Issuu

Type:

Solutions Manual

Resource:

College Physics

Edition:

11th Edition

Author(s):

Raymond Serway Chris Vuille

Topic 1

Topic 1 Units, Trigonometry, and Vectors

QUICK QUIZZES 1.1 Choice (c). The largest possible magnitude of the resultant occurs when the two vectors are in the same direction. In this case, the magnitude of   the resultant is the sum of the magnitudes of !A and !B : R = A + B = 20

units. The smallest possible magnitude of the resultant occurs when the two vectors are in opposite directions, and the magnitude is the   difference of the magnitudes of !A and !B : R = | A − B | = 4 units.

1.2

Vector

x-component

y-component

 !A

−

 !B

−

  !A + B

−

 1.3 Vector !B . The range of the inverse tangent function includes only the first

and fourth quadrants (i.e., angles in the range −π/2 < θ < π/2). Only

Topic 1

 vector !B has an orientation in this range.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 1.2

Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks.

1.4

1.6

(a)

~0.5 lb ≈ 0.25 kg or ~10−1 kg

(b)

∼4 lb ≈ 0.25 kg or ~100 kg

(c)

~4 000 lb ≈ 2 000 kg or ~103 kg

Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom is of the order of 10−30 m3. (More precisely, volume = 4π r 3 /3 .) Therefore, since 1 cm3 = 10−6 m3, the number of atoms in the 1 cm3 solid is on the order of 10−6 / 10−30 = 1024 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10.

1.8

Choice (d). For an angle θ from 0° to 360°, the sine and cosine functions take the values

−1 ≤ sin θ ≤ 1 and − 1 ≤ cosθ ≤ 1 1.10

In the metric system, units differ by powers of ten, so it’s very easy

Topic 1

and accurate to convert from one unit to another. 1.12

Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions.

1.14

The correct answer is (a). The second measurement is more precise but, given the number of reported significant figures, each measurement is consistent with the other.

1.16

The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two axes along which the components lie.

ANSWERS TO EVEN NUMBERED PROBLEMS L / T2

1.2

(a)

1.4

All three equations are dimensionally incorrect.

1.6

(a)

1.8

1.10 1.12

1.14

(b)

kg ⋅ m/s

(b)

Ft = p

(a)

22.6

(b)

22.7

(a)

3.00 × 108 m/s

(b)

2.997 9 × 108 m/s

(c)

2.997 925 × 108 m/s

(a)

346 m2 ± 13 m3

(b)

66.0 m ± 1.3 m

Topic 1

1.16

(a)

797

(b)

1.1

1.18

132 m2

1.20

3.09 cm/s

1.22

(a)

5.60 × 102 km = 5.60 × 105 m = 5.60 × 107 cm

(b)

0.491 2 km = 4.912 × 104 cm

(c)

6.192 km = 6.192 × 103 m = 6.192 × 105 cm

(d)

2.499 km = 2.499 × 103 m = 2.499 × 105 cm

1.24

10.6 km/L

1.26

9.2 nm/s

1.28

2.9 × 102 m3 = 2.9 × 108 cm3

1.30

2.57 × 106 m3

1.32

~108 steps

1.34

~108 people with colds on any given day

1.36

(a)

1.38

10−14 kg

1.40

2.2 m

1.42

8.1 cm

1.44

Δs = r12 + r22 − 2r1r2 cos(θ1 − θ 2 )

4.2 × 10−18 m3

(b)

~10−1 m3

(c)

17.66

(c)

~1016 cells

Topic 1

1.46

2.33 m

1.48

(a)

1.50

8.60 m

1.52

1.44 × 103 m

1.54

(a) 6.1 units at θ = +113°

(b) 15 units at θ = +23°

1.56

(a) 484 km

(b) 18.1° N of W

1.50 m

(b)

2.60 m

 R ! = 9.5 units at!!57° above!!the!+ x2axis

1.60

(a) 13.4 m

1.62

1.31 km northward, 2.81 km eastward

1.64

(a) 10.0 m

1.66

42.7 yards

1.68

788 mi at 48.0° N of E

1.70

(a) 185 N at 77.8° from the x-axis

(b) 185 N at 258° from the x-axis

1.72

(a) 1.609 km/h

(b) 88 km/h

1.74

(a)

(b) 2.70 × 10−4 m3/s

(b) 19.9 m

(b) 15.7 m

7.14 × 10−2 gal/s

Topic 1

1.76

(a)

A2/A1 = 4

(b) V2/V1 = 8

1.78

(a)

600 yr

(b)

1.80

1 × 106

7 × 104 times

PROBLEM SOLUTIONS 1.1

Substituting dimensions into the given equation T = 2π ℓ/ g , and recognizing that 2π is a dimensionless constant, we have ℓ [T ] = [[ g]]

or T =

L = T2 = T L/T 2

Thus, the dimensions are consistent. 1.2

(a)

From x = Bt2, we find that B = x/t2. Thus, B has units of [B] = [x]/[t2] = L/T2 .

(b) If x = A sin (2π ft), then [A] = [x]/[sin (2π ft)] But the sine of an angle is a dimensionless ratio. Therefore, [A] = [x] = L. 1.3

(a)

The units of volume, area, and height are: [V] = L3, [A] = L2, and [h] = L. We then observe that L3 = L2L or [V] = [A][h]. Thus, the equation V

Topic 1

= Ah is dimensionally correct.

(

)

2 2 (b) Vcylinder = π R h = π R h = Ah , where A = π R 2 .

Vrectangular box = ℓwh = ( ℓw ) h = Ah , where A = ℓw = length × width . 1.4

(a)

2 2 In the equation 12 mv = 12 mv0 + mgh ,

2 ML2 ⎛ L⎞ [mv ] = [mv ] = M ⎜ ⎟ = 2 while ⎡ mgh ⎤ = M ⎛⎜ L ⎞⎟ L = M L . ⎝ T⎠ ⎣ ⎦ T ⎝ T2 ⎠ T

2 0

Thus, the equation is dimensionally incorrect.

(b) In v = v0 + at , [v] = [v0 ] = 2

L ⎛ L⎞ 2 2 2 but [at ] = [a][t ] = ⎜ 2 ⎟ ( T ) = L . ⎝T ⎠ T

Hence, this equation is dimensionally incorrect.

(c)

⎛ L ⎞ ML In the equation ma = v2, we see that [ma] = [m][a] = M ⎜ 2 ⎟ = 2 , ⎝T ⎠ T 2

L2 ⎛ L⎞ while [v ] = ⎜ ⎟ = 2 . Therefore, this equation is also ⎝ T⎠ T 2

dimensionally incorrect. 1.5

From the universal gravitation law, the constant G is G = Fr2/Mm. Its units are then

[ F ] ⎡⎣ r ⎤⎦ = ( kg ⋅ m/s )( m ) = m G = [ ] [ M ][ m] kg ⋅s kg ⋅ kg 2

1.6

(a)

Solving KE = p2/2m for the momentum, p, gives p = 2m ( KE)

Topic 1

where the numeral 2 is a dimensionless constant. Dimensional analysis gives the units of momentum as:

[ p] = [ m][ KE] = M ( M ⋅ L /T ) = M ⋅ L /T = M ( L/T ) 2

Therefore, in the SI system, the units of momentum are kg ⋅ (m/s). (b)

Note that the units of force are kg ⋅ m/s2 or [F] = ML/T2. Then, observe that [F][t] = (ML/T2)T = M(L/T) = [p] From this, it follows that force multiplied by time is proportional to momentum: Ft = p. (See the impulse-momentum theorem in Topic 6, which says that a constant force F multiplied by a duration of time Δt equals the change in momentum, Δp.)

1.7

The rectangular airstrip’s area is its length times its width. The width contains four significant figures while the length contains two. Therefore the answer should be rounded to two significant figures: A = LW = ( 210 m ) ( 32.30 m ) A = 6.8 × 103 m 2 (rounded to two significant figures)

1.8

The number 15 contains no significant figures after the decimal, so the sum should be rounded to no digits after the decimal:

Topic 1

21.4 + 15 + 17.17 + 4.003 = 58 (rounded to no digits past the decimal)

1.9

The area A of the rectangular room is A = LW = (9.72 m)(5.3 m) = 52 m2 .

1.10 (a)

Computing yields

(b)

( 8 ) without rounding the intermediate result 3

( 8 ) = 22.6 to three significant figures. 3

Rounding the intermediate result to three significant figures yields 8 = 2.8284 → 2.83 . Then, we obtain

( 8 ) = ( 2.83) = 22.7 3

to three significant figures. (c)

The answer 22.6 is more reliable because rounding in part (b) was carried out too soon.

1.11 (a)

78.9 ± 0.2 has 3 significant figures with the uncertainty in the tenths position.

(b) 3.788 × 109 has 4 significant figures 4 significant figures (c) 246 × 10−6 has 3 significant figures (d) 0.003 2 = 3.2 × 10−3 has 2 significant figures. The two zeros were originally included only to position the decimal. 1.12

c = 2.997 924 58 × 108 m/s (a) Rounded to 3 significant figures: c = 3.00 × 108 m/s

Topic 1

(b) Rounded to 5 significant figures: c = 2.997 9 × 108 m/s (c) Rounded to 7 significant figures: c = 2.997 925 × 108 m/s 1.13

For a length of L = (2.0 ± 0.2) m and a width of W = (1.5 ± 0.1) m: (a) The maximum value for the area, A = LW, is found using the maximum values of L and W: Amax = (2.2 m)(1.6 m) = 3.52 m2 The minimum area is found using the minimum values of L and W: Amin = (1.8 m)(1.4 m) = 2.52 m2 The area using numbers in the middle of the uncertainty range is Amiddle = LW = (2.0 m)(1.5 m) = 3.0 m2 An uncertainty of about ± 0.5 m2 is needed to cover the range from Amax to Amin so that A = ( 3.0 ± 0.5 ) m . 2

(b) The perimeter of a rectangle is P = 2L + 2W = 2(L + W). As before, the maximum, minimum, and middle values are: Pmax = 2(2.2 m + 1.6 m) = 7.6 m Pmin = 2(1.8 m + 1.4 m) = 6.4 m Pmiddle = 2(2.0 m + 1.5 m) = 7.0 m An uncertainty of ±0.6 is needed to cover the range from Pmax to

Topic 1

Pmin so that P = ( 7.0 ± 0.6 ) m . 1.14

(a)

A = π r2 = π (10.5 m ± 0.2 m)2 = π [(10.5 m)2 ± 2(10.5 m)(0.2 m) + (0.2 m)2]. Recognize that the last term in the brackets is insignificant in comparison to the other two. Thus, we have A = π [110 m2 ± 4.2 m2] = 346 m2 ± 13 m2

(b) C = 2πr = 2π (10.5 m ± 0.2 m) = 66.0 m ± 1.3 m 1.15

The least accurate dimension of the box has two significant figures. Thus, the volume (product of the three dimensions) will contain only two significant figures. V = ℓ ⋅ w ⋅ h = ( 29 cm ) (17.8 cm ) (11.4 cm ) = 5.9 × 10 3 cm 3

1.16

(a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. (b) 0.003 2 × 356.3 = (3.2 × 10−3) × 356.3 = 1.140 16 must be rounded to 1.1 because 3.2 × 10−3 has only two significant figures. (c) 5.620 × π must be rounded to 17.66 because 5.620 has only four significant figures.

1.17

Use the conversion factor 1 cubitus = 0.445 m to find the basketball player’s height:

Topic 1

⎛ 1 cubitus ⎞ 2.00 m = 2.00 m ⎜ = 4.49 cubiti ⎝ 0.445 m ⎟⎠ 1.18

Use the conversion factor 1 m = 3.281 ft to find 2

⎛ 1m ⎞ 1 420 ft 2 = 1 420 ft 2 ⎜ = 132 m 2 ⎝ 3.281 ft ⎟⎠

1.19

⎛ 5 280 ft ⎞ ⎛ 1 fathom ⎞ d = (250 000 mi) ⎜ = 2 × 10 8 fathoms ⎟ ⎜ ⎟ ⎝ 1.000 mi ⎠ ⎝ 6 ft ⎠ The answer is limited to one significant figure because of the accuracy to which the conversion from fathoms to feet is given.

