA  (0.400 m/s2 )(cos 60.0î  sin 60.0ˆj)  (0.200 m/s2 )iˆ  (0.346 m/s2 )ˆj, and aPA  aPB . Thus, we have | aPB | 0.400 m/s 2 . (d) The direction is 60.0 north of east (or 30.0 east of north). 79. THINK This problem involves analyzing the relative motion of two ships sailing in different directions. EXPRESS Given that  A  45  , and  B  40  , as defined in the figure, the velocity vectors (relative to the shore) for ships A and B are given by vA   (vA cos 45) î  (vA sin 45) ˆj v   (v sin 40) î  (v cos 40) ˆj, B

with vA = 24 knots and vB = 28 knots. We take east as  i and north as j . The velocity of ship A relative to ship B is simply given by vAB  vA  vB . ANALYZE (a) The relative velocity is

165 vA B  vA  vB  (vB sin 40  vA cos 45)iˆ  (vB cos 40  vA sin 45) ˆj  (1.03 knots)iˆ  (38.4 knots)ˆj

the magnitude of which is | vA B |  (1.03 knots)2  (38.4 knots)2  38.4 knots.  (b) The angle  AB which v A B makes with north is given by

v   vAB , y   

 1.03 knots    1.5  38.4 knots 

 AB  tan 1  AB , x   tan 1 

 which is to say that v A B points 1.5° east of north.

(c) Since the two ships started at the same time, their relative velocity describes at what rate the distance between them is increasing. Because the rate is steady, we have t

| rAB | 160 nautical miles   4.2 h. | vAB | 38.4 knots

 (d) The velocity v A B does not change with time in   this problem, and rA B is in the same direction as v A B since they started at the same time. Reversing the   points of view, we have v A B  vB A so that   rA B  rB A (i.e., they are 180° opposite to each other). Hence, we conclude that B stays at a bearing of 1.5° west of south relative to A during the journey (neglecting the curvature of Earth).

LEARN The relative velocity is depicted in the figure on the right. When analyzing relative motion in two dimensions, a vector diagram such as the one shown can be very helpful. 80. This is a classic problem involving two-dimensional relative motion. We align our coordinates so that east corresponds to +x and north corresponds to +y. We write the     vector addition equation as vBG  vBW  vWG . We have vWG  (2.00 ) in the magnitude angle notation (with the unit m/s understood), or vWG  2.0i in unit-vector notation. We  also have vBW  (8.0120 ) where we have been careful to phrase the angle in the ˆ ˆ m/s. ‘standard’ way (measured counterclockwise from the +x axis), or v  (4.0i+6.9j) BW

 (a) We can solve the vector addition equation for v BG :

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ˆ ˆ m/s  (2.0 m/s)iˆ  (6.9 m/s) ˆj. vBG  vBW  vWG  (2.0 m/s) ˆi  ( 4.0i+6.9j)  Thus, we find | vBG |  7.2 m/s.

(b) The direction of vBG is   tan 1[(6.9 m/s) /( 2.0 m/s)]  106 (measured counterclockwise from the +x axis), or 16° west of north. (c) The velocity is constant, and we apply y – y0 = vyt in a reference frame. Thus, in the ground reference frame, we have (200 m)  (7.2 m/s)sin(106)t  t  29 s. Note: If a student obtains “28 s,” then the student has probably neglected to take the y component properly (a common mistake). 81. Here, the subscript W refers to the water. Our coordinates are chosen with +x being east and +y being north. In these terms, the angle specifying east would be 0° and the angle specifying south would be –90° or 270°. Where the length unit is not displayed, km is to be understood.    (a) We have v A W  v A B  vB W , so that

 v A B = (22  – 90°) – (40  37°) = (56  – 125°)

in the magnitude-angle notation (conveniently done with a vector-capable calculator in polar mode). Converting to rectangular components, we obtain vA B  (32km/h) ˆi  (46 km/h) ˆj .

Of course, this could have been done in unit-vector notation from the outset. (b) Since the velocity-components are constant, integrating them to obtain the position is    straightforward (r  r0  v dt )

r  (2.5  32t ) ˆi  (4.0  46t ) ˆj

with lengths in kilometers and time in hours.  (c) The magnitude of this r is r  (2.5  32t ) 2  (4.0  46t ) 2 . We minimize this by taking a derivative and requiring it to equal zero — which leaves us with an equation for t

dr 1 6286t  528  0 dt 2 (2.5  32t ) 2  (4.0  46t ) 2

which yields t = 0.084 h.

167 (d) Plugging this value of t back into the expression for the distance between the ships (r), we obtain r = 0.2 km. Of course, the calculator offers more digits (r = 0.225…), but they are not significant; in fact, the uncertainties implicit in the given data, here, should make the ship captains worry. 82. We construct a right triangle starting from the clearing on the south bank, drawing a line (200 m long) due north (upward in our sketch) across the river, and then a line due west (upstream, leftward in our sketch) along the north bank for a distance (82 m)  (1.1 m/s)t , where the t-dependent contribution is the distance that the river will carry the boat downstream during time t. The hypotenuse of this right triangle (the arrow in our sketch) also depends on t and on the boat’s speed (relative to the water), and we set it equal to the Pythagorean “sum” of the triangle’s sides:

b4.0gt  200  b82  11. t g 2

which leads to a quadratic equation for t 46724  180.4t  14.8t 2  0.

