Física para Ciencias e Ingeniería. Volumen I - Douglas C. Giancoli - 4ta edición [Solucionario]

SOLUTION MANUAL FOR

http://www.elsolucionario.blogspot.com

LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA

VISITANOS PARA
DESARGALOS GRATIS.

CHAPTER 1: Introduction, Measurement, Estimating

Responses to Questions

1. (a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general
public; not invariable (could change size with age, time of day, etc.); not indestructible.

(b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have
different size feet); not invariable (could change size with age, time of day, etc.); not
indestructible.

Neither of these options would make a good standard.

2. The number of digits you present in your answer should represent the precision with which you
know a measurement; it says very little about the accuracy of the measurement. For example, if you
measure the length of a table to great precision, but with a measuring instrument that is not
calibrated correctly, you will not measure accurately.

3. The writers of the sign converted 3000 ft to meters without taking significant figures into account.
To be consistent, the elevation should be reported as 900 m.

4. The distance in miles is given to one significant figure and the distance in kilometers is given to five
significant figures! The figure in kilometers indicates more precision than really exists or than is
meaningful. The last digit represents a distance on the same order of magnitude as the car’s length!

5. If you are asked to measure a flower bed, and you report that it is “four,” you haven’t given enough
information for your answer to be useful. There is a large difference between a flower bed that is 4 m
long and one that is 4 ft long. Units are necessary to give meaning to the numerical answer.

6. Imagine the jar cut into slices each about the thickness of a marble. By looking through the bottom
of the jar, you can roughly count how many marbles are in one slice. Then estimate the height of the
jar in slices, or in marbles. By symmetry, we assume that all marbles are the same size and shape.
Therefore the total number of marbles in the jar will be the product of the number of marbles per
slice and the number of slices.

7. You should report a result of 8.32 cm. Your measurement had three significant figures. When you
multiply by 2, you are really multiplying by the integer 2, which is exact. The number of significant
figures is determined by your measurement.

8. The correct number of significant figures is three: sin 30.0º = 0.500.

9. You only need to measure the other ingredients to within 10% as well.

10. Useful assumptions include the population of the city, the fraction of people who own cars, the
average number of visits to a mechanic that each car makes in a year, the average number of weeks a
mechanic works in a year, and the average number of cars each mechanic can see in a week.
(a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of
these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a
year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be
seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a
week. The final estimate, then, is 800 car mechanics in San Francisco.
(b) Answers will vary.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

11. One common way is to observe Venus at a Sun Venus

time when a line drawn from Earth to Venus

is perpendicular to a line connecting Venus

to the Sun. Then Earth, Venus, and the Sun

are at the vertices of a right triangle, with

Venus at the 90º angle. (This configuration

will result in the greatest angular distance

between Venus and the Sun, as seen from Earth
Earth.) One can then measure the distance to

Venus, using radar, and measure the angular distance between Venus and the Sun. From this

information you can use trigonometry to calculate the length of the leg of the triangle that is the

distance from Earth to the Sun.

12. No. Length must be included as a base quantity.

Solutions to Problems

1. (a) 14 billion years = 1.4 ×1010 years

( ) ( )(b) 1.4 ×1010 y 3.156 ×107s 1 y = 4.4 ×1017s

2. (a) 214 3 significant figures

(b) 81.60 4 significant figures

(c) 7.03 3 significant figures

(d) 0.03 1 significant figure

(e) 0.0086 2 significant figures

(f) 3236 4 significant figures

(g) 8700 2 significant figures

3. (a) 1.156 = 1.156 ×100
(b) 21.8 = 2.18 ×101
(c) 0.0068 = 6.8×10−3
(d) 328.65 = 3.2865 ×102
(e) 0.219 = 2.19 ×10−1
(f) 444 = 4.44 ×102

4. (a) 8.69 ×104 = 86, 900
(b) 9.1×103 = 9,100
(c) 8.8 ×10−1 = 0.88

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2

Chapter 1 Introduction, Measurement, Estimating

(d) 4.76 ×102 = 476
(e) 3.62 ×10−5 = 0.0000362

5. % uncertainty = 0.25 m × 100% = 4.6%
5.48 m

6. (a) % uncertainty = 0.2 s ×100% = 4%
5s

(b) % uncertainty = 0.2 s ×100% = 0.4%
50 s

(c) % uncertainty = 0.2 s ×100% = 0.07%
300 s

7. To add values with significant figures, adjust all values to be added so that their exponents are all the
same.

( ) ( ) ( ) ( ) ( ) ( )9.2 ×103s + 8.3×104 s + 0.008×106 s = 9.2 ×103s + 83×103s + 8×103s

= (9.2 + 83 + 8) ×103 s = 100.2 ×103 s = 1.00 ×105 s

When adding, keep the least accurate value, and so keep to the “ones” place in the last set of
parentheses.

( ) ( )8. 2.079 ×102 m 0.082 ×10−1 = 1.7 m . When multiplying, the result should have as many digits as

the number with the least number of significant digits used in the calculation.

9. θ (radians) sin(θ ) tan(θ ) Keeping 2 significant figures in the angle, and
0 0.00 0.00 expressing the angle in radians, the largest angle that has
0.10 0.10 0.10
the same sine and tangent is 0.24 radians . In degrees,
0.12 0.12 0.12
the largest angle (keeping 2 significant figure) is 12°.
0.20 0.20 0.20 The spreadsheet used for this problem can be found on
0.24 0.24 0.24 the Media Manager, with filename
0.25 0.25 0.26 “PSE4_ISM_CH01.XLS,” on tab “Problem 1.9.”