1.20 v=

ℓ 186 furlongs ⎛ 1 fortnight ⎞ ⎛ 1 days ⎞ ⎛ 220 yds ⎞ ⎛ 3 ft ⎞ ⎛ 100 cm ⎞ = t 1 fortnight ⎜⎝ 14 days ⎟⎠ ⎜⎝ 8.64 × 10 4 s ⎟⎠ ⎜⎝ 1 furlongs ⎟⎠ ⎜⎝ 1 yd ⎟⎠ ⎜⎝ 3.281 ft ⎟⎠

giving v = 3.09 cm/s

1.21

⎛ 9 gal ⎞ ⎛ 3.786 L ⎞ ⎛ 10 3 cm 3 ⎞ ⎛ 1 m 3 ⎞ 6.00 firkins = 6.00 firkins ⎜ 3⎟ ⎟⎜ ⎟⎜ ⎟⎜ 6 ⎝ 1 firkin ⎠ ⎝ 1 gal ⎠ ⎝ 1 L ⎠ ⎝ 10 cm ⎠ = 0.204 m 3

1.22

⎛ 1.609 km ⎞ ℓ = (348 mi) ⎜ = 5.60 × 10 2 km = 5.60 × 10 5 m ⎟ ⎝ 1.000 mi ⎠ (a) = 5.60 × 10 7 cm ⎛ 1.609 km ⎞ h = (1 612 ft) ⎜ = 0.491 2 km = 491.2 m ⎝ 5 280 ft ⎟⎠ (b) = 4.912 × 10 4 cm

Topic 1

⎛ 1.609 km ⎞ h = (20 320 ft) ⎜ = 6.192 km = 6.192 × 10 3 m ⎟ 5 280 ft ⎝ ⎠ (c) = 6.192 × 10 5 cm ⎛ 1.609 km ⎞ d = (8 200 ft) ⎜ = 2.499 km = 2.499 × 10 3 m ⎟ ⎝ 5 280 ft ⎠ (d) = 2.499 × 10 5 cm In (a), the answer is limited to three significant figures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four significant figures because of the accuracy to which the kilometers-to-feet conversion factor is given.

1.23

v = 38.0

m ⎛ 1 km ⎞ ⎛ 1 mi ⎞ ⎛ 3 600 s ⎞ = 85.0 mi/h s ⎜⎝ 10 3 m ⎟⎠ ⎜⎝ 1.609 km ⎟⎠ ⎜⎝ 1 h ⎟⎠

Yes, the driver is exceeding the speed limit by 10.0 mi/h.

1.24

efficiency = 25.0

1.25

(a) r =

mi ⎛ 1 km ⎞ ⎛ 1 gal ⎞ = 10.6 km/L gal ⎜⎝ 0.621 mi ⎟⎠ ⎜⎝ 3.786 L ⎟⎠

diameter 5.36 in ⎛ 2.54 cm ⎞ = = 6.81 cm 2 2 ⎜⎝ 1 in ⎟⎠

(b) A = 4π r2 = 4π (6.81 cm)2 = 5.83 × 102 cm2

1.26

4 3 4 3 π r = π ( 6.81 cm ) = 1.32 × 10 3 cm 3 3 3

⎛ 1 in ⎞ ⎛ 1 day ⎞ ⎛ 1 h ⎞ ⎛ 2.54 cm ⎞ ⎛ 10 9 nm ⎞ rate = ⎜ ⎟⎜ ⎟⎜ ⎟ = 9.2 nm / s ⎟⎜ ⎟⎜ 2 ⎝ 32 day ⎠ ⎝ 24 h ⎠ ⎝ 3 600 s ⎠ ⎝ 1.00 in ⎠ ⎝ 10 cm ⎠

Topic 1

This means that the proteins are assembled at a rate of many layers of atoms each second!

1.27

⎛ m ⎞ ⎛ 3 600 s ⎞ ⎛ 1 km ⎞ ⎛ 1 mi ⎞ c = ⎜ 3.00 × 10 8 = 6.71× 10 8 mi/h 3 ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ s ⎠ ⎝ 1 h ⎠ ⎝ 10 m ⎠ ⎝ 1.609 km ⎠ ⎝

⎛ 2.832 × 10 −2 m 3 ⎞ Volume of house = (50.0 ft)(26 ft)(8.0 ft) ⎜ ⎟⎠ 1 ft 3 ⎝ ⎛ 100 cm ⎞ = 2.9 × 10 m = (2.9 × 10 m ) ⎜ ⎝ 1 cm ⎟⎠

1.28

= 2.9 × 10 8 cm 3

1.29

2 2 ⎛ 1 m ⎞ ⎡⎛ 43 560 ft ⎞ ⎛ 1 m ⎞ ⎤ Volume = 25.0 acre ft ⎜ ⎢⎜ ⎥ ⎟ ⎝ 3.281 ft ⎟⎠ ⎢⎣⎝ 1 acre ⎠ ⎜⎝ 3.281 ft ⎟⎠ ⎥⎦

(

)

= 3.08 × 10 4 m 3

1.30

1 Volume of pyramid = (area of base)(height) 3 1 = [(13.0 acres)(43560 ft 2 / acre)](481 ft) = 9.08 × 10 7 ft 3 3 ⎛ 2.832 × 10 −2 m 3 ⎞ = (9.08 × 10 3 ft 3 ) ⎜ = 2.57 × 10 6 m 3 3 ⎟ 1 ft ⎝ ⎠ 1.31

Volume of cube = L3 = 1 quart (Where L = length of one side of the cube.) ⎛ 1 gallon ⎞ ⎛ 3.786 liter ⎞ ⎛ 1000 cm 3 ⎞ = 947 cm 3 Thus, L = 1 quart ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 quarts ⎠ ⎝ 1 gallon ⎠ ⎝ 1 liter ⎠ 3

(

)

Topic 1

and L = 3 947 cm 3 = 9.82 cm 1.32

We estimate that the length of a step for an average person is about 18 inches, or roughly 0.5 m. Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be

or 1.33

Circumference 2π RE 2π (6.38 × 10 6 m) = = ≈ 8 × 10 7 steps Step Length Step Length 0.5 m / step

N ~ 103 steps

We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is ⎛ 3.156 × 10 7 s ⎞ ⎛ 1 min ⎞ ⎛ breaths ⎞ n = ⎜ 10 70 yr ⎜ = 4 × 10 8 breaths ⎟ ⎟ ⎜ ⎟ ⎝ min ⎠ 1 yr ⎝ ⎠ ⎝ 60 s ⎠

(

or 1.34

)

n ~ 108 breaths

We assume that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each year (52 weeks). The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or

Topic 1

probability of sickness =

2 weeks 1 = 52 weeks 26

The population of the Earth is approximately 7 billion. The number of people expected to have a cold on any given day is then

⎛ 1⎞ Number sick = ( population)( probability of sickness) = (7 × 10 9 ) ⎜ ⎟ = 3 × 10 8 or 10 8 ⎝ 26 ⎠

1.35

Earth’s current population is about 7.3 billion people. If six people can stand within a 1-m2 area, the total area required to fit all 7.3 billion people is: ⎛ 1 m2 ⎞ Apeople = 7.3 × 10 people ⎜ = 1.2 × 109 m 2 ⎟ ⎝ 6 people ⎠ 9

The required percentage of the habitable part of Earth’s surface is: ⎛ 1.2 × 109 m 2 ⎞ Occupied percent = ⎜ 100% = 2 × 10−3% 12 2⎟ ⎝ 60 × 10 m ⎠

1.36

(a) Vcell =

4 3 4 π r = π (1.0 × 10 −6 m)3 = 4.2 × 10 −18 m 3 3 3

(b) Consider your body to be a cylinder having a radius of about 6 inches (or 0.15 m) and a height of about 1.5 meters. Then, its volume is Vbody = Ah = (π r2)h = π (0.15 m)2 (1.5 m) = 0.11 m3 or ~10−1 m3 (c) The estimate of the number of cells in the body is then

Topic 1

1.37

Vbody 0.11 m 3 = = 2.6 × 1016 or ~ 1016 Vcell 4.2 × 10 −18 m 3

A reasonable guess for the diameter of a tire might be 3 ft, with a circumference (C = 2π r = πD = distance traveled per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is

1.38

total distance traveled (50 000 mi)(5 280 ft/mi) = = 3 × 10 7 rev, or ~ 10 7 rev distance per revolution 9 ft/rev

An average body mass is about 70 kg and, taking the center of the given range, approximately 2% or 1.4 kg of this mass is made of microorganisms. If there are 100 trillion microorganisms in a human body, the average mass of a microorganism is about

mmicroorganism ≈

1.39

1.4 kg ≈ 1 × 10−14 kg 100 × 10 microorganisms 12

The x coordinate is found as x = r cos θ = (2.5 m)cos 35° = 2.0 m and the y coordinate y = r sin θ = (2.5 m)sin 35° = 1.4 m

1.40

The x distance out to the fly is 2.0 m and the y distance up to the fly is 1.0 m. Thus, we can use the Pythagorean theorem to find the distance from the origin to the fly as

d = x 2 + y 2 = (2.0 m)2 + (1.0 m)2 = 2.2 m 1.41

The distance from the origin to the fly is r in polar coordinates, and this

Topic 1

was found to be 2.2 m in Problem 40. The angle θ is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as

tan θ =

y 1.0 m = = 0.50 x 2.0 m

and θ = tan−1 (0.50) = 27°

The polar coordinates are r = 2.2 m and θ = 27° 1.42

The x distance between the two points is |Δx| = |x2 − x1| = |−3.0 cm − 5.0 cm|= 8.0 cm and the y distance between them is |Δy| = |y2 − y1| = | −3.0 cm − 4.0 cm| = 1.0 cm. The distance between them is found from the Pythagorean theorem:

1.43

Δx + Δy = (8.0 cm)2 + (1.0 cm)2 = 65 cm 2 = 8.1 cm 2

Refer to the Figure given in the solution to Problem 1.44 below. The Cartesian coordinates for the two given points are: x1 = r1 cos θ1 = (2.00 m) cos 50.0° = 1.29 m y1 = r1 sin θ1 = (2.00 m) sin 50.0° = 1.53 m x2 = r2 cos θ2 = (5.00 m) cos (−50.0°) = 3.21 m y2 = r2 sin θ2 = (5.00 m) sin (−50.0°) = 3.83 m The distance between the two points is then:

Δs = (Δx)2 + (Δy)2 = (1.29 m − 3.21 m)2 + (1.53 m + 3.83 m)2 = 5.69 m

Topic 1

1.44

Consider the Figure shown below. The Cartesian coordinates for the two points are: x1 =r1 cos θ1

x2 = r2 cos θ2

y1 = r1 sin θ1

y2 = r2 sin θ2

The distance between the two points is the length of the hypotenuse of the shaded triangle and is given by Δs = (Δx)2 + (Δy)2 = (x1 − x2 )2 + (y1 − y2 )2

Δs = (r12 cos 2 θ1 + r22 cos 2 θ 2 − 2r1r2 cosθ1 cosθ 2 ) + (r12 sin 2 θ1 + r22 sin 2 θ 2 − 2r1r2 sin θ1 sin θ 2 ) = r12 (cos 2 θ1 + sin 2 θ1 ) + r22 (cos 2 θ 2 + sin 2 θ 2 ) − 2r1r2 (cosθ1 cosθ 2 + sin θ1 sin θ 2 )

Applying the identities cos2 θ + sin2 θ = 1 and cos θ1 cos θ2 + sin θ1 sin θ2 = cos(θ1 − θ2), this reduces to Δs = r12 + r22 − 2r1r2 (cosθ1 cosθ 2 + sin θ1 sin θ 2 ) =

1.45

(a)

r12 + r22 − 2r1r2 cos(θ1 − θ 2 )

With a = 6.00 m and b being two sides of this right triangle having hypotenuse c = 9.00 m, the Pythagorean theorem gives the

Topic 1

unknown side as b = c 2 − a 2 = (9.00 m)2 − (6.00 m)2 = 6.71 m .

1.46

(b)

tan θ =

a 6.00 m = = 0.894 b 6.71 m

(c)

sin φ =

b 6.71 m = = 0.746 c 9.00 m

From the diagram, cos(75.0°) = d/L Thus, d = L cos(75.0°) = (9.00 m) cos(75.0°) = 2.33 m

1.47

The circumference of the fountain is C = 2πr, so the radius is

C 15.0 m = = 2.39 m 2π 2π

Topic 1

Thus, tan(55.0°) =

h h = which gives r 2.39 m

h = (2.39 m) tan (55.0°) = 3.41 m.

1.48

1.49

1.50

(a) sin θ =

opposite side so, opposite side = (3.00 m) sin (30.0°) = 1.50 m hypotenuse

(b) cosθ =

adjacent side so, adjacent side = (3.00 m) cos (30.0°) = 2.60 m hypotenuse

(a) The side opposite θ = 3.00

(c)

cosθ =

4.00 = 0.800 5.00

(e)

tan φ =

4.00 = 1.33 3.00

(b)

(d)

The side adjacent to φ = 3.00

sin φ =

4.00 = 0.800 5.00

Using the diagram below, the Pythagorean theorem yields

c = (5.00 m)2 + (7.00 m)2 = 8.60 m

Topic 1

1.51

From the diagram given in Problem 1.50 above, it is seen that

tan θ =

1.52

5.00 = 0.714 7.00

and

θ = tan−1 (0.714) = 35.5°

Use the diagram below to equate two expressions for the mountain’s height (units and significant digits have been suppressed for clarity):

y = (x − 1)tan(14.0) = (x)tan(12.0°) Solve for the unknown distance x to find

( x − 1) tan (14.0°) = x tan (12.0°) x ( tan14° − tan12° ) = tan14° x=

tan14° = 6.78 km tan14° − tan12°

Substitute the known value of x into y = (x)tan(12.0°) to find the mountain’s height: y = x tan (12° ) = ( 6.78 km ) tan (12° ) = 1.44 km = 1.44 × 103 m

1.53

Using the sketch below:

Topic 1

vector diagram and should be

(b)

  A + B = 5.0 units!!at!!− 53°

     To find the vector difference!A − B = A + (−B) , form the vector !−B   (same magnitude as !B , opposite direction) and add it to vector !A

as shown in the diagram. You should find that

1.58

  A − B = 5.0 units!!at!!+ 53°

    Find the resultant !R = F1 + F2 graphically by placing the tail of !F2 at the  head of !F1 in a scale drawing as shown below. The resultant is the   vector drawn from the tail of !F1 to the head of !F2 to close up the triangle. Measuring the length and orientation of this vector shows

 R that = 9.5 units at 57° above the + x2axis . !