(b) We solve for t first and find a positive value: t = 62.6 s. (a) The angle between the northward (200 m) leg of the triangle and the hypotenuse (which is measured “west of north”) is then given by

  tan 1

FG 82  11. t IJ  tan FG 151IJ  37 . H 200K H 200 K 1

83. We establish coordinates with i pointing to the far side of the river (perpendicular to the current) and j pointing in the direction of the current. We are told that the magnitude (presumed constant) of the velocity of the boat relative to the water is | vbw | = 6.4 km/h. Its angle, relative to the x axis is . With km and h as the understood units, the velocity ˆ of the water (relative to the ground) is vwg  (3.2 km/h)j. (a) To reach a point “directly opposite” means that the velocity of her boat relative to ground must be vbg = vbg ˆi where vbg  0 is unknown. Thus, all j components must cancel    in the vector sum v + v = v , which means the v sin  = (–3.2 km/h) j , so bw

 = sin–1 [(–3.2 km/h)/(6.4 km/h)] = –30°.

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(b) Using the result from part (a), we find vbg = vbw cos = 5.5 km/h. Thus, traveling a distance of  = 6.4 km requires a time of (6.4 km)/(5.5 km/h) = 1.15 h or 69 min. (c) If her motion is completely along the y axis (as the problem implies) then with vwg = 3.2 km/h (the water speed) we have ttotal =

D D + = 1.33 h vbw + vwg vbw  vwg

where D = 3.2 km. This is equivalent to 80 min. (d) Since

D D D D    vbw +vwg vbw  vwg vbw  vwg vbw  vwg

the answer is the same as in the previous part, that is, ttotal = 80 min . (e) The shortest-time path should have   0. This can also be shown by noting that the case of general  leads to

vbg  vbw  vwg  vbwcos ˆi  (vbwsin  + vwg ) ˆj

 where the x component of vbg must equal l/t. Thus, t =

l vbwcos 

which can be minimized using dt/d = 0. (f) The above expression leads to t = (6.4 km)/(6.4 km/h) = 1.0 h, or 60 min. ^

84. Relative to the sled, the launch velocity is v0rel = vox i + voy j . Since the sled’s motion is in the negative direction with speed vs (note that we are treating vs as a positive ^ number, so the sled’s velocity is actually –vs i ), then the launch velocity relative to the ^ ^ ground is v0 = (vox – vs) i + voy j . The horizontal and vertical displacement (relative to the ground) are therefore xland – xlaunch = xbg = (vox – vs) tflight 1

yland – ylaunch = 0 = voy tflight + 2 (g)(tflight)2 . Combining these equations leads to

169

xbg =

2v0 x v0 y g

 2v0 y    vs .  g 

The first term corresponds to the “y intercept” on the graph, and the second term (in parentheses) corresponds to the magnitude of the “slope.” From the figure, we have

 xbg  40  4vs . This implies voy = (4.0 s)(9.8 m/s2)/2 = 19.6 m/s, and that furnishes enough information to determine vox. (a) vox = 40g/2voy = (40 m)(9.8 m/s2)/(39.2 m/s) = 10 m/s. (b) As noted above, voy = 19.6 m/s. (c) Relative to the sled, the displacement xbs does not depend on the sled’s speed, so xbs = vox tflight = 40 m. (d) As in (c), relative to the sled, the displacement xbs does not depend on the sled’s speed, and xbs = vox tflight = 40 m. 85. Using displacement = velocity × time (for each constant-velocity part of the trip), along with the fact that 1 hour = 60 minutes, we have the following vector addition exercise (using notation appropriate to many vector-capable calculators): (1667 m  0º) + (1333 m  90º) + (333 m  180º) + (833 m  90º) + (667 m  180º) + (417 m  90º) = (2668 m  76º). (a) Thus, the magnitude of the net displacement is 2.7 km. (b) Its direction is 76 clockwise (relative to the initial direction of motion). 86. We use a coordinate system with +x eastward and +y upward. (a) We note that 123° is the angle between the initial position and later position vectors, so that the angle from +x to the later position vector is 40° + 123° = 163°. In unit-vector notation, the position vectors are

r1 = (360 m)cos(40) ˆi + (360 m)sin(40) ˆj = (276 m)iˆ + (231 m) ˆj r = (790 m) cos(163) ˆi + (790 m) sin(163) ˆj = (  755 m )iˆ + (231 m )ˆj 2

respectively. Consequently, we plug into Eq. 4-3

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r = [(  755 m)  (276 m)]iˆ +(231 m  231 m) ˆj  (1031 m) ˆi.