10. To find the approximate uncertainty in the volume, calculate the volume for the minimum radius and
the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volume
is then half this variation in volume.

43 0.84 m 3 = 2.483 m3
3 specified
( )V = π r = πspecified 4
3

π r4 3 0.80 m 3 = 2.145 m3

3 min
( )Vmin= = 4 π
3

π r4 3 0.88 m 3 = 2.855 m3

3 max
( )Vmax= = 4 π
3

( )( )ΔV
= 1 Vmax − Vmin = 1 2.855 m3 − 2.145 m3 = 0.355 m3
2 2

The percent uncertainty is ΔV = 0.355 m3 ×100 = 14.3 ≈ 14 % .
Vspecified 2.483 m3

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

11. (a) 286.6 mm 286.6 ×10−3 m 0.286 6 m
(b) 85 μV 85 ×10−6 V 0.000 085 V
(c) 760 mg 760 ×10−6 kg 0.000 76 kg (if last zero is not significant)
(d) 60.0 ps 60.0 ×10−12 s 0.000 000 000 060 0 s
(e) 22.5 fm 22.5 ×10−15 m 0.000 000 000 000 022 5 m
(f) 2.50 gigavolts 2.5×109 volts 2, 500, 000, 000 volts

12. (a) 1×106 volts 1 megavolt = 1 Mvolt

(b) 2 ×10−6 meters 2 micrometers = 2μm
(c) 6 ×103 days 6 kilodays = 6 kdays
18 hectobucks = 18 hbucks or 1.8 kilobucks
(d) 18×102 bucks 80 nanoseconds = 80 ns
(e) 8 ×10−8seconds

13. Assuming a height of 5 feet 10 inches, then 5'10" = (70 in) (1 m 39.37 in) = 1.8 m . Assuming a
weight of 165 lbs, then (165 lbs) (0.456 kg 1 lb) = 75.2 kg . Technically, pounds and mass

measure two separate properties. To make this conversion, we have to assume that we are at a
location where the acceleration due to gravity is 9.80 m/s2.

( )14. (a) 93 million miles = 93×106 miles (1610 m 1 mile) = 1.5 ×1011 m

(b) 1.5 ×1011 m = 150 ×109 m = 150 gigameters or 1.5 ×1011 m = 0.15 ×1012 m = 0.15 terameters

( )15. 0.111 yd2 .
(a) 1 ft2 = 1 ft2 (1 yd 3 ft)2 = 0.111 yd2 , and so the conversion factor is 1 ft2

( )(b) 1 m2 = 1 m2 (3.28 ft 1 m)2 = 10.8 ft2 , and so the conversion factor is 10.8 ft2 .
1m2

16. Use the speed of the airplane to convert the travel distance into a time. d = vt , so t = d v .

t=d v = 1.00 km ⎛⎝⎜ 1 h ⎞⎟⎠ ⎛⎝⎜ 3600 s ⎟⎞⎠ = 3.8 s
950 km 1h

( )17. (a) 1.0 ×10−10 m = 1.0 ×10−10 m (39.37 in 1 m ) = 3.9 ×10−9 in

(b) (1.0 cm ) ⎛ 1m ⎞ ⎛ 1 atom ⎞ = 1.0 ×108 atoms
⎝⎜ 100 cm ⎟⎠ ⎝⎜ 1.0 ×10−10 m ⎠⎟

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4

Chapter 1 Introduction, Measurement, Estimating

18. To add values with significant figures, adjust all values to be added so that their units are all the
same.

1.80 m + 142.5 cm + 5.34 ×105 μm = 1.80 m + 1.425 m + 0.534 m = 3.759 m = 3.76 m

When adding, the final result is to be no more accurate than the least accurate number used. In this
case, that is the first measurement, which is accurate to the hundredths place when expressed in
meters.

19. (a) (1km h) ⎜⎛⎝ 0.621 mi ⎟⎞⎠ = 0.621 mi h , and so the conversion factor is 0.621mi h .
1 km 1km h

(b) (1 m s ) ⎛ 3.28 ft ⎞ = 3.28 ft s , and so the conversion factor is 3.28ft s .
⎝⎜ 1m ⎠⎟ 1m s

(c) (1 km h ) ⎛ 1000 m ⎞⎛ 1 h ⎞ = 0.278 m s, and so the conversion factor is 0.278 m s .
⎜⎝ 1 km ⎠⎟ ⎜⎝ 3600 s ⎟⎠ 1km h

20. One mile is 1.61×103 m . It is 110 m longer than a 1500-m race. The percentage difference is

calculated here.

110 m ×100% = 7.3%
1500 m

21. (a) Find the distance by multiplying the speed times the time.

( ) ( )1.00 ly = 2.998 × 108 m s 3.156 × 107 s = 9.462 × 1015 m ≈ 9.46 × 1015 m

(b) Do a unit conversion from ly to AU.