Topic 1

1.59

Your sketch should be drawn to scale, and be similar to that pictured  below. The length of !R and the angle θ can be measured to find,

with use of your scale factor, the magnitude and direction of the resultant displacement. The result should be approximately 421 ft 3° below the horizontal.

1.60

(a) The x-component is x = (24.0 m)cos(56.0°) = 13.4 m. (b) The y-component is y = (24.0 m)sin(56.0°) = 19.9 m.

1.61

(a) The vector’s magnitude is ! A = A = Ax2 + Ay2 =

( −5.00 m ) + ( 9.00 m ) 2

= 10.3 m

(b) The vector is in the second quadrant so its direction is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 1

⎛ Ay ⎞ ⎛ 9.00 m ⎞ θ = tan −1 ⎜ ⎟ + 180° = tan −1 ⎜ + 180° ⎝ −5.00 m ⎟⎠ ⎝ Ax ⎠ = 119°

1.62

The person undergoes a displacement ! A = 3.10 km at 25.0° north of east. Choose a coordinate system with

origin at the starting point, positive x-direction oriented eastward, and positive y-direction oriented northward. Then the components of her displacement are Ax = A cos θ = (3.10 km)cos 25.0° = 2.81 km eastward and Ay = A sin θ = (3.10 km)sin 25.0° = 1.31 km northward

1.63

 The x- and y-components of vector !A are its projections on lines

parallel to the x- and y-axes, respectively, as shown in the sketch. The magnitude of these components can be computed using the sine and cosine functions as shown below:

  Ax = A cos 325° = + A cos 35° = (35.0)cos 35° = 28.7 units ! and

  Ay = A sin 325° = − A sin 35° = −(35.0)sin 35° = −20.1 units !

Topic 1

1.64

(a)

 The skater’s displacement vector, !d , extends in a straight line from her starting point A to the end point B. When she has coasted half way around a circular path as shown in the sketch below, the displacement vector coincides with the diameter of the

 circle and has magnitude d = 2r = 2(5.00 m) = 10.0 m . !

(b)

The actual distance skated, s, is one half the circumference of the

1 circular path of radius r. Thus s = (2π r) = π (5.00 m) = 15.7 m . ! 2 (c)

When the skater skates all the way around the circular path, her end point, B, coincides with the start point, A. Thus, the

 displacement vector has zero length, or d = 0 . ! 1.65

(a)

Her net x (east-west) displacement is −3.00 + 0 + 6.00 = +3.00 blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is

Topic 1

R = (∑ x)2 + (∑ y)2 = (3.00)2 + (4.00)2 = 5.00 blocks ! and the angle the resultant makes with the x-axis (eastward direction) is

⎛ ∑y⎞ ⎛ 4.00 ⎞ θ = tan −1 ⎜ = tan −1 ⎜ = tan −1 (1.33) = 53.1° ⎟ ⎟ ⎝ 3.00 ⎠ ⎝ ∑x⎠ ! The resultant displacement is then 5.00 blocks at 53.1° N of E. (b) 1.66

The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks.

  Let !A be the vector corresponding to the 10.0 yd run, !B to the 15.0 yd

 run, and !C to the 50.0 yd pass. Also, we choose a coordinate system with the + y-direction downfield, and the + x-direction toward the sideline to which the player runs. The components of the vectors are then Ax = 0

Ay = −10.0 yds

Bx = 15.0 yds

By = 0

Cx = 0

Cy = +50.0 yds

From these, Rx = ∑x = 15.0 yds, and Ry = ∑y = 40.0 yds, and 1.67

R = Rx2 + Ry2 = (15.0 yds)2 + (40.0 yds)2 = 42.7 yards !

Ax = −25.0

Ay = 40.0

Topic 1

A = Ax2 + Ay2 = (−25.0)2 + (40.0)2 = 47.2 units ! From the triangle, we find that ⎛ Ay ⎞ ⎛ 40.0 ⎞ φ = tan −1 ⎜ = tan −1 ⎜ = 58.0° , so θ = 180° − φ = 122° ⎟ ⎝ 25.0 ⎟⎠ ⎝ Ax ⎠ !  Thus, A = 47.2 units at!!122°!!counterclockwise!!from!!the!!+x8axis !

1.68

Choose the positive x-direction to be eastward and positive y as northward. Then, the components of the resultant displacement from Dallas to Chicago are Rx = ∑x = (730 mi)cos 5.00° − (560 mi)sin 21.0° = 527 mi and Ry = ∑y = (730 mi)sin 5.00° + (560 mi)cos 21.0° = 586 mi

R = Rx2 + Ry2 = (527 mi)2 + (586 mi)2 = 788 mi ! ⎛ ∑y⎞ −1 θ = tan −1 ⎜ ⎟⎠ = tan (1.11) = 48.0° ⎝ ∑ x ! Thus, the displacement from Dallas to Chicago is

Topic 1

1.69

After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are:

Displacement

x-component (eastward)

y-component (northward)

123 km

−61.5 km

+107 km

37.5 km

+37.5 km

Resultant

−61.5 km

144 km

Therefore, the eye of the hurricane is now R = (−61.5 km)2 + (144 km)2 = 157 km!!from!!the!!island !

1.70

(a)

F1 = 120 N F1x = (120 N)cos 60.0° = 60.0 N F1y = (120 N)sin 60.0° = 104 N

Topic 1

F2 = 80.0 N F2x = −(80 N)cos 75.0° = −20.7 N F2y = (80.0 N)sin 75.0° = 77.3 N

F = (∑ Fx )2 + (∑ Fy )2 = (39.3 N)2 + (181 N)2 = 185 N !R −1 ⎛ 181 N ⎞ = tan −1 (4.61) = 77.8° and θ = tan ⎜ ⎝ 39.3 N ⎟⎠ !

 The resultant force is FR = 185 N at 77.8°!!from!!the!!x0axis ! (b) To have zero net force on the mule, the resultant above must be cancelled by a force equal in magnitude and oppositely directed. Thus, the required force is 185 N at 258° from the x-axis 1.71

   The components of the displacements !a , !b , and !c are

ax = a⋅cos 30.0° = +152 km bx = b⋅cos 110° = −51.3 km cx = c⋅cos 180° = −190 km and ay = a⋅sin 30.0° = +87.5 km by = b⋅sin 110° = +141 km cy = c⋅sin 180° = 0 Thus, Rx = ax + bx + cx = −89.3 km, and Ry = ay + by + cy = 229 km

Topic 1

R = Rx2 + Ry2 = 246 km , and !

(

)

θ = tan −1 |Rx |/Ry = tan −1 (0.390) = 21.3° ! City C is 246 km at 21.3° W of N from the starting point.

1.72

(a) 1

mi ⎛ mi ⎞ ⎛ 1.609 km ⎞ km = ⎜1 ⎟ ⎜ = 1.609 h ⎝ h ⎠ ⎝ 1 mi ⎟⎠ h

(b) vmax = 55

(c)

1.73

mi ⎛ mi ⎞ ⎛ 1.609 km/h ⎞ km = ⎜ 55 ⎟ ⎜ = 88 ⎟ ⎝ ⎠ h h ⎝ 1 mi / h ⎠ h

Δvmax = 65

mi mi ⎛ mi ⎞ ⎛ 1.609 km/h ⎞ km − 55 = ⎜ 10 ⎟ ⎜ = 16 ⎟ ⎝ ⎠ h h h ⎝ 1 mi/h ⎠ h

The term s has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the equation, s = kam tn with k being dimensionless, has dimensions of L = (LT−2)m (T)n

L1T0 = Lm Tn−2m

The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 Likewise, equating powers of T, we see that n − 2m = 0, or n = 2m = 2

Topic 1

Dimensional analysis cannot determine the value of k, a dimensionless constant. 1.74

(a) The rate of filling in gallons per second is

rate =

30.0 gal ⎛ 1 min ⎞ = 7.14 × 10 −2 gal/s 7.00 min ⎜⎝ 60 s ⎟⎠

1L ⎞ 3 2 3 6 3 ⎛ = 10 3 L . Thus, (b) Note that 1 m = (10 cm) = (10 cm ) ⎜ 3 3⎟ 10 cm ⎝ ⎠

rate = 7.14 × 10 −2

(c)

1.75

gal ⎛ 3.786 L ⎞ ⎛ 1 m 3 ⎞ = 2.70 × 10 −4 m 3 /s s ⎜⎝ 1 gal ⎟⎠ ⎜⎝ 10 3 L ⎟⎠

⎛ 1h ⎞ Vfilled 1.00 m 3 = = 3.70 × 10 3 s ⎜ = 1.03 h −4 3 rate 2.70 × 10 m /s ⎝ 3600 s ⎟⎠

The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus,

V 3.79 × 10 −3 m 3 h= = = 1.52 × 10 −4 m = 152 × 10 −6 m = 152 µm 2 A 25.0 m 1.76

(a) For a sphere, A = 4πR2. In this case, the radius of the second sphere is twice that of the first, or R2 = 2R1.

Hence,

A2 4π R22 R22 (2 R1 )2 = = = = 4 A1 4π R12 R12 R12

(b) For a sphere, the volume is V =

4 π R3 3

Topic 1

Thus,

1.77

V2 (4 / 3)π R23 R23 (2 R1 )3 = = = = 8 V1 (4 / 3)π R13 R13 R13

The estimate of the total distance cars are driven each year is d = (cars in use)(distance traveled per car) = (100 × 106 cars)(104 mi/car) = 1 × 1012 mi At a rate of 20 mi/gal, the fuel used per year would be

d 1× 1012 mi V1 = = = 5 × 1010 gal rate1 20 mi/gal If the rate increased to 25 mi/gal, the annual fuel consumption would be

V2 =

d 1× 1012 mi = = 4 × 1010 gal rate2 25 mi/gal

and the fuel savings each year would be savings = V1 − V2 = 5 × 1010 gal − 4 × 1010 gal = 1 × 1010 gal 1.79

(a)

The Sun’s radius is about 6.96 × 105 km and Earth’s radius is about 6.38 × 103 km. The number of Earths that could fit inside the Sun, NES, is approximately equal to the ratio of the Sun’s volume to Earth’s volume:

N ES =

4 VSun π R3 = 43 3Sun ≈ 1 × 106 VEarth 3 π REarth

Topic 1

The Sun’s radius is about 102 times Earth’s radius, so its volume (which scales as radius cubed) is about 106 times Earth’s volume. Approximately a million Earths could fit inside the Sun. (b)

Similarly, the Moon’s radius is about 1.74 × 103 km so that NME, the number of Moons that could fit inside the Earth, is

N ME =

3 VEarth 43 π REarth = 4 3 ≈ 50 VMoon 3 π RMoon

Approximately 50 Moons could fit inside the Earth. 1.80

Assume a sneeze lasts about 3 seconds. One day equals 8.64 × 104 s which can be divided into 8.64/3 × 104 = 2.88 × 104 3-second intervals. If each of Earth’s 7.3 billion people sneezes 3 times per day, then 3(7.3 billion) = 21.9 billion sneezes occur each day. The number of sneezes occurring during any particular 3-second interval is then (rounded to 1 significant figure):

21.9 × 109 sneezes/day ≈ 1 × 106 sneezes/(3-sec interval) 4 2.88 × 10 (3-sec intervals)/day During any given 3-second interval, approximately one million people from around the world will sneeze. 1.81

The volume of the Milky Way galaxy is roughly

⎛ πd2 ⎞ π VG = At = ⎜ t = (10 21 m)2 (1019 m) or VG ~ 10 61 m 3 ⎟ 4 ⎝ 4 ⎠

Topic 1

If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius r = 3 × 1018 m, then the galactic volume per neutron star is

4 4 V1 = π r 3 = π (3 × 1018 m)3 = 1× 10 56 m 3 or V1 ~ 10 56 m 3 3 3 The order of magnitude of the number of neutron stars in the Milky Way is then

VG 10 61 m 3 ~ 56 3 or n  10 5 neutron stars V1 10 m

Topic 2

Topic 2 Motion in One Dimension

QUICK QUIZZES 2.1

(a) 200 yd

(b)

(c)

2.2

(a) False. The car may be slowing down, so that the direction of its acceleration is opposite the direction of its velocity. (b) True. If the velocity is in the direction chosen as negative, a positive acceleration causes a decrease in speed. (c) True. For an accelerating particle to stop at all, the velocity and acceleration must have opposite signs, so that the speed is decreasing. If this is the case, the particle will eventually come to rest. If the acceleration remains constant, however, the particle must begin to move again, opposite to the direction of its original velocity. If the particle comes to rest and then stays at rest, the acceleration has become zero at the moment the motion stops. This is the case for a braking car—the acceleration is negative and goes to zero as the car comes to rest.