The magnitude of the displacement  r is |  r |  1031 m. (b) The direction of  r is  î , or westward. 87. THINK This problem deals with the projectile motion of a baseball. Given the information on the position of the ball at two instants, we are asked to analyze its trajectory. EXPRESS The trajectory of the baseball is shown in the figure on the right. According to the problem statement, at t1  3.0 s, the ball reaches it maximum height ymax , and at t2  t1  2.5 s  5.5 s , it barely clears a fence at x2  97.5 m . Eq. 2-15 can be applied to the vertical (y axis) motion related to reaching the maximum height (when t1 = 3.0 s and vy = 0): 1 ymax – y0 = vyt – 2gt2 . ANALYZE (a) With ground level chosen so y0 = 0, this equation gives the result ymax 

1 2 1 gt1  (9.8 m/s 2 )(3.0 s) 2  44.1 m 2 2

(b) After the moment it reached maximum height, it is falling; at t2  t1  2.5 s  5.5 s , it will have fallen an amount given by Eq. 2-18:

1 yfence  ymax  0  g (t2  t1 )2 . 2 Thus, the height of the fence is 1 1 yfence  ymax  g (t2  t1 )2  44.1 m  (9.8 m/s 2 )(2.5 s) 2  13.48 m . 2 2

(c) Since the horizontal component of velocity in a projectile-motion problem is constant (neglecting air friction), we find from 97.5 m = v0x(5.5 s) that v0x = 17.7 m/s. The total flight time of the ball is T  2t1  2(3.0 s)  6.0 s . Thus, the range of the baseball is R  v0 xT  (17.7 m/s)(6.0 s)  106.4 m

which means that the ball travels an additional distance

171

 x  R  x2  106.4 m  97.5 m  8.86 m

beyond the fence before striking the ground. LEARN Part (c) can also be solved by noting that after passing the fence, the ball will strike the ground in 0.5 s (so that the total "fall-time" equals the "rise-time"). With v0x = 17.7 m/s, we have x = (17.7 m/s)(0.5 s) = 8.86 m. 88. When moving in the same direction as the jet stream (of speed vs), the time is

t1 

d , v ja  vs

where d = 4000 km is the distance and vja is the speed of the jet relative to the air (1000 km/h). When moving against the jet stream, the time is

t2 

d , v ja  vs

70 where t2 – t1 = 60 h . Combining these equations and using the quadratic formula to solve gives vs = 143 km/h.

89. THINK We have a particle moving in a two-dimensional plane with a constant acceleration. Since the x and y components of the acceleration are constants, we can use Table 2-1 for the motion along both axes.  EXPRESS Using vector notation with r0  0 , the position and velocity of the particle as 1 a function of time are given by r (t )  v0t  at 2 and v (t )  v0  at , respectively. Where 2 units are not shown, SI units are to be understood.

ANALYZE (a) Given the initial velocity v0  (8.0 m/s)ˆj and the acceleration a  (4.0 m/s 2 )iˆ  (2.0 m/s 2 )ˆj , the position vector of the particle is r  v0t 

 



 

1 2 1 at  8.0 ˆj t  4.0 ˆi  2.0 ˆj t 2  2.0t 2 ˆi + 8.0t + 1.0t 2 ˆj. 2 2

 



Therefore, the time that corresponds to x = 29 m can be found by solving the equation 2.0t2 = 29, which leads to t = 3.8 s. The y coordinate at that time is y = (8.0 m/s)(3.8 s) + (1.0 m/s2)(3.8 s)2 = 45 m.

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   (b) The velocity of the particle is given by v  v0  at . Thus, at t = 3.8 s, the velocity is





v  (8.0 m/s) ˆj  (4.0 m/s 2 ) ˆi  (2.0 m/s 2 ) ˆj 3.8 s   (15.2 m/s) ˆi  ( 15.6 m/s) ˆj which has a magnitude of v  vx2  vy2  (15.2 m/s)2  (15.6 m/s)2  22 m/s. LEARN Instead of using the vector notation, we can also deal with thex- and the ycomponents individually. 90. Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the initial speed: x g v0 = cos  0 2 ( x tan  0  y ) which yields v0 = 23 ft/s for g = 32 ft/s2, x = 13 ft, y = 3 ft and 0 = 55°. 91. We make use of Eq. 4-25. (a) By rearranging Eq. 4-25, we obtain the initial speed:

v0 

x cos 0

g 2( x tan  0  y )

which yields v0 = 255.5  2.6  102 m/s for x = 9400 m, y = –3300 m, and 0 = 35°. (b) From Eq. 4-21, we obtain the time of flight: t

x 9400 m   45 s. v0 cos 0 (255.5 m/s) cos 35

(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target. 92. We apply Eq. 4-34 to solve for speed v and Eq. 4-35 to find the period T. (a) We obtain

v  ra 

b5.0 mgb7.0gc9.8 m / s h  19 m / s. 2

(b) The time to go around once (the period) is T = 2r/v = 1.7 s. Therefore, in one minute (t = 60 s), the astronaut executes