(1.00 ly ) ⎛ 9.462 ×1015 m ⎞ ⎛ 1 AU ⎞ = 6.31×104 AU
⎜ 1.00 ly ⎟ ⎝⎜ 1.50 ×1011m ⎠⎟
⎝ ⎠

( )(c) ⎛ 1 AU ⎞ ⎛ 3600 s ⎞
2.998×108 m s ⎝⎜ 1.50 ×1011 ⎟⎠ ⎝⎜ 1 hr ⎠⎟ = 7.20 AU h
m

( )22. 1char 1 min 1 hour 1day 1 year
82 ×109 bytes × 1 byte × 180 char × 60 min × 8 hour × 365.25 days = 2598 years ≈ 2600 years

23. The surface area of a sphere is found by A = 4π r2 = 4π (d 2)2 = π d 2.

3.48 ×106 m 2 = 3.80 ×1013 m2
( )(a)
AMoon = π D2 =π
Moon

AEarth 2 ⎛ DEarth ⎞2 ⎛ REarth ⎞2 ⎛ 6.38 ×106 m ⎞ 2
AMoon ⎜ DMoon ⎟ ⎜ RMoon ⎟ ⎝⎜ 1.74 ×106 m ⎠⎟
(b) π DEarth = ⎝ ⎠ = ⎝ ⎠ = = 13.4
=2

π DMoon

24. (a) 2800 = 2.8 ×103 ≈ 1×103 = 103

(b) 86.30 ×102 = 8.630 ×103 ≈ 10 ×103 = 104

(c) 0.0076 = 7.6 ×10−3 ≈ 10 ×10−3 = 10−2

(d) 15.0 ×108 = 1.5 ×109 ≈ 1×109 = 109

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

25. The textbook is approximately 25 cm deep and 5 cm wide. With books on both sides of a shelf, the

shelf would need to be about 50 cm deep. If the aisle is 1.5 meter wide, then about 1/4 of the floor
space is covered by shelving. The number of books on a single shelf level is then

( )1 ⎛ 1 book ⎞
⎜ ⎟
4 ⎝ m) (0.05 ⎠
3500 m2 (0.25 m ) = 7.0 × 104 books. With 8 shelves of books, the total number of

books stored is as follows.

⎛⎝⎜ 7.0 ×104 books ⎞⎟⎠ (8 shelves) ≈ 6 ×105 books
shelf level

26. The distance across the United States is about 3000 miles.

(3000 mi) (1 km 0.621 mi) (1 hr 10 km) ≈ 500 hr

Of course, it would take more time on the clock for the runner to run across the U.S. The runner
could obviously not run for 500 hours non-stop. If they could run for 5 hours a day, then it would
take about 100 days for them to cross the country.

27. A commonly accepted measure is that a person should drink eight 8-oz. glasses of water each day.
That is about 2 quarts, or 2 liters of water per day. Approximate the lifetime as 70 years.

(70 y) (365 d 1 y)(2 L 1 d) ≈ 5×104 L

28. An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide,
which is about 110 meters by 50 meters, or 5500 m2. The mower has a cutting width of 0.5 meters.
Thus the distance to be walked is as follows.

d = area = 5500 m2 = 11000 m = 11 km
width 0.5 m

At a speed of 1 km/hr, then it will take about 11 h to mow the field.

29. In estimating the number of dentists, the assumptions and estimates needed are:

the population of the city

the number of patients that a dentist sees in a day

the number of days that a dentist works in a year

the number of times that each person visits the dentist each year

We estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that

each person visits the dentist twice per year.

(a) For San Francisco, the population as of 2001 was about 1.7 million, so we estimate the

population at two million people. The number of dentists is found by the following calculation.

( )2 ×106 people ⎛ 2 visits ⎞ ⎛ 225 1 yr ⎞ ⎛ 1 dentist ⎞ ≈ 1800 dentists
⎜ year ⎟ ⎝⎜ workdays ⎠⎟ ⎜ 10 visits ⎟
⎜ ⎟ ⎜ ⎟
⎝⎜⎜ 1 person ⎟⎠⎟ ⎝⎜⎜ workday ⎟⎟⎠

(b) For Marion, Indiana, the population is about 50,000. The number of dentists is found by a

similar calculation to that in part (a), and would be 45 dentists . There are about 50 dentists

listed in the 2005 yellow pages.

30. Assume that the tires last for 5 years, and so there is a tread wearing of 0.2 cm/year. Assume the
average tire has a radius of 40 cm, and a width of 10 cm. Thus the volume of rubber that is
becoming pollution each year from one tire is the surface area of the tire, times the thickness per year

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6

Chapter 1 Introduction, Measurement, Estimating

that is wearing. Also assume that there are 1.5 ×108 automobiles in the country – approximately one
automobile for every two people. And there are 4 tires per automobile. The mass wear per year is

given by the following calculation.

⎛ mass ⎞ = ⎛ surface area ⎞ ⎛ thickness wear ⎞ (density of rubber ) ( # of tires)
⎝⎜ year ⎟⎠ ⎝⎜ tire ⎠⎟ ⎝⎜ year ⎟⎠

⎡ 2π (0.4 m) (0.1m) ⎤
⎢⎣ ⎥⎦
1 tire
( )( )=
( 0.002 m y) 1200 kg m3 6.0 ×108 tires = 4 ×108 kg y

31. Consider the diagram shown (not to scale). The balloon is a distance h above the dh
surface of the Earth, and the tangent line from the balloon height to the surface of r
the earth indicates the location of the horizon, a distance d away from the balloon.
Use the Pythagorean theorem. r

(r + h)2 = r2 + d 2 → r2 + 2rh + h2 = r2 + d 2

2rh + h2 = d 2 → d = 2rh + h2

( )d = 2 6.4 ×106 m (200 m) + (200 m)2 = 5.1×104 m ≈ 5 ×104 m (≈ 80 mi)

32. At $1,000 per day, you would earn $30,000 in the 30 days. With the other pay method, you would

( ) ( )get $0.01 2t−1 on the tth day. On the first day, you get $0.01 21−1 = $0.01 . On the second day,
( ) ( )you get $0.01 22−1 = $0.02 . On the third day, you get $0.01 23−1 = $0.04 . On the 30th day, you
( )get $0.01 230−1 = $5.4 ×106 , which is over 5 million dollars. Get paid by the second method.