Topic 2

2.3

The velocity-vs-time graph (a) has a constant slope, indicating a constant acceleration, which is represented by the acceleration-vs.-time graph (e). Graph (b) represents an object whose speed always increases, and does so at an ever-increasing rate. Thus, the acceleration must be increasing, and the acceleration-vs-time graph that best indicates this behaviour is (d). Graph (c) depicts an object which first has a velocity that increases at a constant rate, which means that the object’s acceleration is constant. The motion then changes to one at constant speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this situation is graph (f).

2.4

Choice (b). According to graph b, there are some instants in time when the object is simultaneously at two different x-coordinates. This is physically impossible.

2.5

(a) The blue graph of Figure 2.14b best shows the puck’s position as a function of time. As seen in Figure 2.14a, the distance the puck has traveled grows at an increasing rate for approximately three time intervals, grows at a steady rate for about four time intervals, and then grows at a diminishing rate for the last two intervals. (b) The red graph of Figure 2.14c best illustrates the speed (distance

Topic 2

traveled per time interval) of the puck as a function of time. It shows the puck gaining speed for approximately three time intervals, moving at constant speed for about four time intervals, then slowing to rest during the last two intervals. (c) The green graph of Figure 2.14d best shows the puck’s acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant velocity (zero acceleration) for about four time intervals, and then loses velocity (negative acceleration) for roughly the last two time intervals. 2.6

Choice (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceleration is the free-fall acceleration, g = 9.80 m/s2.

2.7

Choice (c). As it travels upward, its speed decreases by 9.80 m/s during each second of its motion. When it reaches the peak of its motion, its speed becomes zero. As the ball moves downward, its speed increases by 9.80 m/s each second.

2.8

Choices (a) and (f). The first jumper will always be moving with a higher velocity than the second. Thus, in a given time interval, the first jumper covers more distance than the second, and the separation distance

Topic 2

between them increases. At any given instant of time, the velocities of the jumpers are definitely different, because one had a head start. In a time interval after this instant, however, each jumper increases his or her velocity by the same amount, because they have the same acceleration. Thus, the difference in velocities stays the same.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.2

Yes. The particle may stop at some instant, but still have an acceleration, as when a ball thrown straight up reaches its maximum height.

2.4

(a) No. They can be used only when the acceleration is constant. (b) Yes. Zero is a constant.

2.6

(a) In Figure (c), the images are farther apart for each successive time interval. The object is moving toward the right and speeding up. This means that the acceleration is positive in Figure (c). (b) In Figure (a), the first four images show an increasing distance traveled each time interval and therefore a positive acceleration. However, after the fourth image, the spacing is decreasing, showing that the object is now slowing down (or has negative acceleration).

Topic 2

(c) In Figure (b), the images are equally spaced, showing that the object moved the same distance in each time interval. Hence, the velocity is constant in Figure (b). 2.8

(a) At the maximum height, the ball is momentarily at rest (i.e., has zero velocity). The acceleration remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the downward direction. (b) The acceleration of the ball remains constant in magnitude and direction throughout the ball’s free flight, from the instant it leaves the hand until the instant just before it strikes the ground. The acceleration is directed downward and has a magnitude equal to the freefall acceleration g.

2.10

Once the ball has left the thrower's hand, it is a freely falling body with a constant, nonzero, acceleration of a = −g. Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) through (d) are all false and the correct response is (e).

2.12

The initial velocity of the car is v0 = 0 and the velocity at the time t is v. The constant acceleration is therefore given by

Topic 2

Δv v − v0 v − 0 v = = = Δt t t t

and the average velocity of the car is

( v + v0 ) = ( v + 0 ) = v 2

The distance traveled in time t is Δx = vt = vt/2. In the special case where a = 0 (and hence v = v0 = 0), we see that statements (a), (b), (c), and (d) are all correct. However, in the general case (a ≠ 0, and hence v ≠ 0) only statements (b) and (c) are true. Statement (e) is not true in either case.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.2

(a)

2 × 104 mi

(b)

Δx/2RE = 2.4

2.4

(a)

8.33 yards/s

(b)

2.78 yards/s

2.6

(a)

5.00 m/s

(b)

1.25 m/s

(d)

−3.33 m/s

(e)

(a)

+4.0 m/s

(b)

−0.50 m/s

(d)

(a)

2.3 min

(b)

64 mi

2.8

2.10

(c)

−2.50 m/s

(c)

−1.0 m/s

Topic 2

2.12

(a)

L/t1

(b)

−L/t2

(d)

2L/(t1 + t2)

2.14

(a)

2.16 2.18

(c)

1.3 × 102 s

(b)

13 m

(a)

37.1 m/s

(b)

1.30 × 10−5 m

(a)

Some data points that can be used to plot the graph are as given below:

x (m)

5.75

16.0

35.3

68.0

119

192

t (s)

1.00

2.00

3.00

4.00

5.00

6.00

(b)

41.0 m/s, 41.0 m/s, 41.0 m/s

(c)

17.0 m/s, much smaller than the instantaneous velocity at t = 4.00 s

2.20

(a)

2.00 m/s, 5.0 m/s (b)

2.22

0.391 s

2.24

(i)

(a)

(b)

1.6 m/s2

(c)

0.80 m/s2

(ii)

(a)

(b)

1.6 m/s2

(c)

2.26

263 m

The curves intersect at t = 16.9 s.

Topic 2

2.28

a = 2.74 × 105 m/s2 = (2.79 × 104)g

2.30

(a)

(c)

a = (v 2f − vi2 )/ ( 2 Δx)

(b)

13.5 m (c)

(b)

v 2f = vi2 + 2a(Δx)

(e)

8.00 s

(a)

13.5 m

(d)

22.5 m

(a)

20.0 s

(b)

No, it cannot land safely on the 0.800 km runway.

2.36

(a)

5.51 km

(b)

20.8 m/s, 41.5 m/s, 20.8 m/s, 38.7 m/s

2.38

(a)

107 m

(b)

1.49 m/s2

2.40

(a)

v = a1t1

(b)

Δx =

(c)

Δxtotal =

1 2 1 a1t1 + a1t1t 2 + a2t 22 2 2

2.32

2.34

(d)

1.25 m/s2

13.5 m

1 2 a1t1 2

Topic 2

2.42

8.9 months

2.44

29.1 s

2.46

1.79 s

2.48

(a)

Yes.

(c)

 Δv downward = 2.39 m/s

(d)

! No, Δv upward = 3.71 m/s. The two rocks have the same acceleration,

(b)

vtop = 3.69 m/s

but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. 2.50

(a)

21.1 m/s

(b)

19.6 m

2.52

(a)

v = | −v0 − gt| = |v0 + gt|

(c)

v = v0 − gt , d =

2.54

(a)

29.4 m/s

(b)

44.1 m

2.56

(a)

−202 m/s2

(b)

198 m

2.58

(a)

4.53 s

(b)

14.1 m/s

2.60

8.4 m/s

2.62

See Solutions Section for Motion Diagrams.

(c)

1.81 m/s, 19.6 m

(b)

1 2 gt 2

Topic 2

2.64

(a)

v = v02 + 2gh

(b)

Δt = 2v0/g

2.66

(a)

2.45 × 10−2 m

(b)

4.67 × 10−2 s

2.68

(a)

3.00 s

(b)

v0,2 = −15.2 m/s

(c)

v1 = −31.4 m/s, v2 = −34.8 m/s

(a)

2.2 s

2.70

(b)

−21 m/s

PROBLEM SOLUTIONS 2.1

We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then

Δt =

2.2

Δx 2m = = 2 × 10 −2 s = 0.02 s v 100 m/s

(a) At constant speed, c = 3 × 108 m/s, the distance light travels in 0.1 s is Δx = c(Δt) = (3 × 10 8 m/s)(0.1 s) ⎛ 1 mi ⎞ ⎛ 1 km ⎞ = (3 × 10 7 m ) ⎜ = 2 × 10 4 mi ⎝ 1.609 km ⎟⎠ ⎜⎝ 10 3 m ⎟⎠

(b) Comparing the result of part (a) to the diameter of the Earth, DE, we find

Δx Δx 3.0 × 10 7 m = = ≈ 2.4 DE 2RE 2(6.38 × 10 6 m )

(with RE = Earth's radius)

Topic 2

2.3

Distances traveled between pairs of cities are Δx1 = v1(Δt1) = (80.0 km/h)(0.500 h) = 40.0 km Δx2 = v2(Δt2) = (100.0 km/h)(0.200 h) = 20.0 km Δx3 = v3(Δt3) = (40.0 km/h)(0.750 h) = 30.0 km Thus, the total distance traveled is Δx = (40.0 + 20.0 + 30.0) km = 90.0 km, and the elapsed time is Δt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.

(a)

Δx 90.0 km = = 52.9 km/h Δt 1.70 h

(b) Δx = 90.0 km (see above) 2.4

(a) The player runs 100 yards from his own goal line to the opposing team’s goal line. Then he runs an additional 50 yards back to the fiftyyard line, all in 18.0 s. Substitute values into the definition of average speed to find

Average speed =

path length 100 yards + 50 yards = elapsed time 18.0 s

= 8.33 yards/s (b) After returning to the fifty-yard line, the player’s displacement is Δx = xf − xi = 50.0 yards − 0 yards = 50.0 yards. Substitute values into the definition of average velocity to find © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Δx 50.0 yards = Δt 18.0 s

= 2.78 yards/s 2.5

(a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h. Boat B requires 2.0 h to cross the lake at which time the race is over. Boat A wins, being 60 km ahead of B when the race ends. (b) Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero.

2.6

The average velocity over any time interval is

Δx x f − xi = Δt t f − ti

Δx 10.0 m − 0 = = 5.00 m/s Δt 2.00 s − 0

(b) v =

Δx 5.00 m − 0 = = 1.25 m/s Δt 4.00 s − 0

(a)

Δx 5.00 m − 10.0 m = = −2.50 m/s Δt 4.00 s − 2.00 s

(d) v =

Δx −5.00 m − 5.00 m = = −3.33 m/s Δt 7.00 s − 4.00 s

(c)

Topic 2

(e)

2.7

Δx x2 − x1 0−0 = = = 0 Δt t 2 − t1 8.00s − 0

(a) ⎛ 1h ⎞ Displacement = Δx = (8.50 km/h)(35.0 min) ⎜ + 130 km = 180 km ⎝ 60.0 min ⎟⎠

(b) The total elapsed time is ⎛ 1h ⎞ Δt = (35.0 min + 15.0 min) ⎜ + 2.00 h = 2.83 h ⎝ 60.0 min ⎟⎠

so,

2.8

Δx 180 km = = 63.6 km Δt 2.84 h

The average velocity over any time interval is

(a)

Δx x f − xi = Δt t f − ti

Δx 4.0 m − 0 = = +4.0 m / s Δt 1.0 s − 0

Topic 2

(b) v =

Δx −2.0 m − 0 = = −0.50 m / s Δt 4.0 s − 0

Δx 0 − 4.0 m = = −1.0 m / s Δt 5.0 s − 1.0 s

(c)

(d) v =

2.9

Δx 0−0 = = 0 Δt 5.0 s − 0

The plane starts from rest (v0 = 0) and maintains a constant acceleration of a = +1.3 m/s2. Thus, we find the distance it will travel before reaching the required takeoff speed (v = 75 m/s), from v 2 = v02 + 2a(Δx) , as

Δx =

v 2 − v02 (75 m/s)2 = = 2.2 × 10 3 m = 2.2 km 2 2a 2(1.3 m/s )

Since this distance is less than the length of the runway, the plane takes off safely.