173 t 60 s   35 T 1.7 s

revolutions. Thus, 35 rev/min is needed to produce a centripetal acceleration of 7g when the radius is 5.0 m. (c) As noted above, T = 1.7 s. 93. THINK This problem deals with the two-dimensional kinematics of a desert camel moving from oasis A to oasis B. EXPRESS The journey of the camel is illustrated in the figure on the right. We use a ‘standard’ coordinate system with +x East and +y North. Lengths are in kilometers and times are in hours. Using vector notation, we write the displacements for the first two segments of the trip as: r1  (75 km)cos(37) ˆi  (75 km) sin(37) ˆj r  (  65 km) ˆj 2

The net displacement is r12  r1  r2 . As can be seen from the figure, to reach oasis B requires an additional displacement  r3 . ANALYZE (a) We perform the vector addition of individual displacements to find the net displacement of the camel: r12  r1  r2  (60 km) ˆi  (20 km)ˆj. Its corresponding magnitude is | r12 |  (60 km)2  (  20 km)2  63 km.

(b) The direction of  r12 is 12  tan 1[( 20 km) /(60 km)]   18  , or 18 south of east. (c) To calculate the average velocity for the first two segments of the journey (including rest), we use the result from part (a) in Eq. 4-8 along with the fact that t12  t1  t2  trest  50 h  35 h  5.0 h  90 h.

In unit vector notation, we have v12,avg 

(60 ˆi  20 ˆj) km = (0.67 ˆi  0.22 ˆj) km/h. 90 h

This leads to | v12,avg |  0.70 km/h. (d) The direction of v12,avg is 12  tan 1[( 0.22 km/h) /(0.67 km/h)]  18 , or 18 south of east.

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(e) The average speed is distinguished from the magnitude of average velocity in that it depends on the total distance as opposed to the net displacement. Since the camel travels 140 km, we obtain (140 km)/(90 h) = 1.56 km/h  1.6 km/h. (f) The net displacement is required to be the 90 km East from A to B. The displacement from the resting place to B is denoted r3 . Thus, we must have

r1 + r2 + r3 = (90 km) ˆi which produces r3  (30 km)iˆ  (20 km)jˆ in unit-vector notation, or (36  33 ) in magnitude-angle notation. Therefore, using Eq. 4-8 we obtain 36 km  1.2 km/h. (120  90) h (g) The direction of v3,avg is the same as  r3 (that is, 33° north of east). | v3,avg | 

LEARN With a vector-capable calculator in polar mode, we could perform the vector addition of the displacements as (75  37)  (65   90)  (63   18) . Note the distinction between average velocity and average speed. 94. We compute the coordinate pairs (x, y) from x = (v0 cost and y  v0 sin  t  12 gt 2 for t = 20 s and the speeds and angles given in the problem. (a) We obtain  xA , yA   10.1 km, 0.556 km   xC , yC   14.3 km, 2.68 km 

 xB , yB   12.1 km, 1.51 km   xD , yD   16.4 km, 3.99 km 

and (xE, yE) = (18.5 km, 5.53 km) which we plot in the next part. (b) The vertical (y) and horizontal (x) axes are in kilometers. The graph does not start at the origin. The curve to “fit” the data is not shown, but is easily imagined (forming the “curtain of death”).

95. (a) With x = 8.0 m, t = t1, a = ax , and vox = 0, Eq. 2-15 gives

175 1

8.0 m = 2 ax(t1)2 , and the corresponding expression for motion along the y axis leads to 1

y = 12 m = 2 ay(t1)2 . Dividing the second expression by the first leads to ay / ax  3/ 2 = 1.5. (b) Letting t = 2t1, then Eq. 2-15 leads to x = (8.0 m)(2)2 = 32 m, which implies that its x coordinate is now (4.0 + 32) m = 36 m. Similarly, y = (12 m)(2)2 = 48 m, which means its y coordinate has become (6.0 + 48) m = 54 m. 96. We assume the ball’s initial velocity is perpendicular to the plane of the net. We choose coordinates so that (x0, y0) = (0, 3.0) m, and vx > 0 (note that v0y = 0). (a) To (barely) clear the net, we have y  y0  v0 y t 

1 2 1 gt  2.24 m  3.0 m  0   9.8 m/s 2  t 2 2 2

which gives t = 0.39 s for the time it is passing over the net. This is plugged into the xequation to yield the (minimum) initial velocity vx = (8.0 m)/(0.39 s) = 20.3 m/s. (b) We require y = 0 and find time t from the equation y  y0  v0 yt  12 gt 2 . This value

(t  2 3.0 m /(9.8 m/s )  0.78 s) is plugged into the x-equation to yield the 2

(maximum) initial velocity

vx = (17.0 m)/(0.78 s) = 21.7 m/s.