33. In the figure in the textbook, the distance d is perpendicular to the vertical radius. Thus there is a
right triangle, with legs of d and R, and a hypotenuse of R+h. Since h R , h2 2Rh.

d 2 + R2 = ( R + h)2 = R2 + 2Rh + h2 → d 2 = 2Rh + h2 → d 2 ≈ 2Rh →

R= d2 = (4400 m)2 = 6.5 ×106 m
2h 2(1.5 m)

A better measurement gives R = 6.38 ×106 m.

34. To see the Sun “disappear,” your line of sight to the top To 1st sunset dh
of the Sun is tangent to the Earth’s surface. Initially, you A
are lying down at point A, and you see the first sunset. θ
Then you stand up, elevating your eyes by the height h. R
While standing, your line of sight is tangent to the To 2nd sunset B R
Earth’s surface at point B, and so that is the direction to
the second sunset. The angle θ is the angle through θ
which the Sun appears to move relative to the Earth
during the time to be measured. The distance d is the Earth center
distance from your eyes when standing to point B.

Use the Pythagorean theorem for the following
relationship.

d 2 + R2 = ( R + h)2 = R2 + 2Rh + h2 → d 2 = 2Rh + h2

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The distance h is much smaller than the distance R, and so h2 2Rh which leads to d 2 ≈ 2Rh. We
also have from the same triangle that d R = tanθ , and so d = R tanθ. Combining these two

relationships gives d2 ≈ 2Rh = R2 tan2 θ , and so R = 2h .
tan2 θ
The angle θ can be found from the height change and the radius of the Earth. The elapsed time
between the two sightings can then be found from the angle, knowing that a full revolution takes 24

hours.

2h 2h 2 (1.3m) o
tan2 θ R
6.38 ×106 m
( )R
= → θ = tan−1 = tan −1 = 3.66 ×10−2

θ = t sec →
360o
24 h × 3600 s
1h

( )t ⎛ o ⎞
θ ⎛ 3600 s ⎞ ⎜ 3.66 ×10−2 ⎟ ⎛ 3600 s ⎞
= ⎛ 360o ⎞ ⎜⎝ 24 h × 1h ⎟⎠ = ⎝⎜ 360o ⎟⎠ ⎜⎝ 24 h × 1h ⎟⎠ = 8.8 s
⎜⎝ ⎟⎠

35. Density units = mass units = ⎡M ⎤
volume units ⎢⎣ L3 ⎥⎦

36. (a) For the equation v = At3 − Bt , the units of At3 must be the same as the units of v . So the units
of A must be the same as the units of v t3 , which would be L T 4 . Also, the units of Bt
must be the same as the units of v . So the units of B must be the same as the units of v t ,
which would be L T 2 .

(b) For A, the SI units would be m s4 , and for B, the SI units would be m s2 .

( )37. (a) The quantity vt2 has units of (m s) s2 = mis , which do not match with the units of meters

( )for x. The quantity 2at has units m s2 (s) = m s , which also do not match with the units of

meters for x. Thus this equation cannot be correct .

( )( )(b)
The quantity v0t has units of (m s) (s) = m, and 1 at 2 has units of m s2 s2 = m. Thus,
2

since each term has units of meters, this equation can be correct .

( ) ( )(c) The quantity v0t has units of (m s) (s) = m, and 2at2 has units of m s2 s2 = m. Thus,

since each term has units of meters, this equation can be correct .

⎡ L3 ⎤ ⎡ ML2 ⎤
⎣⎢ MT 2 ⎦⎥ ⎣⎢ T ⎥⎦ =
38. tP = Gh → ⎡ L ⎤5 ⎡ L3L2T 5M ⎤ ⎡T5 ⎤ ⎣⎡T 2 ⎤⎦ = [T ]
c5 ⎢⎣ MT 3L5 ⎥⎦ = ⎣⎢T 3 ⎦⎥ =

⎢⎣T ⎥⎦

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8

Chapter 1 Introduction, Measurement, Estimating

39. The percentage accuracy is 2 2m m × 100% = 1 ×10−5% . The distance of 20,000,000 m needs to
× 107

be distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the

distance measurements.

40. Multiply the number of chips per wafer times the number of wafers that can be made from a
cylinder.

⎝⎜⎛100 chips ⎞⎛ 1 wafer ⎞ ⎛ 250 mm ⎞ = 83, 000 chips
wafer ⎠⎟ ⎜⎝ 0.300 mm ⎟⎠ ⎝⎜ 1 cylinder ⎟⎠ cylinder

41. (a) # of seconds in 1.00 y: 1.00 y = (1.00 y ) ⎛ 3.156 × 107 s ⎞ = 3.16 ×107s
⎝⎜ 1 y ⎠⎟

(b) # of nanoseconds in 1.00 y: 1.00 y = (1.00 y ) ⎛ 3.156 × 107s ⎞ ⎛ 1×109 ns ⎞ = 3.16 ×1016ns
⎜⎝ 1 y ⎟⎠ ⎝⎜ 1s ⎟⎠

(c) # of years in 1.00 s: 1.00 s = (1.00 s ) ⎛ 1 y 107 s ⎞ = 3.17 ×10−8y
⎝⎜ 3.156 × ⎟⎠

42. Since the meter is longer than the yard, the soccer field is longer than the football field.

Lsoccer − Lfootball = 100 m × 1.09 yd − 100 yd = 9 yd
1m

Lsoccer − Lfootball = 100 m − 100 yd × 1m = 8m
1.09 yd

Since the soccer field is 109 yd compare to the 100-yd football field, the soccer field is 9% longer
than the football field.