2.10

(a) The time for a car to make the trip is t =

Δx . Thus, the difference in v

the times for the two cars to complete the same 10 mile trip is

Topic 2

Δt = t1 − t 2 =

Δx Δx ⎛ 10 mi 10 mi ⎞ ⎛ 60 min ⎞ − =⎜ − = 2.3 min v1 v2 ⎝ 55 mi/h 70 mi/h ⎟⎠ ⎜⎝ 1 h ⎟⎠

(b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15.0 min. This distance is given by Δx1 = v1(Δt) = (55 mi/h)(15 min). The faster car pulls ahead of the slower car at a rate of vrelative = 70 mi/h − 55 mi/h = 15 mi/h Thus, the time required for it to get distance Δx1 ahead is

Δt =

Δx1 (55 mi/h)(15 min) = = 15 min vrelative 15.0 mi/h

Finally, the distance the faster car has traveled during this time is ⎛ 1h ⎞ Δx2 = v2 (Δt) = (70 mi/h)(55 min) ⎜ = 64 mi ⎝ 60 min ⎟⎠

2.11

(a) From v f + vi + 2a(Δx) , with vi = 0, vf = 72 km/h, and Δx = 45 m, the 2

acceleration of the cheetah is found to be

⎡⎛ km ⎞ ⎛ 10 3 m ⎞ ⎛ 1 h ⎞ ⎤ 72 ⎢ ⎥−0 h ⎟⎠ ⎜⎝ 1 km ⎟⎠ ⎜⎝ 3600 s ⎟⎠ ⎦ v 2f − vi2 ⎣⎜⎝ a= = = 4.4 m/s 2 2(Δx) 2(45 m) (b) The cheetah’s displacement 3.5 s after starting from rest is

Topic 2

1 1 Δx = vi t + at 2 = 0 + (4.4 m/s 2 )(3.5 s) = 27 m 2 2

2.12

v1 =

(Δx)1 +L = = +L /t1 (Δt)1 t1

(b) v2 =

(Δx)2 −L = = −L /t 2 (Δt)2 t2

(a)

(c)

vtotal =

(Δx)total (Δx)1 + (Δx)2 +L − L 0 = = = = 0 (Δt)total t1 + t 2 t1 + t 2 t1 + t 2

(d)

(ave speed)trip =

2.13

+L + −L totaldistance traveled (Δx)1 + (Δx)2 2L = = = (Δt)total t1 + t 2 t1 + t 2 t1 + t 2

(a) The total time for the trip is ttotal = t1 + 22.0 min = t1 + 0.367 h, where t1 is the time spent traveling at v1 = 89.5 km/h. Thus, the distance traveled is Δx = v1t1 = vt total , which gives (89.5 km/h)t1 = (77.8 km/h)(t1 + 0.367 h) = (77.8 km/h)t1 + 28.5 km or, (89.5 km/h − 77.8 km/h)t1 = 28.5 km From which, t1 = 2.44 h for a total time of ttotal = t1 + 0.367 h = 2.81 h (b) The distance traveled during the trip is Δx = v1t1 = vt total , giving Δx = vt total = (77.8 km/h)(2.81 h) = 219 km

Topic 2

2.14

(a) At the end of the race, the tortoise has been moving for time t and the hare for a time t − 2.0 min = t − 120 s. The speed of the tortoise is vi = 0.100 m/s, and the speed of the hare is vh = 20 vt = 2.0 m/s. The tortoise travels distance xt, which is 0.20 m larger than the distance xh traveled by the hare. Hence, xt = xh + 0.20 m which becomes or

vtt = vh(t − 120 s) + 0.20 m

(0.100 m/s)t = (2.0 m/s)(t − 120 s) + 0.20 m

This gives the time of the race as t = 1.30 × 102 s (b) xt = vtt = (0.100 m/s)(1.3 × 102 s) = 13 m 2.15

The maximum allowed time to complete the trip is

t total =

total distance 1600 m ⎛ 1 km/h ⎞ = = 23.0 s required averagespeed 250 km/h ⎜⎝ 0.278 m/s ⎟⎠

The time spent in the first half of the trip is

t1 =

half distance 800 m ⎛ 1 km/h ⎞ = = 12.5 s v1 230 km/h ⎜⎝ 0.278 m/s ⎟⎠

Thus, the maximum time that can be spent on the second half of the trip is

Topic 2

t2 = ttotal − t1 = 23.0 s − 12.5 s = 10.5 s and the required average speed on the second half is

v2 =

2.16

⎛ 1 km/h ⎞ half distance 800 m = = 76.2 m/s ⎜ = 274 km/h t2 10.5 s ⎝ 0.278 m/s ⎟⎠

(a) From the first kinematic equation with v0 = 0, t = 7.00 × 10-7 s, and a = 5.30 × 107 m/s2, the maximum speed is

v = v0 + at = (5.30 × 107 m/s2 )(7.00 × 10−7 s) v = 37.1 m/s (b) Use Δx = v0t + 12 at with v0 = 0 to find the distance travelled during 2

the acceleration:

(

)(

Δx = 12 at 2 = 12 5.30 × 107 m/s2 7.00 × 10−7 s

)

Δx = 1.30 × 10−5 m 2.17

The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We compute this slope by using two points on a straight segment of the curve, one point on each side of the point of interest.

(a)

vt=1.00 s =

10.0 m − 0 = 5.00 m/s 2.00 s − 0

Topic 2

(b) vt=3.00 s =

(c)

vt=4.50 s =

(d) vt=7.50 s =

2.18

(5.00 − 10.0) m = −2.50 m/s (4.00 − 2.00) s (5.00 − 5.00) m = 0 (5.00 − 4.00) s 0 − (−5.00) m = 5.00 m/s (8.00 − 7.00) s

(a) A few typical values are t (s)

x (m)

1.00

5.75

2.00

16.0

3.00

35.3

4.00

68.0

5.00

119

6.00

192

Topic 2

(b) We will use a 0.400 s interval centered at t = 400 s. We find at t = 3.80 s, x = 60.2 m and at t = 4.20 s, x = 76.6 m. Therefore,

Δx 16.4 m = = 41.0 m/s Δt 0.400 s

Using a time interval of 0.200 s, we find the corresponding values to be: at t = 3.90 s, x = 64.0 m and at t = 4.10 s, x = 72.2 m. Thus,

Δx 8.20 m = = 41.0 m/s Δt 0.200 s

For a time interval of 0.100 s, the values are: at t = 3.95 s, x = 66.0 m, and at t = 4.05 s, x = 70.1 m. Therefore,

Δx 4.10 m = = 41.0 m/s Δt 0.100 s

Δx 68.0 m − 0 = = 17.0 m/s Δt 4.00 s − 0

Topic 2

This value is much less than the instantaneous velocity at t = 4.00 s. 2.19

Choose a coordinate axis with the origin at the flagpole and east as the 1 2 positive direction. Then, using x = x0 + v0t at with a = 0 for each runner, 2

the x-coordinate of each runner at time t is xA = −4.0 mi + (6.0 mi/h)t and xB = 3.0 mi + (−5.0 mi/h)t When the runners meet, xA = xB giving −4.0 mi + (60 mi/h)t = 3.0 mi + (−5.0 mi/h)t or

(6.0 mi/h + 5.0 mi/h)t = 3.0 mi + 4.0 mi. This gives the elapsed time

when they meet as t = (7.0 mi)/(11.0 mi/h) = 0.64 h. At this time, xA = xB = −0.18 mi. Thus, they meet 0.18 mi west of the flagpole. 2.20

From the figure below, observe that the motion of this particle can be broken into three distinct time intervals, during each of which the particle has a constant acceleration. These intervals and the associated accelerations are 0 ≤ t ≤ 10.0 s, a = a1 = +2.00 m/s2 10 ≤ t ≤ 15.0 s, a = a2 = 0

Topic 2

and

15.0 ≤ t ≤ 20.0 s, a = a3 = −3.00 m/s2

(a) Applying vf = vi + a(Δt) to each of the three time intervals gives for 0 ≤ t ≤ 10.0 s, v10 + v0 + a1(Δt1) = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s for 10.0 s ≤ t ≤ 15.0 s, v15 + v10 + a2(Δt2) = 20.0 m/s + 0 = 20.0 m/s for 15.0 s ≤ t ≤ 20.0 s, v20 + v15 + a3(Δt3) = 20.0 m/s + (−3.00 m/s2)(5.00 s) = 5.00 m/s 1 2 (b) Applying Δx = vi (Δt) + a(Δt) to each of the time intervals gives 2

for 0 ≤ t ≤ 10.0 s, 1 1 Δx1 = v0 Δt1 + a1 (Δt1 )2 = 0 + (2.00 m/s 2 )(10.0 s)2 = 1.00 × 10 2 m 2 2

for 10.0 s ≤ t ≤ 15.0 s,

Topic 2

1 Δx2 = v10 Δt 2 + a2 (Δt 2 )2 = (20.0 m/s)(5.00 s) + 0 = 1.00 × 10 2 m 2

for 15.0 s ≤ t ≤ 20.0 s,

1 Δx = v15 Δt 3 + a3 (Δt 3 )2 2 1 = (20.0 m/s)(5.00 s) + (−3.00 m/s 2 )(5.00 s)2 = 62.5 m 2 Thus, the total distance traveled in the first 20.0 s is Δxtotal = Δx1 + Δx2 + Δx3 = 100 m + 100 m + 62.5 m = 263 m 2.21

We choose the positive direction to point away from the wall. Then, the initial velocity of the ball is vi = −25.0 m/s and the final velocity is vf = +22.0 m/s. If this change in velocity occurs over a time interval of Δt = 3.50 ms (i.e., the interval during which the ball is in contact with the wall), the average acceleration is

2.22

Δv v f − vi +22.0 m/s − (−25.0 m/s) = = = 1.34 × 10 4 m/s 2 −3 Δt Δt 3.50 × 10 s

From a = Δv/Δt, the required time is

Δt =

2.23

From a =

⎞ ⎛ 0.447 m/s ⎞ Δv ⎛ 60.0 mi/h − 0 ⎞ ⎛ 1g =⎜ = = 0.391 s 2⎟ ⎟ ⎜ ⎠ ⎝ 9.80 m/s ⎠ ⎜⎝ 1 mi/h ⎟⎠ a ⎝ 7g

Δv Δv (60 − 55) mi/h ⎛ 0.447 m/s ⎞ = = 3.7 s , we have Δt = a 0.60 m/s 2 ⎜⎝ 1 mi/h ⎟⎠ Δt

Topic 2

2.24

(i)

(a)

From t = 0 to t = 5.0 s,

v f − vi −8.0 m/s − (−8.0 m/s) = = 0 t f − ti 5.0 s − 0

(b) From t = 5.0 s to t = 15 s,

8.0 m/s − (−8.0 m/s) = 1.6 m/s 2 15 s − 5.0 s

8.0 m/s − (−8.0 m/s) = 0.80 m/s 2 20 s − 0

(ii) At any instant, the instantaneous acceleration equals the slope of the line tangent to the v vs. t graph at that point in time. (a) At t = 2.0 s, the slope of the tangent line to the curve is 0. (b) At t = 10 s, the slope of the tangent line is 1.6 m/s2 . (c) At t = 18 s, the slope of the tangent line is 0.

2.25

(a) a =

Δv 175 mi/h − 0 = = 70.0 mi/h ⋅s Δt 2.5 s

Topic 2

⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ a = ⎜ 70.0 = = 31.3 m/s 2 h ⋅s ⎟⎠ ⎜⎝ 1 mi ⎟⎠ ⎜⎝ 3600 s ⎟⎠ ⎝

Alternatively,

m⎞ 1g ⎛ a = ⎜ 31.3 2 ⎟ = = 3.19g ⎝ s ⎠ 9.80 m/s 2

1 2 (b) If the acceleration is constant, Δx = v0t + at 2

1⎛ m⎞ Δx = 0 + ⎜ 3.13 2 ⎟ (2.50 s)2 = 97.8 m 2⎝ s ⎠

2.26

⎛ 3.281 ft ⎞ Δx = (97.8 m) ⎜ = 321 ft ⎝ 1 m ⎟⎠

As in the algebraic solution to Example 2.5, we let t represent the time the trooper has been moving.

We graph xcar = 24.0 m + (24.0 m/s)t and xtrooper = (1.50 m/s2)t2

Topic 2

The curves intersect at t = 16.9 s

2.27

1 2 Apply Δx = v0 + at to the 2.00-second time interval during which the 2

object moves from xi = 3.00 cm to xf = −5.00 cm. With v0 = 12.0 cm/s, this yields an acceleration of 2 ⎡(x f − xi ) − v0t ⎤⎦ 2[(−5.00 − 3.00) cm − (12.0 cm/s)(2.00 s)] a= ⎣ = t2 (2.00 s)2

or 2.28

a = −16.0 cm/s2

From v 2 = v02 + 2a(Δx) , we have (10.97 × 103 m/s)2 = 0 + 2a(220 m) so that

v 2 − v02 (10.97 × 10 3 m/s)2 − 0 = = 2.74 × 10 5 m/s 2 2(Δx) 2(220 m)

⎛ ⎞ 1g = (2.74 × 10 5 m/s 2 ) ⎜ = 2.79 × 10 4 times g! 2⎟ ⎝ 9.80 m/s ⎠ 2.29

(a)

Δx = v(Δt) = [(v + v0 ) / 2]Δt becomes

⎛ 2.80 m/s + v0 ⎞ 40.0 m= ⎜ ⎟⎠ (8.50 s) ⎝ 2

which yields

(b) a =

v0 =

2 (40.0 m) − 2.80 m/s = 6.61 m/s 8.50 s

v − v0 2.80 m/s − 6.61 m/s = = −0.448 m/s 2 Δt 8.50 s

Topic 2

2.30

(a)

(b) The known quantities are initial velocity, final velocity, and displacement. The kinematics equation that relates these quantities to acceleration is v f = vi + 2a(Δx) 2

(c)

v 2f − vi2 a= 2(Δx)

(d) a =

v 2f − vi2 (30.0 m/s)2 − (20.0 m/s)2 = = 1.25 m/s 2 2(Δx) 2(2.00 × 10 2 m)

(e) Using a = Δv/Δt, we find that

Δt =

2.31

Δv v f − vi 30.0 m/s − 20.0 m/s = = = 8.00 s a a 1.25 m/s 2

(a) With v = 120 km/h, v 2 = v02 + 2a(Δx) yields v 2 − v02 ⎡⎣(120 km/h) − 0 ⎤⎦ ⎛ 0.278 m/s ⎞ a= = = 2.32 m/s 2 ⎜ ⎟ 2(Δx) 2(240 m) ⎝ 1 km/h ⎠ 2