97. THINK A bullet fired horizontally from a rifle strikes the target at some distance below its aiming point. We’re asked to find its total flight time and speed. EXPRESS The trajectory of the bullet is shown in the figure on the right (not to scale). Note that the origin is chosen to be at the firing point. With this convention, the y coordinate of the bullet is given by y   21 gt 2 . Knowing the coordinates (x, y) at the target allows us to calculate the total flight time and speed of the bullet. ANALYZE (a) If t is the time of flight and y = – 0.019 m indicates where the bullet hits the target, then

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t

2  0.019 m  2 y   6.2  102 s. 2 g 9.8 m/s

(b) The muzzle velocity is the initial (horizontal) velocity of the bullet. Since x = 30 m is the horizontal position of the target, we have x = v0t. Thus, v0 

x 30 m   4.8  102 m/s. t 6.3  102 s

LEARN Alternatively, we may use Eq. 4-25 to solve for the initial velocity. With 0  0 and y0  0 , the equation simplifies to y  

v0  

gx 2 , from which we find 2v02

gx 2 (9.8 m/s 2 )(30 m) 2    4.8 102 m/s , 2y 2( 0.019 m)

in agreement with what we calculated in part (b). 

98. For circular motion, we must have v with direction perpendicular to r and (since the speed is constant) magnitude v  2 r / T where r  (2.00 m)2  (3.00 m)2 and 

T  7.00 s . The r (given in the problem statement) specifies a point in the fourth quadrant, and since the motion is clockwise then the velocity must have both components negative. Our result, satisfying these three conditions, (using unit-vector notation which makes it easy to double-check that r  v  0 ) for v = (–2.69 m/s)i^ + (–1.80 m/s)j^.

99. Let vo = 2(0.200 m)/(0.00500 s)  251 m/s (using Eq. 4-35) be the speed it had in circular motion and o = (1 hr)(360º/12 hr [for full rotation]) = 30.0º. Then Eq. 4-25 leads to (9.8 m/s2 )(2.50 m)2 y  (2.50 m) tan 30.0   1.44 m 2(251 m/s)2 (cos30.0)2 which means its height above the floor is 1.44 m + 1.20 m = 2.64 m.  100. Noting that v2  0 , then, using Eq. 4-15, the average acceleration is





ˆ ˆ v 0  6.30 i  8.42 j m/s aavg    2.1iˆ  2.8 ˆj m/s 2 t 3s



101. Using Eq. 2-16, we obtain v 2  v02  2 gh , or h  (v02  v 2 ) / 2 g.



177 (a) Since v  0 at the maximum height of an upward motion, with v0  7.00 m/s , we have h  (7.00 m/s)2 / 2(9.80 m/s 2 )  2.50 m. (b) The relative speed is vr  v0  vc  7.00 m/s  3.00 m/s  4.00 m/s with respect to the floor. Using the above equation we obtain h  (4.00 m/s)2 / 2(9.80 m/s2 )  0.82 m. (c) The acceleration, or the rate of change of speed of the ball with respect to the ground is 9.80 m/s2 (downward). (d) Since the elevator cab moves at constant velocity, the rate of change of speed of the ball with respect to the cab floor is also 9.80 m/s2 (downward). 102. (a) With r = 0.15 m and a = 3.0  1014 m/s2, Eq. 4-34 gives

v  ra  6.7  106 m / s. (b) The period is given by Eq. 4-35: T

2r  14 .  107 s. v

 103. (a) The magnitude of the displacement vector r is given by

| r |  (21.5 km)2  (9.7 km)2  (2.88 km)2  23.8 km.

Thus, | vavg | 

| r | 23.8 km   6.79 km/h. t 3.50 h

(b) The angle  in question is given by



   6.96.  (21.5 km)2  (9.7 km) 2   

  tan 1 

2.88 km

104. The initial velocity has magnitude v0 and because it is horizontal, it is equal to vx the horizontal component of velocity at impact. Thus, the speed at impact is

v02  v y2  3v0 where vy  2 gh and we have used Eq. 2-16 with x replaced with h = 20 m. Squaring both sides of the first equality and substituting from the second, we find

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b g

v02  2 gh  3v0

which leads to gh  4v02 and therefore to v0  (9.8 m/s2 )(20 m) / 2  7.0 m/s.  105. We choose horizontal x and vertical y axes such that both components of v0 are positive. Positive angles are counterclockwise from +x and negative angles are clockwise from it. In unit-vector notation, the velocity at each instant during the projectile motion is

v  v0 cos 0 ˆi   v0 sin 0  gt  ˆj. (a) With v0 = 30 m/s and 0 = 60°, we obtain v  (15iˆ +6.4 ˆj) m/s , for t = 2.0 s. The magnitude of v is | v | (15 m/s)2  (6.4 m/s)2  16 m/s. (b) The direction of v is

  tan 1[(6.4 m/s) /(15 m/s)]  23 ,

measured counterclockwise from +x. (c) Since the angle is positive, it is above the horizontal. (d) With t = 5.0 s, we find v  (15iˆ  23ˆj) m/s , which yields | v | (15 m/s)2  ( 23 m/s)2  27 m/s.