43. Assume that the alveoli are spherical, and that the volume of a typical human lung is about 2 liters,

which is .002 m3. The diameter can be found from the volume of a sphere, 4 π r 3.
3

π( )4r3 = 4 π d 2 3 = πd3
3 6
3

( ) ( )3×108 π d 3 = 2 ×10−3 m3 ⎤1/ 3
6 m3 ⎥ = 2 ×10−4 m
⎡ 6 2 ×10−3
→ d =⎢ ⎥⎦
⎢⎣ 3 × 108 π

44. 1 hectare = (1 hectare) ⎛ 1.000 ×104m2 ⎞ ⎛ 3.281ft ⎞2 ⎝⎜⎛ 1acre ft 2 ⎟⎠⎞ = 2.471acres
⎜⎝ 1 hectare ⎟⎠ ⎜⎝ 1m ⎠⎟ 4.356 ×104

45. There are about 3 ×108 people in the United States. Assume that half of them have cars, that they
each drive 12,000 miles per year, and their cars get 20 miles per gallon of gasoline.

( )3 ×108 people ⎛ 1 automobile ⎞ ⎛ 12, 000 mi auto ⎞ ⎛ 1 gallon ⎞ ≈ 1 ×1011 gal y
⎜⎝ 2 people ⎠⎟ ⎝⎜ ⎟⎠ ⎝⎜ 20 mi ⎟⎠
1y

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9

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

46. (a) ⎛ 10−15 kg ⎞⎛1 proton or neutron ⎞ = 1012 protons or neutrons
(b) ⎜⎝ 1 bacterium ⎠⎟ ⎜⎝ 10−27 kg ⎠⎟
(c)
(d) ⎛ 10−17 kg ⎞⎛1 proton or neutron ⎞ = 1010 protons or neutrons
⎝⎜ 1 DNA molecule ⎟⎠ ⎜⎝ 10−27 kg ⎟⎠

⎛ 102 kg ⎞⎛1 proton or neutron ⎞ = 1029 protons or neutrons
⎝⎜ 1 human ⎟⎠ ⎜⎝ 10−27 kg ⎟⎠

⎛ 1041 kg ⎞⎛1 proton or neutron ⎞ = 1068 protons or neutrons
⎜ ⎟ ⎜⎝ 10−27 kg ⎠⎟
⎝ 1 galaxy ⎠

47. The volume of water used by the people can be calculated as follows:

( )4 ×104 people ⎛ 1200 L day ⎞⎛ 365 day ⎞ ⎛ 1000 cm3 ⎞ ⎛ 1km 3 = 4.38 ×10−3 km3 y
⎜⎝ 4 people ⎠⎟ ⎝⎜ 1y ⎟⎠ ⎝⎜ 1L ⎟⎠ ⎝⎜ 105cm
⎞
⎠⎟

The depth of water is found by dividing the volume by the area.

d V = 4.38 ×10−3 km3 y = ⎛ 8.76 × 10−5 km ⎞ ⎛ 105 cm ⎞ = 8.76 cm y≈ 9 cm y
=A 50 km2 ⎜⎝ y ⎟⎠ ⎜⎝ 1 km ⎟⎠

48. Approximate the gumball machine as a rectangular box with a square cross-sectional area. In
counting gumballs across the bottom, there are about 10 in a row. Thus we estimate that one layer
contains about 100 gumballs. In counting vertically, we see that there are about 15 rows. Thus we

estimate that there are 1500 gumballs in the machine.

49. Make the estimate that each person has 1.5 loads of laundry per week, and that there are 300 million
people in the United States.

( )300 ×106 people 1.5 loads week × 52 weeks × 0.1 kg = 2.34 ×109 kg ≈ 2 × 109 kg
× 1 person 1y 1load y y

⎝⎜⎛ 3V ⎞1 / 3
4π ⎟⎠
50. The volume of a sphere is given by V = 4 π r 3 , and so the radius is r = . For a 1-ton rock,
3

the volume is calculated from the density, and then the diameter from the volume.

V = (1 T ) ⎛ 2000 lb ⎞ ⎛ 1ft3 ⎞ = 10.8 ft 3
⎝⎜ 1 T ⎠⎟ ⎜⎝ 186 lb ⎠⎟

( )d= 2r = 2 ⎛ 3V ⎞1/ 3 = ⎡3 10.8 ft 3 ⎤1/ 3 3 ft
⎝⎜ 4π ⎠⎟ 2⎢ 4π ⎥ = 2.74 ft ≈
⎥⎦
⎢⎣

( )51. 8 bits 1sec 1min
783.216 ×106 bytes × 1byte × 1.4 ×106 bits × 60 sec = 74.592 min ≈ 75 min

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10

Chapter 1 Introduction, Measurement, Estimating

52. A pencil has a diameter of about 0.7 cm. If held about 0.75 m from the eye, it can just block out the
Moon. The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moon
distance. From the diagram, we have the following ratios.

Pencil Moon

Pencil Moon
Distance Distance

Pencil diameter Moon diameter
Pencil distance = Moon distance →

Pencil diameter 7 ×10−3 m
Pencil distance 0.75 m
( )Moon
diameter = ( Moon distance) = 3.8 ×105 km ≈ 3500 km

The actual value is 3480 km.

53. To calculate the mass of water, we need to find the volume of water, and then convert the volume to

( )mass. The volume of water is the area of the city 40 km2 times the depth of the water (1.0 cm).