(b) The required time is

Δt =

2.32

v − v0 (120 km/h − 0) ⎛ 0.278 m/s ⎞ = = 14.4 s a 2.32 m/s 2 ⎜⎝ 1 km/h ⎟⎠

(a) From v f = vi + 2a(Δx) , with vi = 6.00 m/s and vf = 12.0 m/s, we find 2

Topic 2

Δx =

v 2f − vi2 (120 m/s)2 − (6.00 m/s)2 = = 13.5 m 2a 2(4.00 m/s 2 )

(b) In this case, the object moves in the same direction for the entire time interval and the total distance traveled is simply the magnitude or absolute value of the displacement. That is, d = |Δx| = 13.5 m (c) Here, vi = −6.00 m/s and vf = 12.0 m/s, and we find

v 2f − vi2 Δx = = 13.5m 2a

[the same as in part (a)]

(d) In this case, the object initially slows down as it travels in the negative x-direction, stops momentarily, and then gains speed as it begins traveling in the positive x-direction. We find the total distance traveled by first finding the displacement during each phase of this motion. While coming to rest (vi = −6.00 m/s, vf = 0), v 2f − vi2 (0)2 − (−6.00 m/s)2 Δx1 = = = −4.50 m 2a 2(4.00 m/s2 )

After reversing direction (vi = 0 m/s, vf = 12.0 m/s), v 2f − vi2 (12.0 m/s)2 − (0)2 Δx2 = = = 18.0 m 2a 2(4.00 m/s 2 ) © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Note that the net displacement is Δx = Δx1 + Δx2 = −4.50 m + 18.0 m = 13.5 m, as found in part (c) above. However, the total distance traveled in this case is d = |Δx1| + |Δx2| = |−4.50 m| + |18.0 m| = 22.5 m

2.33

v − v0 24.0 m/s 2 − 0 = = 8.14 m / s 2 (a) a = Δt 2.95 s (b) From a = Δv/Δt, the required time is

Δt =

v f − vi 20.0 m/s − 10.0 m/s = = 1.23 s a 8.15 m/s 2

(c) Yes. For uniform acceleration, the change in velocity Δv generated in time Δt is given by Δv = a(Δt). From this, it is seen that doubling the length of the time interval Δt will always double the change in velocity Δv. A more precise way of stating this is: “When acceleration is constant, velocity is a linear function of time.” 2.34

(a) The time required to stop the plane is

v − v0 0 − 100 m/s = = 20.0 s a −5.00 m/s 2

(b) The minimum distance needed to stop is

⎛ v + v0 ⎞ ⎛ 0 + 100 m/s ⎞ Δx = v = ⎜ t=⎜ ⎟ ⎟⎠ (20.0 s) = 1 000 m = 1.00 km ⎝ 2 ⎠ ⎝ 2 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Thus, the plane requires a minimum runway length of 1.00 km. It cannot land safely on a 0.800 km runway. 2.35

We choose x = 0 and t = 0 at the location of Sue’s car when she first spots the van and applies the brakes. Then, the initial conditions for Sue’s car are x0S = 0 and v0S = 30.0 m/s. Her constant acceleration for aS = −2.00 m/s2. The initial conditions for the van are x0V = 155 m, v0V = 5.00 m/s, and its 1 2 constant acceleration is aV = 0. We then use Δx = x − x0 = v0t + at to write 2

an equation for the x-coordinate of each vehicle for t ≥ 0. This gives Sue’s Car: 1 xS − 0 = (30.0 m/s)t + (−2.00 m/s 2 )t 2 or xS = (30.0 m/s)t − (−1.00 m/s 2 )t 2 2 xV − 155 m = (5.00 m/s)t +

Van:

1 (0)t 2 or xV = 155 m + (5.00 m/s)t 2

In order for a collision to occur, the two vehicles must be at the same location (i.e., xS = xV). Thus, we test for a collision by equating the two equations for the x-coordinates and see if the resulting equation has any real solutions. xS = xV

⇒

(30.0 m/s)t − (1.00 m/s2)t2 = 155 m + (5.00 m/s)t

(1.00 m/s2)t2 − (25.00 m/s) + 155 m = 0

Topic 2

Using the quadratic formula yields −(−25.00 m/s) ± (−25.00 m/s 2 ) − 4(1.00 m/s 2 )(155 m) t= = 13.6s or 11.4 s 2(1.00 m/s 2 )

The solutions are real, not imaginary, so a collision will occur. The smaller of the two solutions is the collision time. (The larger solution tells when the van would pull ahead of the car again if the vehicles could pass harmlessly through each other.) The x-coordinate where the collision occurs is given by xcollision = xS|t= 11.4 s = xV|t= 11.4 s = 155 m + (5.00 m/s)(11.4 s) = 212 m 2.36

The velocity at the end of the first interval is v = v0 + at = 0 + (2.77 m/s2)(15.0 s) = 41.6 m/s This is also the constant velocity during the second interval and the initial velocity for the third interval. Also, note that the duration of the second interval is t2 = (2.05 min)(60.0 s/1 min) = 123 s. 1 2 (a) From Δx = v0t + at , the total displacement is 2

(Δx)total = (Δx)1 + (Δx)2 + (Δx)3 1 ⎡ ⎤ − ⎢ 0 + (2.77 m/s 2 )(15.0 s)2 ⎥ + [(2.77 m/s 2 )(15.0 s)(123 s) + 0] 2 ⎣ ⎦ 1 ⎡ ⎤ + ⎢(2.77 m/s 2 )(15.0 s)(4.39 s) + (−9.47 m/s 2 )(4.39 s)2 ⎥ 2 ⎣ ⎦

Topic 2

(Δx)total = 312 m + 5.11 × 103 m + 91.2 m = 5.51 × 103 m = 5.51 km

(b) v1 =

(Δx)1 312 m = = 20.8 m/s t1 15.0 s

v2 =

(Δx)2 5.11× 10 3 m = = 41.5 m/s t2 123 s

v3 =

(Δx)3 91.2 m = = 20.8 m/s , and the average velocity for the t3 4.39 s

(Δx)total 5.51× 10 3 m = = 38.7 m/s total trip is vtotal = t total (15.0 + 123 + 4.39) s

2.37

1 2 Using the uniformly accelerated motion equation Δx + v0t + at for the 2 1 2 2 full 40 s interval yields Δx = (20 m/s)(40 s) + (−1.0 m/s )(40 s) = 0 , which 2

is obviously wrong. The source of the error is found by computing the time required for the train to come to rest. This time is

v − v0 0 − 20 m/s = = 20 s. a −1.0 m/s 2

Thus, the train is slowing down for the first 20 s and is at rest for the last 20 s of the 40 s interval. The acceleration is not constant during the full 40 s. It is, however, constant during the first 20 s as the train slows to rest. Application of

Topic 2

1 Δx = v0t + at 2 to this interval gives the stopping distance as 2 1 Δx = (20 m/s)(20 s) + (−1.0 m/s 2 )(20 s)2 = 200 m 2

2.38

mi ⎞ ⎛ 0.447 m/s ⎞ ⎛ v0 = 0 and v f = ⎜ 400 ⎟ ⎜ = 17.9 m/s ⎝ h ⎠ ⎝ 1 mi/h ⎟⎠

(a) To find the distance traveled, we use

⎛ v + v0 ⎞ ⎛ 17.9 m/s + 0 ⎞ Δx = vt = ⎜ f t=⎜ ⎟⎠ (12.0 s) = 107 m ⎝ 2 ⎟⎠ ⎝ 2

(b) The constant acceleration is

2.39

v f − v0 17.9 m/s − 0 = = 1.49 m/s 2 t 12.0 s

At the end of the acceleration period, the velocity is v = v0 + ataccel = 0 + (1.5 m/s2)(5.0 s) = 7.5 m/s This is also the initial velocity for the braking period. (a) After braking,

vf = v + atbrake = 7.5 m/s + (−2.0 m/s2)(3.0 s) = 1.5 m/s

(b) The total distance traveled is

⎛ v + v⎞ ⎛ v + v0 ⎞ Δxtotal = (Δx)accel + (Δx)brake = (vt)accel + (vt)brake = ⎜ t accel + ⎜ f t ⎟ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ brake ⎛ 7.5 m/s +0 ⎞ ⎛ 1.5 m/s +7.5 m/s ⎞ Δx total = ⎜ (5.0 s) + ⎜ ⎟ ⎟⎠ (3.0 s) = 32 m ⎝ ⎠ ⎝ 2 2

Topic 2

2.40

For the acceleration period, the parameters for the car are: initial velocity = via = 0, acceleration = aa = a1, elapsed time = (Δt)a = t1, and final velocity = vfa. For the braking period, the parameters are: initial velocity = vib = final velocity of acceleration period = vfa, acceleration = ab = a2, and elapsed time = (Δt)b = t2. (a) To determine the velocity of the car just before the brakes are engaged, we apply vf = vi + a(Δt) to the acceleration period and find vib = vfa = via + aa(Δt)a = 0 + a1t1

vib = a1t1

1 2 (b) We may use Δx = vi (Δt) + a(Δt) to determine the distance traveled 2

during the acceleration period (i.e., before the driver begins to brake). This gives 1 1 (Δx)a = via (Δt)a + aa (Δt)2a = 0 + a1t12 2 2

(Δx)a =

1 2 a1t1 2

Thus, the total displacement of the car during the two intervals combined is

Topic 2

(Δx)total = (Δx)a + (Δx)b =

2.41

1 2 1 a1t1 + a1t1t 2 + a2t 22 2 2

The time the Thunderbird spends slowing down is

Δt1 =

Δx1 2(Δx1 ) 2(250 m) = = = 6.99 s v1 v + v0 0 + 71.5 m/s

The time required to regain speed after the pit stop is

Δt 2 =

Δx2 2(Δx2 ) 2(350 m) = = = 9.79 s. v2 v + v0 71.5 m/s + 0

Thus, the total elapsed time before the Thunderbird is back up to speed is Δt = Δt1 + 5.00 s + Δt2 = 6.99 s + 5.00 s + 9.79 s = 2.18 s During this time, the Mercedes has traveled (at constant speed) a distance ΔxM = v0(Δt) = (71.5 m/s)(21.8 s) = 1 559 m and the Thunderbird has fallen behind a distance d = ΔxM − Δx1 − Δx2 = 1 559 m − 250 m − 350 m = 959 m 2.42

The initial bank account balance is xi = $1.0 × 104 (to two significant figures) and the bank account is empty when xf = 0 with a change of Δx = xf − xi = $(−1.0 × 104). Use Δx = v0t + 12 at 2

Topic 2

with v0 = 0 and a = −2.5 × 102 $/month to find the time, t:

Δx = 12 at 2 → t =

2Δx a

(

)

2.43

2 $(−1.0 × 10 4 )

( −2.5 × 10 $/month ) 2

= 8.9 months

(a) Take t = 0 at the time when the player starts to chase his opponent. At this time, the opponent is distance d = (12 m/s)(30 s) = 36 m in front of the player. At time t > 0, the displacements of the players from their initial positions are 1 1 Δxplayer = (v0 )player t + aplayer t 2 = 0 + (4.0 m/s 2 )t 2 2 2 1 Δxopponent = (v0 )opponent t + aopponent t 2 = (12 m/s)t + 0 2

and

When the players are side-by-side, Δxplayer = Δxopponent + 36 m

[1]

[2] [3]

Substituting Equations [1] and [2] into Equation [3] gives 1 (4.0 m/s 2 )t 2 = (12 m/s)t + 36 m 2

t2 + (−6.0 s)t + (−18 s2) = 0

Applying the quadratic formula to this result gives

−(−6.0 s) ± (−6.0 s)2 − 4(1)(−18 s 2 ) 2(1)

Topic 2

which has solutions of t = −2.2 s and t = +8.2 s. Since the time must be greater than zero, we must choose t = 8.2 s as the proper answer. 1 1 2 2 2 2 (b) Δxplayer =(v0 )player t + aplayer t = 0 + (4.0 m/s )(8.2 s) = 1.3 × 10 m 2 2

2.44

The initial velocity of the train is v0 = 82.4 and the final velocity is v = 16.4 km/h. The time required for the 400 m train to pass the crossing is found from Δx = vt = [(v + v0 )/2]t as

2.45

⎛ 3600 s ⎞ 2(Δx) 2(0.400 km) = = (8.10 × 10 −3 h) ⎜ = 29.1 s v + v0 (82.4 + 16.4) km/h ⎝ 1 h ⎟⎠

(a) From v 2 = v02 + 2a(Δy) with v = 0, we have

v 2 − v02 0 − (25.0 m/s)2 (Δy)max = = = 31.9 m 2a 2(−9.80 m/s 2 ) (b) The time to reach the highest point is

v 2 − v02 0 − 25.0 m/s t up = = = 2.55 s a −9.80 m/s 2 (c) The time required for the ball to fall 31.9 m, starting from rest, is found from 1 Δy = (0)t + at 2 as t = 2

2(Δy) = a

2(−31.9 m) = 2.55 s −9.80 m/s 2

(d) The velocity of the ball when it returns to the original level (2.55 s

Topic 2

after it starts to fall from rest) is v = v0 + at = 0 + (−9.80 m/s2)(2.55 s) = −25.0 m/s 2.46