(e) The direction of v is   tan 1[( 23 m/s) /(15 m/s)]   57  , or 57 measured clockwise from +x. (f) Since the angle is negative, it is below the horizontal. 106. We use Eq. 4-2 and Eq. 4-3.   (a) With the initial position vector as r1 and the later vector as r2 , Eq. 4-3 yields r  [(  2.0 m)  5.0 m]iˆ  [(6.0m)  (  6.0 m)]jˆ  (2.0 m  2.0 m) kˆ  (7.0 m) ˆi  (12 m) ˆj

for the displacement vector in unit-vector notation. (b) Since there is no z component (that is, the coefficient of k̂ is zero), the displacement vector is in the xy plane.

179

107. We write our magnitude-angle results in the form R   with SI units for the magnitude understood (m for distances, m/s for speeds, m/s2 for accelerations). All angles  are measured counterclockwise from +x, but we will occasionally refer to angles  , which are measured counterclockwise from the vertical line between the circle-center and the coordinate origin and the line drawn from the circle-center to the particle location (see r in the figure). We note that the speed of the particle is v = 2r/T where r = 3.00 m and T = 20.0 s; thus, v = 0.942 m/s. The particle is moving counterclockwise in Fig. 4-56. (a) At t = 5.0 s, the particle has traveled a fraction of t 5.00 s 1   T 20.0 s 4

of a full revolution around the circle (starting at the origin). Thus, relative to the circlecenter, the particle is at 1   (360 )  90 4 measured from vertical (as explained above). Referring to Fig. 4-56, we see that this position (which is the “3 o’clock” position on the circle) corresponds to x = 3.0 m and y = 3.0 m relative to the coordinate origin. In our magnitude-angle notation, this is expressed as  R     4.2  45 . Although this position is easy to analyze without resorting to trigonometric relations, it is useful (for the computations below) to note that these values of x and y relative to coordinate origin can be gotten from the angle  from the relations x  r sin  ,

y  r  r cos  .

Of course, R  x 2  y 2 and  comes from choosing the appropriate possibility from tan–1 (y/x) (or by using particular functions of vector-capable calculators). (b) At t = 7.5 s, the particle has traveled a fraction of 7.5/20 = 3/8 of a revolution around the circle (starting at the origin). Relative to the circle-center, the particle is therefore at  = 3/8 (360°) = 135° measured from vertical in the manner discussed above. Referring to Fig. 4-56, we compute that this position corresponds to x = (3.00 m)sin 135° = 2.1 m y = (3.0 m) – (3.0 m)cos 135° = 5.1 m relative to the coordinate origin. In our magnitude-angle notation, this is expressed as (R  ) = (5.5  68°). (c) At t = 10.0 s, the particle has traveled a fraction of 10/20 = 1/2 of a revolution around the circle. Relative to the circle-center, the particle is at  = 180° measured from vertical (see explanation above). Referring to Fig. 4-56, we see that this position corresponds to x

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180

= 0 and y = 6.0 m relative to the coordinate origin. In our magnitude-angle notation, this is expressed as  R     6.0  90  . (d) We subtract the position vector in part (a) from the position vector in part (c):

 6.0  90    4.2  45    4.2 135  using magnitude-angle notation (convenient when using vector-capable calculators). If we wish instead to use unit-vector notation, we write R  (0  3.0 m) ˆi  (6.0 m  3.0 m) ˆj  (3.0 m)iˆ  (3.0 m)ˆj

which leads to | R | 4.2 m and  = 135°. (e) From Eq. 4-8, we have vavg  R / t . With t  5.0 s , we have

vavg  (0.60 m/s) ˆi  (0.60 m/s) ˆj in unit-vector notation or (0.85  135°) in magnitude-angle notation. (f) The speed has already been noted (v = 0.94 m/s), but its direction is best seen by referring again to Fig. 4-56. The velocity vector is tangent to the circle at its “3 o’clock  position” (see part (a)), which means v is vertical. Thus, our result is  0.94  90 . (g) Again, the speed has been noted above (v = 0.94 m/s), but its direction is best seen by referring to Fig. 4-56. The velocity vector is tangent to the circle at its “12 o’clock  position” (see part (c)), which means v is horizontal. Thus, our result is  0.94  180 . (h) The acceleration has magnitude a = v2/r = 0.30 m/s2, and at this instant (see part (a)) it is horizontal (toward the center of the circle). Thus, our result is  0.30  180 . (i) Again, a = v2/r = 0.30 m/s2, but at this instant (see part (c)) it is vertical (toward the center of the circle). Thus, our result is  0.30  270 . 108. Equation 4-34 describes an inverse proportionality between r and a, so that a large acceleration results from a small radius. Thus, an upper limit for a corresponds to a lower limit for r. (a) The minimum turning radius of the train is given by

181

216 km / h v2   7.3  103 m. rmin  2 amax 0.050 9.8 m / s

(b) The speed of the train must be reduced to no more than

v  amax r  0.050  9.8 m/s 2 1.00  103 m   22 m/s which is roughly 80 km/h. 109. (a) Using the same coordinate system assumed in Eq. 4-25, we find

y  x tan  0 

gx 2

2 v0 cos 0

 2

gx 2 2v02

if  0  0.