( )⎡ ⎛ 105cm ⎞2 ⎤ ⎛ 10−3 kg ⎞ ⎛ 1 metric ton ⎞
⎝⎜ 1 km ⎠⎟ ⎥ ⎜⎝ 1 cm3 ⎟⎠ ⎜⎝ 103 kg ⎟⎠
⎢ ⎥⎦
⎢⎣
4 ×101 km2 (1.0 cm) = 4 ×105 metric tons

To find the number of gallons, convert the volume to gallons.

( )⎡ ⎛ 105cm ⎞ 2 ⎤ ⎛ 1L ⎞ ⎛ 1 gal ⎞
⎝⎜ 1 km ⎠⎟ ⎥ ⎜⎝ 103 cm3 ⎟⎠ ⎝⎜ 3.78 L ⎠⎟
⎢ ⎦⎥ = 1.06 ×108 gal 1×108 gal
⎢⎣
4 ×101 km2 (1.0 cm ) 1 × ≈

54. A cubit is about a half of a meter, by measuring several people’s forearms. Thus the dimensions of
Noah’s ark would be 150 m long , 25 m wide, 15 m high . The volume of the ark is found by
multiplying the three dimensions.

V = (150 m) (25 m) (15 m) = 5.625 ×104 m3 ≈ 6 ×104 m3

55. The person walks 4 km h , 10 hours each day. The radius of the Earth is about 6380 km, and the

distance around the Earth at the equator is the circumference, 2π REarth. We assume that the person

can “walk on water,” and so ignore the existence of the oceans.

2π (6380 km ) ⎛ 1h ⎞ ⎛ 1d ⎞ = 1 ×103d
⎜⎝ 4 km ⎟⎠ ⎝⎜ 10 h ⎟⎠

56. The volume of the oil will be the area times the thickness. The area is π r2 = π (d 2)2 , and so

V 1000 cm3 ⎝⎜⎛ 1m ⎞3
πt 100 cm ⎠⎟
( )V = π (d 2)2 t → d =2 =2 = 3 ×103 m .
π 2 ×10−10 m

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11

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

57. Consider the diagram shown. Let l represent is the distance she walks upstream, which

is about 120 yards. Find the distance across the river from the diagram.

tan 60o = d → d = l tan 60o = (120 yd) tan 60o = 210 yd d

l

(210 yd ) ⎛ 3 ft ⎞⎛ 0.305 m ⎞ = 190 m 60o
⎜⎝ 1 yd ⎟⎠ ⎝⎜ 1ft ⎠⎟ l

58. ⎛ 8 s ⎞ ⎛ 1 y 107s ⎞ × 100% = 3 × 10−5%
⎝⎜ 1 y ⎠⎟ ⎜⎝ 3.156 × ⎟⎠

59. (a) 1.0 o = ⎝⎛⎜1.0 o ⎠⎞⎟ ⎛ 10−10 m ⎞ ⎝⎜⎛ 1 nm ⎠⎟⎞ = 0.10 nm
⎝⎜⎜ ⎠⎟⎟ 10−9 m
A A o

1A

(b) o = ⎝⎜⎛1.0 o ⎟⎠⎞ ⎛ 10−10 m ⎞ ⎛ 1 fm ⎞ = 1.0 ×105 fm
⎜⎝⎜ ⎠⎟⎟ ⎜⎝ 10−15 m ⎟⎠
1.0 A A o

1A

(c) 1.0 m = (1.0 m ) ⎛ o ⎞ = 1.0 ×1010 o
⎜ ⎟
⎜⎝ 1A ⎟⎠ A
10−10 m

(d) 1.0 ly = (1.0 ly ) ⎛ 9.46 ×1015 m ⎞ ⎛ o ⎞ = o
⎜ 1 ly ⎟ ⎜ ⎟
⎝ ⎠ ⎜⎝ 1A ⎠⎟ 9.5 ×1025 A
10−10 m

60. The volume of a sphere is found by V = 4 π r 3 .
3

( )π πV = R =Moon 3 2.21×1019 m3
4 3 4 1.74 ×106 m
3 Moon 3 =

VEarth = π R4 3 = ⎛ REarth ⎞3 = ⎛ 6.38 ×106 m ⎞3 = 49.3
VMoon ⎜ RMoon ⎟ ⎝⎜ 1.74 ×106 m ⎠⎟
3 Earth ⎝ ⎠

π R4 3

3 Moon

Thus it would take about 49.3 Moons to create a volume equal to that of the Earth.

61. (a) Note that sin15.0o = 0.259 and sin15.5o = 0.267, and so Δ sinθ = 0.267 − 0.259 = 0.008.

⎛ Δθ ⎞⎟⎠100 = ⎛ 0.5o ⎟⎠⎞100 = 3% ⎛ Δ sinθ ⎠⎟⎞100 = ⎛ 8 ×10−3 ⎞⎠⎟100 = 3%
⎜⎝ θ ⎜⎝ 15.0o ⎝⎜ sinθ ⎝⎜ 0.259

(b) Note that sin 75.0o = 0.966 and sin 75.5o = 0.968, and so Δ sinθ = 0.968 − 0.966 = 0.002.