We take upward as the positive y-direction and y = 0 at the point where the ball is released. Then, v0y = −8.00 m/s, ay = −g = −9.80 m/s2, and Δy = −30.0 m when the ball reaches the ground. From v y = v0 y + 2ay (Δy) , the 2

velocity of the ball just before it hits the ground is

v y = v02 y + 2ay (Δy) = − (8.00 m/s)2 + 2(−9.80 m/s 2 )(−30.0 m) = −25.5 m/s Then, vy = v0y + ayt gives the elapsed time as

2.47

v y − v0 y −25.5 m/s − (−8.00 m/s) = = 1.79 s ay −9.80 m/s 2

(a) The velocity of the object when it was 30.0 m above the ground can be 1 2 determined by applying Δy = v0t + at to the last 1.50 s of the fall. 2

This gives

1⎛ m⎞ −30.0 m = v0 (1.50 s) + ⎜ −9.80 2 ⎟ (1.50 s)2 2⎝ s ⎠

v0 = −12.7 m/s

(b) The displacement the object must have undergone, starting from rest, to achieve this velocity at a point 30.0 m above the ground is given by

v 2 = v02 + 2a(Δy) as

Topic 2

(Δy)1 =

v 2 − v02 (−12.7 m/s)2 − 0 = = −8.23 m 2a 2(−9.80 m/s 2 )

The total distance the object drops during the fall is then |(Δy)total|=|(−8.23 m) + (−30.0 m)| = 38.2 m 2.48

(a) Consider the rock’s entire upward flight, for which v0 = +7.40 m/s, vf = 0, a = −g = −9.80 m/s2, yi = 1.55 m, (taking y = 0 at ground level), and yf = hmax = maximum altitude reached. Then applying v 2f = vi2 + 2a(Δy) to this upward flight gives

0 = (7.40 m/s)2 + 2(−9.80 m/s2)(hmax − 1.55 m) Solving for the maximum altitude of the rock gives

hmax = 1.55 m +

(7.40 m/s)2 = 4.34 m 2(9.80 m/s 2 )

Since hmax > 3.65 m (height of the wall), the rock does reach the top of the wall. (b) To find the velocity of the rock when it reaches the top of the wall, we use v f = vi + 2a(Δy) and solve for vf when yf = 3.65 m (starting with vi 2

= +7.40 m/s at yi = 1.55 m). This yields

v f = vi2 + 2a(y f − yi ) = (7.40 m/s)2 + 2(−9.80 m/s 2 )(3.65 m − 1.55 m) = 3.69 m/s © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

(c) A rock thrown downward at a speed of 7.40 m/s (vi = −7.40 m/s) from the top of the wall undergoes a displacement of (Δy) = yf − yi = 1.55 m − 3.65 m = −2.10 m before reaching the level of the attacker. Its velocity when it reaches the attacker is

v f = − vi2 + 2a(Δy) = − (−7.40 m/s)2 + 2(−9.80 m/s 2 )(−2.10 m) = −9.79 m/s so the change in speed of this rock as it goes between the 2 points located at the top of the wall and the attacker is given by Δ(speed)down = ||vf| − |vi|| = ||−9.79 m/s| − |−7.40 m/s|| = 2.39 m/s. (d) Observe that the change in speed of the ball thrown upward as it went from the attacker to the top of the wall was Δ(speed)up = ||vf| − |vi||=|3.69 m/s − 7.40 m/s| = 3.71 m/s The two rocks do not undergo the same magnitude speed change. The rocks have the same acceleration, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. 2.49

The velocity of the child’s head just before impact (after falling a distance of 0.40 m, starting from rest) is given by v 2 = v02 + 2a(Δy) as v I = v02 + 2a(Δy) = − 0 + 2(−9.8 m/s 2 )(−0.40 m) = −2.8 m/s

Topic 2

If, upon impact, the child’s head undergoes an additional displacement Δy = −h before coming to rest, the acceleration during the impact can be found from v 2 = v02 + 2a(Δy) to be a = (0 − v I2 )/2(−h) = v I2 /2h . The duration of the impact is found from v = v0 + at as

t = Δv/a = −v I /(v I2 /2h), or t = −2h/v I . Applying these results to the two cases yields Hardwood Floor (h = 2.0 × 10 −3 m) : a =

and

vI2 (−2.8 m/s)2 = = 2.0 × 10 3 m/s 2 2h 2(2.0 × 10 −3 m)

−2h −2(2.0 × 10 −3 m) = = 1.4 × 10 −3 s = 1.4 ms vI −2.8 m/s

Carpeted Floor (h = 1.0 × 10

and

2.50

−2

m): a =

vI2 (−2.8 m/s)2 = = 3.9 × 10 2 m/s 2 −2 2h 2(1.0 × 10 m)

−2h −2(1.0 × 10 −2 m) = = 7.1× 10 −3 s = 7.1 ms vI −2.8 m/s

(a) After 2.00 s, the velocity of the mailbag is vbag = v0 + at = −1.50 m/s + (−9.80 m/s2)(2.00 s) = −21.1 m/s The negative sign tells us that the bag is moving downward and the magnitude of the velocity gives the speed as 21.1 m/s. (b) The displacement of the mailbag after 2.00 s is

Topic 2

⎛ v + v0 ⎞ ⎡ −21.1 m/s + ( − 1.50 m/s) ⎤ (Δy)bag = ⎜ t=⎢ ⎟ ⎥⎦ (2.00 s) = − 22.6 m ⎝ 2 ⎠ 2 ⎣ During this time, the helicopter, moving downward with constant velocity, undergoes a displacement of 1 (Δy)copter = v0t + at 2 = (−1.5 m/s)(2.00 s) + 0 = −3.00 m 2

The distance separating the package and the helicopter at this time is then d = |(Δy)P − (Δy)h| = |−22.6 m − (−3.00 m)| = |−19.6 m| = 19.6 m (c) Here, (v0)bag = (v0)copter = +1.50 m/s and abag = −9.80 m/s2 while acopter = 0. After 2.00 s, the velocity of the mailbag is

vbag = 1.50

m ⎛ m⎞ m m + ⎜ −9.80 2 ⎟ (2.00 s) = −18.1 and its speed is vbag = 18.1 s ⎝ s ⎠ s s

In this case, the displacement of the helicopter during the 2.00 s interval is Δycopier = (+1.50 m/s)(2.00 s) + 0 = +3.00 m Meanwhile, the mailbag has a displacement of

⎛ v + v0 ⎞ ⎡ −18.1 m/s + 1.50 m/s ⎤ (Δy)bag = ⎜ bag t=⎢ ⎟ ⎥⎦ (2.00 s) = −16.6 m ⎝ 2 ⎠ 2 ⎣

Topic 2

The distance separating the package and the helicopter at this time is then d = |(Δy)p − (Δy)h| = |−16.6 m − (+3.00 m)| = |−19.6 m| = 19.6 m 2.51

(a) From the instant the ball leaves the player’s hand until it is caught, the ball is a freely falling body with an acceleration of a = −g = −9.80 m/s2 = 9.80 m/s2 = 9.80 m/s2 (downward) (b) At its maximum height, the ball comes to rest momentarily and then = 0 . begins to fall back downward. Thus, vmax height 1 2 (c) Consider the relation Δy = v0t + at with a = −g. When the ball is at 2

the thrower’s hand, the displacement is Δy = 0, giving 1 0 = v0t − gt 2 2

This equation has two solutions, t = 0, which corresponds to when the ball was thrown, and t = 2v0/g corresponding to when the ball is caught. Therefore, if the ball is caught at t = 2.00 s, the initial velocity must have been

v0 =

gt (9.80 m/s 2 )(2.00 s) = = 9.80 m/s 2 2

Topic 2

(d) From v 2 = v02 + 2a(Δy) , with v = 0 at the maximum height,

(Δy)max =

2.52

v 2 − v02 0 − (9.80 m/s 2 ) = = 4.90 m 2a 2(−9.80 m/s 2 )

(a) Let t = 0 be the instant the package leaves the helicopter, so the package and the helicopter have a common initial velocity of vi = −v0 (choosing upward as positive). At times t > 0, the velocity of the package (in free-fall with constant acceleration ap = −g) is given by v = vi + at as vp = −v0 − gt = −(v0 + gt) and speed = |vp| = v0 + gt. (b) After an elapsed time t, the downward displacement of the package from its point of release will be

1 1 1 ⎞ ⎛ (Δy)P = vi t + aP t 2 = −v0t − gt 2 = − ⎜ v0t + gt 2 ⎟ ⎝ 2 2 2 ⎠ and the downward displacement of the helicopter (moving with constant velocity, or acceleration ah = 0) from the release point at this time is 1 (Δy)h = vi t + aht 2 = −v0t + 0 = −v0t 2

The distance separating the package and the helicopter at this time is then © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

1 1 ⎛ ⎞ d = (Δy) p − (Δy)h = − ⎜ v0t + gt 2 ⎟ − (−v0t) = gt 2 ⎝ ⎠ 2 2

(c) If the helicopter and package are moving upward at the instant of release, then the common initial velocity is vi = +v0. The accelerations of the helicopter (moving with constant velocity) and the package (a freely falling object) remain unchanged from the previous case (ap = −g and ah = 0). In this case, the package speed at time t > 0 is |vP| = |vi + apt| = |v0 − gt| . At this time, the displacements from the release point of the package and the helicopter are given by 1 1 (Δy) p = vi t + a pt 2 = v0t − gt 2 2 2

and

1 (Δy)h = vi t + aht 2 = v0t + 0 = +v0t 2

The distance separating the package and helicopter at time t is now given by 1 1 d = (Δy) p − (Δy)h = v0t − gt 2 − v0t = gt 2 2 2

2.53

(the same as earlier!)

(a) After its engines stop, the rocket is a freely falling body. It continues

Topic 2

upward, slowing under the influence of gravity until it comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls. (b) When it reaches a height of 150 m, the speed of the rocket is v = v02 + 2a(Δy) = (50.0 m/s)2 + 2(2.00 m/s 2 )(150 m) = 55.7 m/s

After the engines stop, the rocket continues moving upward with an initial velocity of v0 = 55.7 m/s and acceleration a = −g = −9.80 m/s2. When the rocket reaches maximum height, v = 0. The displacement of the rocket above the point where the engines stopped (that is, above the 150 m level) is

Δy =

v 2 − v02 0 − (55.7 m/s)2 = = 158 m 2a 2(−9.80 m/s 2 )

The maximum height above ground that the rocket reaches is then given by hmax = 150 m + 158 m = 308 m (c) The total time of the upward motion of the rocket is the sum of two intervals. The first is the time for the rocket to go from v0 = 50.0 m/s at the ground to a velocity of v = 55.7 m/s at an altitude of 150 m. This time is given by

Topic 2

t1 =

(Δy)1 (Δy)1 2(150 m) = = = 2.84 s v1 (v + v0 ) / 2 (55.7 + 50.0) m/s

The second interval is the time to rise 158 m starting with v0 = 55.7 m/s and ending with v = 0. This time is

t2 =

(Δy)2 (Δy)2 2(158 m) = = = 5.67 s v2 (v + v0 ) / 2 0 + 55.7 m/s

The total time of the upward flight is then tap = t1 + t2 = (2.84 + 5.67) s = 8.51 s (d) The time for the rocket to fall 308 m back to the ground, with v0 = 0 1 2 and acceleration a = −g = −9.80 m/s2, is found from Δy = v0t + at as 2

t down =

2(Δy) = a

2(−308 m) = 7.93 s −9.80 m/s 2

so the total time of the flight is tflight = tup + tdown = (8.51 + 7.93) s = 16.4 s. 2.54

(a) For the upward flight of the ball, we have vi = v0, vf = 0, a = −g, and Δt = 3.00 s. Thus, vf = vi + a(Δt) gives the initial velocity as vi = vf − a(Δt) = vf + g(Δt) or v0 = 0 + (9.80 m/s2)(3.00 s) = +29.4 m/s (b) The vertical displacement of the ball during this 3.00-s upward flight

Topic 2

⎛ v + vf ⎞ ⎛ 29.4 m/s + 0 ⎞ Δymax = h = v(Δt) = ⎜ i (Δt) = ⎜ ⎟⎠ (3.00 s) = 44.1 m ⎟ ⎝ ⎝ 2 ⎠ 2 2.55

During the 0.600 s required for the rig to pass completely onto the bridge, the front bumper of the tractor moves a distance equal to the length of the rig at constant velocity of v = 100 km/h. Therefore the length of the rig is ⎡ km ⎛ 0.278 m/s ⎞ ⎤ Lrig = vt = ⎢100 ⎜⎝ 1 km/h ⎟⎠ ⎥ (0.600 s) = 16.7 m h ⎣ ⎦

While some part of the rig is on the bridge, the front bumper moves a distance Δx = Lbridge + Lrig = 100 m + 16.7 m. With a constant velocity of v = 100 km/h, the time for this to occur is