Thus, with v0 = 3.0  106 m/s and x = 1.0 m, we obtain y = –5.4  10–13 m, which is not practical to measure (and suggests why gravitational processes play such a small role in the fields of atomic and subatomic physics). (b) It is clear from the above expression that |y| decreases as v0 is increased. 110. When the escalator is stalled the speed of the person is v p  t , where  is the length of the escalator and t is the time the person takes to walk up it. This is vp = (15 m)/(90 s) = 0.167 m/s. The escalator moves at ve = (15 m)/(60 s) = 0.250 m/s. The speed of the person walking up the moving escalator is v = vp + ve = 0.167 m/s + 0.250 m/s = 0.417 m/s and the time taken to move the length of the escalator is t   / v  (15 m) / (0.417 m / s)  36 s.

If the various times given are independent of the escalator length, then the answer does not depend on that length either. In terms of  (in meters) the speed (in meters per second) of the person walking on the stalled escalator is  90 , the speed of the moving escalator is  60 , and the speed of the person walking on the moving escalator is

v   90   60  0.0278 . The time taken is t   v   0.0278  36 s and is

independent of  . 111. The radius of Earth may be found in Appendix C. (a) The speed of an object at Earth’s equator is v = 2R/T, where R is the radius of Earth (6.37  106 m) and T is the length of a day (8.64  104 s):

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182

v = 2(6.37  106 m)/(8.64  104 s) = 463 m/s. The magnitude of the acceleration is given by

463 m / s v2 a   0.034 m / s2 . R 6.37  106 m (b) If T is the period, then v = 2R/T is the speed and the magnitude of the acceleration is

Thus,

v 2 (2 R / T )2 4 2 R . a   R R T2

R T  2  2 a

6.37  106 m  51 .  103 s = 84 min. 2 9.8 m / s

112. With gB = 9.8128 m/s2 and gM = 9.7999 m/s2, we apply Eq. 4-26:

FG H

IJ K

v02 sin 2 0 v02 sin 2 0 v02 sin 2 0 g B RM  RB    1 gM gB gB gM which becomes

 9.8128 m/s 2  RM  RB  RB   1 2  9.7999 m/s  and yields (upon substituting RB = 8.09 m) RM – RB = 0.01 m = 1 cm. 113. From the figure, the three displacements can be written as d1  d1 (cos 1î  sin 1ˆj)  (5.00 m)(cos 30î  sin 30ˆj)  (4.33 m)iˆ  (2.50 m)ˆj ˆ  (8.00 m)(cos160î  sin160ˆj) d 2  d 2 [cos(180  1   2 )iˆ  sin(180  1   2 )j]  (7.52 m)iˆ  (2.74 m)ˆj ˆ  (12.0 m)(cos 260î  sin 260ˆj) d3  d3[cos(360  3   2  1 )iˆ  sin(360  3   2  1 )j]  (2.08 m)iˆ  (11.8 m)ˆj

where the angles are measured from the +x axis. The net displacement is

d  d1  d2  d3  ( 5.27 m)iˆ  (6.58 m)ˆj. (a) The magnitude of the net displacement is

183

| d | ( 5.27 m)2  ( 6.58 m) 2  8.43 m.

d    6.58 m  (b) The direction of d is   tan 1  y   tan 1    51.3  or 231. d  5.27 m    x We choose 231 (measured counterclockwise from +x) since the desired angle is in the third quadrant. An equivalent answer is  129  (measured clockwise from +x). 114. Taking derivatives of r  2t ˆi  2sin( t / 4)ˆj (with lengths in meters, time in seconds, and angles in radians) provides expressions for velocity and acceleration: dr   t   2iˆ  cos   ˆj dt 2  4 2 dv   t  a   sin   ˆj. dt 8  4 v

Thus, we obtain: time t (s)

0.0

1.0

2.0

3.0

4.0

x (m)

0.0

2.0

4.0

6.0

8.0

y (m)

0.0

1.4

2.0

1.4

0.0

vx(m/s)

2.0

vy (m/s)

1.1

0.0

1.1

ax (m/s2)

0.0

ay (m/s2)

0.87

1.2

0.87



(a)

r position

(b)

v velocity

(c)

a acceleration



115. Since this problem involves constant downward acceleration of magnitude a, similar to the projectile motion situation, we use the equations of §4-6 as long as we substitute a for g. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The initial velocity is horizontal so that v0 y  0 and

v0 x  v0  1.00 109 cm/s. (a) If  is the length of a plate and t is the time an electron is between the plates, then   v0 t , where v0 is the initial speed. Thus t