⎛ Δθ ⎟⎠⎞100 = ⎛ 0.5o ⎞⎟⎠100 = 0.7% ⎛ Δ sinθ ⎞ 100 = ⎛ 2 ×10−3 ⎟⎞⎠100 = 0.2%
⎜⎝ θ ⎝⎜ 75.0o ⎜⎝ sinθ ⎟⎠ ⎜⎝ 0.966

A consequence of this result is that when using a protractor, and you have a fixed uncertainty in the

angle ( ±0.5o in this case), you should measure the angles from a reference line that gives a large
angle measurement rather than a small one. Note above that the angles around 75o had only a 0.2%
error in sin θ , while the angles around 15o had a 3% error in sin θ.

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12

Chapter 1 Introduction, Measurement, Estimating

62. Utilize the fact that walking totally around the Earth along the meridian would trace out a circle
whose full 360o would equal the circumference of the Earth.

( )(1
minute) ⎛ 1o ⎞ ⎛ 2π 6.38 ×103km ⎞ ⎛ 0.621 mi ⎞ = 1.15 mi
⎜⎝ minute ⎟⎠ ⎝⎜⎜ 360o ⎟⎠⎟ ⎝⎜ 1 km ⎟⎠
60

63. Consider the body to be a cylinder, about 170 cm tall (≈ 5′7′′) , and about 12 cm in cross-sectional

radius (which corresponds to a 30-inch waist). The volume of a cylinder is given by the area of the
cross section times the height.

V = π r 2h = π (0.12 m )2 (1.7 m ) = 7.69 ×10−2 m3 ≈ 8 ×10−2 m3

64. The maximum number of buses would be needed during rush hour. We assume that a bus can hold
50 passengers.
(a) The current population of Washington, D.C. is about half a million people. We estimate that
10% of them ride the bus during rush hour.

50, 000 passengers × 1 bus × 1driver ≈ 1000 drivers
50 passengers 1 bus

(b) For Marion, Indiana, the population is about 50,000. Because the town is so much smaller
geographically, we estimate that only 5% of the current population rides the bus during rush
hour.

2500 passengers × 50 1 bus × 1driver ≈ 50 drivers
passengers 1 bus

65. The units for each term must be in liters, since the volume is in liters.

[units of 4.1][m] = [L] → [units of 4.1] = L
m

[units of 0.018][y] = [L] → [units of 0.018] = L
y

[units of 2.69] = L

66. density = mass = 8g = 2.82 g cm3 ≈ 3g cm3
volume 2.8325 cm3

6.38 ×103 km 2
1.74 ×103 km 2 = 13.4
67. (a) ( )πSAEarth4 R2 2 =
(b) ( )πSAMoon=4 Earth
REarth
R2 =2
Moon
RMoon

6.38 ×103 km 3
1.74 ×103 km 3 = 49.3
( )V π R REarth
43 3
3 Earth Earth
= = =4 3 3
Moon
( )V π R RMoon 3 Moon

# atoms 6.02 ×1023 atoms 6.02 ×1023 atoms 1.18 ×109 atoms
m2 π4 R2 4π 6.38 ×106 m 2 m2

Earth
( )68. = = =

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13

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

69. Multiply the volume of a spherical universe times the density of matter, adjusted to ordinary matter.

The volume of a sphere is 4 π r 3 .
3

( ) ( )m = ρV = 1×10−26 kg m3 ⎛ 9.46 ×1015 m ⎞3
4 π ⎜⎝ 13.7 ×109 ly × 1ly ⎟⎠ ( 0.04 )
3

= 3.65 ×1051 kg ≈ 4 ×1051 kg

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14

CHAPTER 2: Describing Motion: Kinematics in One Dimension

Responses to Questions

1. A car speedometer measures only speed, since it gives no indication of the direction in which the car
is traveling.

2. If the velocity of an object is constant, the speed must also be constant. (A constant velocity means
that the speed and direction are both constant.) If the speed of an object is constant, the velocity
CAN vary. For example, a car traveling around a curve at constant speed has a varying velocity,
since the direction of the velocity vector is changing.

3. When an object moves with constant velocity, the average velocity and the instantaneous velocity
are the same at all times.

4. No, if one object has a greater speed than a second object, it does not necessarily have a greater
acceleration. For example, consider a speeding car, traveling at constant velocity, which passes a
stopped police car. The police car will accelerate from rest to try to catch the speeder. The speeding
car has a greater speed than the police car (at least initially!), but has zero acceleration. The police
car will have an initial speed of zero, but a large acceleration.

5. The accelerations of the motorcycle and the bicycle are the same, assuming that both objects travel
in a straight line. Acceleration is the change in velocity divided by the change in time. The
magnitude of the change in velocity in each case is the same, 10 km/h, so over the same time interval
the accelerations will be equal.

6. Yes, for example, a car that is traveling northward and slowing down has a northward velocity and a
southward acceleration.

7. Yes. If the velocity and the acceleration have different signs (opposite directions), then the object is
slowing down. For example, a ball thrown upward has a positive velocity and a negative acceleration
while it is going up. A car traveling in the negative x-direction and braking has a negative velocity
and a positive acceleration.

8. Both velocity and acceleration are negative in the case of a car traveling in the negative x-direction
and speeding up. If the upward direction is chosen as +y, a falling object has negative velocity and
negative acceleration.

9. Car A is going faster at this instant and is covering more distance per unit time, so car A is passing
car B. (Car B is accelerating faster and will eventually overtake car A.)

10. Yes. Remember that acceleration is a change in velocity per unit time, or a rate of change in
velocity. So, velocity can be increasing while the rate of increase goes down. For example, suppose a
car is traveling at 40 km/h and a second later is going 50 km/h. One second after that, the car’s speed
is 55 km/h. The car’s speed was increasing the entire time, but its acceleration in the second time
interval was lower than in the first time interval.