2.56

Lbridge + Lrig 400 m + 16.7 m ⎛ 1 km/h ⎞ = ⎜⎝ 0.278 m/s ⎟⎠ = 15.0 s v 100 km/h

(a) The acceleration experienced as he came to rest is given by v = v0 + at as

v − v0 = t

mi ⎞ ⎛ 0.447 m/s ⎞ ⎛ 0 − ⎜ 632 ⎟ ⎜ ⎝ h ⎠ ⎝ 1 mi/h ⎟⎠ 1.40 s

= −202 m/s 2

(b) The distance traveled while stopping is found from

Topic 2

⎡ ⎛ mi ⎞ ⎛ 0.447 m/s ⎞ ⎤ ⎢ 0 + ⎜⎝ 632 ⎟⎠ ⎜ ⎥ h ⎝ 1 mi/h ⎟⎠ ⎦ ⎛ v + v0 ⎞ ⎣ Δx = vt = ⎜ t= (1.40 s) = 198 m ⎝ 2 ⎟⎠ 2 2.57

(a) The acceleration of the bullet is

v 2 − v02 (300 m/s)2 − (400 m/s)2 = = −3.50 × 10 5 m/s 2 2(Δx) 2(0.100 m)

(b) The time of contact with the board is

2.58

v − v0 (300 − 400) m/s = = 2.86 × 10 −4 s 5 2 a −3.50 × 10 m/s

1 2 (a) From Δx = v0t + at , we have 2 1 100 m = (30.0 m/s)t + (−3.50 m/s 2 )t 2 2

This reduces to 3.50t2 + (−60.0 s)t + (200 s2) = 0, and the quadratic formula gives −(−60.0 s) ± (−60.0 s)2 − 4(3.50)(200 s 2 ) t= 2(3.50)

The desired time is the smaller solution of t = 4.53 s. The larger solution of t = 12.6 s is the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration. (b) The velocity of the boat when it first reaches the buoy is

Topic 2

v = v0 + at = 30.0 m/s + (−350 m/s2)(4.53 s) = 14.1 m/s. 2.59

(a) The keys have acceleration a = −g = −9.80 m/s2 from the release point 1 2 until they are caught 1.50 s later. Thus, Δy = v0t + at gives 2

Δy − at 2 /2 (+4.00 m) − (−9.80 m/s 2 )(1.50 s)2 /2 v0 = = = +10.0 m/s t 1.50 s or

v0 = 10.0 m/s upward

(b) The velocity of the keys just before the catch was v = v0 + at = 10.0 m/s + (−9.80 m/s2)(1.50 s) = −4.70 m/s or 2.60

v = 4.70 m/s downward

(a) While in the air after launching itself from the water, the salmon’s vertical acceleration is ay = −g = −9.80 m/s2. Assume it comes to rest at the top of its vertical leap, a distance Δy = 3.60 m above the bottom of the waterfall. From the time-independent kinematic equation, with the final velocity v = 0, the initial speed v0 is v 2 = v02 − 2g ( Δy )

(

)

v0 = 2g ( Δy ) = 2 9.80 m/s2 ( 3.6 m ) v0 = 8.4 m/s

Topic 2

2.61

(a) From v 2 = v02 + 2a(Δy) , the insect’s velocity after straightening its legs is

v = v02 + 2a(Δy) = 0 + 2(4 000 m/s 2 )(2.0 × 10 −3 m) = 4.0 m/s (b) The time to reach this velocity is

v − v0 4.0 m/s − 0 = = 1.0 × 10 −3 s = 1.0 ms a 4 000 m/s 2

(c) The upward displacement of the insect between when its feet leave the ground and it comes to rest momentarily at maximum altitude is

Δy =

2.62

v 2 − v02 0 − v02 −(4.0 m/s)2 = = = 0.82 m 2a 2(−g) 2(−9.8 m/s 2 )

(a)

For constant speed:

(b)

When speeding up at a constant rate:

Topic 2

(c)

2.63

When slowing down at a constant rate:

The falling ball moves a distance of (15 m − h) before they meet, where h 1 2 is the height above the ground where they meet. Apply Δy = v0t + at , 2

with a = −g, to obtain 1 −(15 m − h) = 0 − gt 2 2

1 h =15 m − gt 2 2

[1]

1 2 Applying Δy = v0t + at to the rising ball gives 2 1 h = (25 m/s)t − gt 2 2

[2]

Topic 2

Combining Equations [1] and [2] gives

(25 m/s)t −

2.64

1 2 1 gt = 15 m − gt 2 2 2

15 m = 0.60 s 25 m/s

(a) When the ball hits the ground, its change in height will be Δy = −h. Solve for the final speed of each ball using the time-independent kinematic equation: v 2 = v02 − 2g ( Δy ) v=

( ±v ) − 2g ( −h) = 2

v02 + 2gh

(b) We’re asked to find an expression for the time difference Δt between the times of flight for the upward- and downward-thrown balls.

For the upward-thrown ball, the path to the ground can be separated into two parts. In the first part the ball rises with initial velocity +v0 and falls back to its original height where its velocity is –v0. In the second part it moves from height h to the ground in exactly the same time it takes the downward-thrown ball to reach the ground. The difference between the two ball’s times of flight is therefore equal to

Topic 2

the time for the first part of the upward-thrown ball’s path. That time is found using the first kinematic equation with v0 = +v0 and v = −v0:

v = v0 − gΔt Δt =

2.65

v − v0 −v0 − v0 2v0 = = −g −g g

From the time-independent kinematic equation with v0 > 0, Δy = 2.00 m, and v = ±1.50 m/s: v 2 = v02 − 2g ( Δy ) v0 = v 2 + 2g ( Δy ) =

2.66

( ±1.50 m/s) + 2g ( 2.00 m ) = 6.44 m/s 2

(a) To find the distance Δx traveled by the blood during the acceleration, apply the time-independent kinematic equation with v0 = 0, v = 1.05 m/s, and a = 22.5 m/s2:

v 2 = v02 + 2aΔx v 2 − v02 (1.05 m/s) − 0 Δx = = = 2.45 × 10−2 m = 2.45 cm 2a 2 22.5 m/s2 2

(

)

(b) Solve for the time t required for the blood to reach its peak speed using the first kinematic equation:

Topic 2

v = v0 + at t=

2.67

v − v0 (1.05 m/s) − 0 = = 4.67 × 10−2 s a 22.5 m/s2

When released from rest (v0 = 0), the bill falls freely with a downward acceleration due to gravity (a = −g = −9.80 m/s2). Thus, the magnitude of its downward displacement during David’s 0.2 s reaction time will be

1 1 Δy = v0t + at 2 = 0 + (−9.80 m/s 2 )(0.2 s)2 = 0.2 m = 20 cm 2 2 This is over twice the distance from the center of the bill to its top edge (≈8 cm), so David will be unsuccessful. 2.68

(a) The velocity with which the first stone hits the water is 2

m⎞ m⎞ m ⎛ ⎛ v1 = − v + 2a(Δy) = − ⎜ −2.00 ⎟ + 2 ⎜ −9.80 2 ⎟ (−5.00 m) = −31.4 ⎝ ⎝ s⎠ s ⎠ s 2 01

The time for this stone to hit the water is

t1 =

v1 − v01 [−31.4 m/s − (−2.00 m/s)] = = 3.00 s a −9.80 m/s 2

(b) Since they hit simultaneously, the second stone, which is released 1.00 s later, will hit the water after an flight time of 2.00 s. Thus,

v02 =

Δy − at 22 /2 −50.0 m − (−9.80 m/s 2 )(2.00 s)2 /2 = = −15.2 m/s t2 2.00 s

Topic 2

(c) From part (a), the final velocity of the first stone is v1 = −31.4 m/s. The final velocity of the second stone is v2 = v02 + at2 = −15.2 m/s + (−9.80 m/s2)(2.00 s) = −34.8 m/s 2.69

When the hare wakes up, the tortoise is a distance L > 0 from the finish line whereas the hare is a distance L + d from the finish line. The hare, running with constant speed v1, reaches the finish line in a time given by thare =

L+d v1

The tortoise, crawling with constant speed v2 < v1, reaches the finish line in a time given by ttortoise =

L v2

The tortoise wins the race if ttortoise < thare, so it follows that L L+d < v2 v1

Rearrange this expression to find a condition on the length L: Lv1 < Lv2 + v2 d → L <

v2 d v1 − v2

The tortoise wins the race if the length to the finish line, L, satisfies that inequality.

Topic 2

2.70

1 2 (a) From Δy = v0t + at with v0 = 0, we have 2

2(Δy) = a

2(−23 m) = 2.2 s −9.80 m/s 2

(b) From the time-independent velocity equation, the final speed is: v 2 = v02 − 2gΔy

(

)

v = 2gΔy = 2 −9.80 m/s2 ( −23 m ) = 21 m/s

Because the man is falling, his final velocity is v = −21 m/s. (c) The time it takes for the sound of the impact to reach the spectator is

tsound =

Δy 23 m = = 6.8 × 10 −2 s vsound 340 m/s

so the total elapsed time is ttotal = 2.2 s + 6.8 × 10–2 s ≈ 2.3 s. 2.71

The time required for the stuntman to fall 3.00 m, starting from rest, is 1 2 found from Δy = v0t + at as 2 1 −3.00 m = 0 + (−9.80 m/s 2 )t 2 2

2(3.00 m) = 0.782 s 9.80 m/s 2

(a) With the horse moving with constant velocity of 10.0 m/s, the horizontal distance is Δx = vhorset = (10.0 m/s)(0.782 s) = 7.82 m © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

(b) The required time is t = 0.782 s as calculated above.

Topic 3

Topic 3 Motion in Two Dimensions

QUICK QUIZZES 3.1

Choice (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may have constant speed (magnitude of velocity) but still be accelerating due to a change in direction of the velocity. If an object is following a curved path, it is accelerating because the velocity is changing in direction.

3.2

Choice (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel produces accelerations by altering the direction of the velocity.

3.3

(a) Average speed is the path length traveled divided by the time interval. In this case, the path length equals πR, half the circumference of the circular path, with R = 10.0 m. Her average speed is (Average speed) = πR/Δt = π(10.0 m)/15.0 s = 2.09 m/s. (b) Her average velocity has magnitude

Topic 3

! Δr 2R 2 (10.0 m ) ! v av = = = = 1.33 m/s Δt Δt 15.0 s 3.4

Choice (c) is correct. A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward and come straight downward. To a stationary observer, the ball follows a parabolic trajectory, moving with the same horizontal velocity as the runner and staying above the runner’s hand.

3.5

Choice (b) is correct. The velocity is always tangential to the path while the acceleration is always directed vertically downward. Thus, the velocity and acceleration are perpendicular only where the path is horizontal. This only occurs at the peak of the path.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

3.2

(a)

3.4

(a)

(b)

The balls will be closest at the instant the second ball is projected.

(b) The first ball will always be going faster than the second ball. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 3

(c)

There will be a one second time interval between their collisions with the ground.

(d) The two move with the same acceleration in the vertical direction. Thus, changing their horizontal velocities can never make them hit at the same time. 3.6

The equations of projectile motion are only valid for objects moving freely under the influence of gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically downward. Of the objects listed, only a and d meet this requirement.

3.8

(a) The passenger sees the ball go into the air and come back in the same way he would if he were at rest on Earth. An observer by the tracks would see the ball follow the path of a projectile. (b) If the train were accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.

3.10

Statement (c) is true. At its highest point, the baseball’s velocity is horizontal and its acceleration is downward.

3.12

Choice (c) is correct. The dolphin is heading due north away from a boat that is moving east at speed v. Relative to a stationary observer, the dolphin is moving both east (at v) and north, so it’s velocity would be

Topic 3

greater than v in a direction north of east.

ANSWERS TO EVEN NUMBERED PROBLEMS 3.2

(a)

3.61 km at 33.7° N of E

(b) 1.44 km/h at 33.7° N of E

(a)

Δx = 0.300 m, Δy = 0.200 m

(b) vav,x = 0.060 0 m/s, vav,y = 0.040 0 m/s (c)

aav,x = 0.020 0 m/s2, aav,y = −0.020 0 m/s2

3.6

(a)

−1.33 m/s2

3.8

2.68 ft

3.10

(a)

76.0°

(b)

−8.00 m/s2

Rmax = 2.13R

3.14

(a)

28.2 m/s

(b)

4.07 s

(c)

v0 and t would increase

(a)

x = 0, y = y0

(b)

v0x = 7.52 m/s, v0y = −2.74 m/s

(c)

x = (7.52 m/s)t, y = y0 − (2.74 m/s)t − (4.90 m/s2)t2

Topic 3

100

(d)

22.6 m

(e)

y0 = 52.3 m

(b)

vCP,y = 0.82 m/s

(f) 1.18 s

3.16

x = 7.22 km, y = 1.69 km

3.18

18.6 m

3.20

(a)

vCP,x = 3.68 m/s

3.22

(a)

57.7 km/h at 60.0° west of vertical

(b)

28.9° km/h vertically downward

3.24

(a)

27.7° east of north

(b)

47.6 min

3.26

(a)

1.26 h

(b)

1.13 h

3.28

(a) 6.80 km

(b)

3.00 km vertically above the impact

point

(c)

66.2°

d tup = v − vs !

(b)

d tdown = v + vs !

3.30

(a)

(c)

1.19 h

See

Solution. (d)

2d tb = v !

(e)

ta > tb as long as vs > 0

3.32

(a)

twoman = L/v1

(b)

L tman = v1 + v2 !

3.34

(a)

hmax = 57.4 m, t = 1.22 s

(b)

Same as for part (a).

3.36

(a) !t = 2h/g

(b)

v = d g/2h !i