2.00 cm  2.00  109 s. 9 1.00  10 cm/s

CHAPTER 4

184 (b) The vertical displacement of the electron is

2 1 1 y   at 2   1.00  1017 cm/s 2  2.00  109 s    0.20 cm   2.00 mm, 2 2

or | y |  2.00 mm. (c) The x component of velocity does not change: vx = v0 = 1.00  109 cm/s = 1.00  107 m/s. (d) The y component of the velocity is

v y  a y t  1.00  1017 cm/s 2  2.00  10 9 s   2.00  108 cm/s  2.00  106 m/s. 116. We neglect air resistance, which justifies setting a = –g = –9.8 m/s2 (taking down as the –y direction) for the duration of the motion of the shot ball. We are allowed to use Table 2-1 (with y replacing x) because the ball has constant acceleration motion. We use primed variables (except t) with the constant-velocity elevator (so v '  10 m/s ), and unprimed variables with the ball (with initial velocity v0  v  20  30 m/s , relative to the ground). SI units are used throughout. (a) Taking the time to be zero at the instant the ball is shot, we compute its maximum height y (relative to the ground) with v2  v02  2 g ( y  y0 ) , where the highest point is characterized by v = 0. Thus, v2 y  yo  0  76 m 2g where yo  yo  2  30 m (where yo  28 m is given in the problem) and v0 = 30 m/s relative to the ground as noted above. (b) There are a variety of approaches to this question. One is to continue working in the frame of reference adopted in part (a) (which treats the ground as motionless and “fixes” the coordinate origin to it); in this case, one describes the elevator motion with y  yo  vt and the ball motion with Eq. 2-15, and solves them for the case where they reach the same point at the same time. Another is to work in the frame of reference of the elevator (the boy in the elevator might be oblivious to the fact the elevator is moving since it isn’t accelerating), which is what we show here in detail: 1 ye  v0e t  gt 2 2



t

v0e  v0 e  2 gye g

185

where v0e = 20 m/s is the initial velocity of the ball relative to the elevator and ye = –2.0 m is the ball’s displacement relative to the floor of the elevator. The positive root is chosen to yield a positive value for t; the result is t = 4.2 s. 117. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the initial position for the football as it begins projectile motion in the sense of §4-5), and we let 0 be the angle of its initial velocity measured from the +x axis. (a) x = 46 m and y = –1.5 m are the coordinates for the landing point; it lands at time t = 4.5 s. Since x = v0xt, x 46 m v0 x    10.2 m/s. t 4.5 s Since y  v0 y t  21 gt 2 ,

v0 y 

y

1 2 1 ( 15 . m)  (9.8 m / s2 )(4.5 s) 2 gt 2 2   217 . m / s. 4.5 s t

The magnitude of the initial velocity is v0  v02 x  v02 y  (10.2 m / s) 2  (217 . m / s) 2  24 m / s.

(b) The initial angle satisfies tan 0 = v0y/v0x. Thus,

0 = tan–1 [(21.7 m/s)/(10.2 m/s) ] = 65°. 118. The velocity of Larry is v1 and that of Curly is v2. Also, we denote the length of the corridor by L. Now, Larry’s time of passage is t1 = 150 s (which must equal L/v1), and Curly’s time of passage is t2 = 70 s (which must equal L/v2). The time Moe takes is therefore L 1 1 t   1  48s. v1  v2 v1 / L  v2 / L 150 s  701 s

 119. The boxcar has velocity vc g  v1 i relative to the ground, and the bullet has velocity  v0b g  v2 cos i  v2 sin  j

relative to the ground before entering the car (we are neglecting the effects of gravity on the bullet). While in the car, its velocity relative to the outside ground is

 vbg  0.8v2 cos i  0.8v2 sin  j

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186

(due to the 20% reduction mentioned in the problem). The problem indicates that the  velocity of the bullet in the car relative to the car is (with v3 unspecified) vb c  v3 j . Now, Eq. 4-44 provides the condition vb g  vb c  vc g 0.8v cos  ˆi  0.8v sin  ˆj  v ˆj  v ˆi 2

so that equating x components allows us to find . If one wished to find v3 one could also equate the y components, and from this, if the car width were given, one could find the time spent by the bullet in the car, but this information is not asked for (which is why the width is irrelevant). Therefore, examining the x components in SI units leads to 1000 m/km  v1  1  85 km/h  3600 s/h    cos     0.8v2   0.8 (650 m/s) 

  cos 1 

 which yields 87° for the direction of vb g (measured from i , which is the direction of motion of the car). The problem asks, “from what direction was it fired?” — which means the answer is not 87° but rather its supplement 93° (measured from the direction of motion). Stating this more carefully, in the coordinate system we have adopted in our solution, the bullet velocity vector is in the first quadrant, at 87° measured counterclockwise from the +x direction (the direction of train motion), which means that the direction from which the bullet came (where the sniper is) is in the third quadrant, at –93° (that is, 93° measured clockwise from +x).

120. (a) Using a  v 2 / R, the radius of the track is R

v 2 (9.20 m/s)2   22.3 m . a 3.80 m/s 2

(b) Using T  2 R / v, the period of the circular motion is T

2 R 2 (22.3 m)   15.2 s v 9.20 m/s

121. (a) With v  c /10  3 107 m/s and a  20 g  196 m/s2 , Eq. 4-34 gives r  v2 / a  4.6 1012 m.

(b) The period is given by Eq. 4-35: T  2 r / v  9.6 105 s. Thus, the time to make a quarter-turn is T/4 = 2.4  105 s or about 2.8 days.