11. If there were no air resistance, the ball’s only acceleration during flight would be the acceleration
due to gravity, so the ball would land in the catcher’s mitt with the same speed it had when it left the
bat, 120 km/h. The path of the ball as it rises and then falls would be symmetric.

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15

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

12. (a) If air resistance is negligible, the acceleration of a freely falling object stays the same as the
object falls toward the ground. (Note that the object’s speed increases, but since it increases at a
constant rate, the acceleration is constant.)

(b) In the presence of air resistance, the acceleration decreases. (Air resistance increases as speed
increases. If the object falls far enough, the acceleration will go to zero and the velocity will
become constant. See Section 5-6.)

13. Average speed is the displacement divided by the time. If the distances from A to B and from B to C

are equal, then you spend more time traveling at 70 km/h than at 90 km/h, so your average speed

should be less than 80 km/h. If the distance from A to B (or B to C) is x, then the total distance

traveled is 2x. The total time required to travel this distance is x/70 plus x/90. Then

v = d = x 2x 90 = 2(90)(70) = 79 km/h.
t 70 + x 90 + 70

14. Yes. For example, a rock thrown straight up in the air has a constant, nonzero acceleration due to
gravity for its entire flight. However, at the highest point it momentarily has a zero velocity. A car, at
the moment it starts moving from rest, has zero velocity and nonzero acceleration.

15. Yes. Anytime the velocity is constant, the acceleration is zero. For example, a car traveling at a
constant 90 km/h in a straight line has nonzero velocity and zero acceleration.

16. A rock falling from a cliff has a constant acceleration IF we neglect air resistance. An elevator
moving from the second floor to the fifth floor making stops along the way does NOT have a
constant acceleration. Its acceleration will change in magnitude and direction as the elevator starts
and stops. The dish resting on a table has a constant acceleration (zero).

17. The time between clinks gets smaller and smaller. The bolts all start from rest and all have the same
acceleration, so at any moment in time, they will all have the same speed. However, they have
different distances to travel in reaching the floor and therefore will be falling for different lengths of
time. The later a bolt hits, the longer it has been accelerating and therefore the faster it is moving.
The time intervals between impacts decrease since the higher a bolt is on the string, the faster it is
moving as it reaches the floor. In order for the clinks to occur at equal time intervals, the higher the
bolt, the further it must be tied from its neighbor. Can you guess the ratio of lengths?

18. The slope of the position versus time curve is the velocity. The object starts at the origin with a
constant velocity (and therefore zero acceleration), which it maintains for about 20 s. For the next 10
s, the positive curvature of the graph indicates the object has a positive acceleration; its speed is
increasing. From 30 s to 45 s, the graph has a negative curvature; the object uniformly slows to a
stop, changes direction, and then moves backwards with increasing speed. During this time interval
its acceleration is negative, since the object is slowing down while traveling in the positive direction
and then speeding up while traveling in the negative direction. For the final 5 s shown, the object
continues moving in the negative direction but slows down, which gives it a positive acceleration.
During the 50 s shown, the object travels from the origin to a point 20 m away, and then back 10 m
to end up 10 m from the starting position.

19. The object begins with a speed of 14 m/s and increases in speed with constant positive acceleration

from t = 0 until t = 45 s. The acceleration then begins to decrease, goes to zero at t = 50 s, and then
goes negative. The object slows down from t = 50 s to t = 90 s, and is at rest from t = 90 s to t = 108
s. At that point the acceleration becomes positive again and the velocity increases from t = 108 s to
t = 130 s.

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16

Chapter 2 Describing Motion: Kinematics in One Dimension

Solutions to Problems

1. The distance of travel (displacement) can be found by rearranging Eq. 2-2 for the average velocity.

Also note that the units of the velocity and the time are not the same, so the speed units will be
converted.

v = Δx → Δx = v Δt = (110 km h ) ⎛ 1h s ⎞ ( 2.0 s) = 0.061 km = 61 m
Δt ⎜⎝ 3600 ⎠⎟

2. The average speed is given by Eq. 2-2.
v = Δx Δt = 235 km 3.25 h = 72.3 km h

3. The average velocity is given by Eq. 2.2.

v = Δx = 8.5 cm − 4.3 cm = 4.2 cm = 0.65 cm s
Δt 6.5 s
4.5 s − (−2.0 s)

The average speed cannot be calculated. To calculate the average speed, we would need to know the

actual distance traveled, and it is not given. We only have the displacement.

4. The average velocity is given by Eq. 2-2.

v = Δx = −4.2 cm − 3.4 cm = −7.6 cm = −3.6 cm s
Δt 5.1s − 3.0 s 2.1s

The negative sign indicates the direction.

5. The speed of sound is intimated in the problem as 1 mile per 5 seconds. The speed is calculated as
follows.

speed = distance = ⎛ 1mi ⎞ ⎛ 1610 m ⎞ = 300 m s
time ⎜ ⎟ ⎜⎝ ⎟⎠
⎝ 5s ⎠ 1 mi

The speed of 300 m s would imply the sound traveling a distance of 900 meters (which is

approximately 1 km) in 3 seconds. So the rule could be approximated as 1 km every 3 seconds .

6. The time for the first part of the trip is calculated from the initial speed and the first distance.

v1 = Δx1 → Δt1 = Δx1 = 130 km = 1.37 h = 82 min
Δt1 v1 95 km h

The time for the second part of the trip is now calculated.

Δt2 = Δttotal − Δt1 = 3.33 h − 1.37 h = 1.96 h = 118 min
The distance for the second part of the trip is calculated from the average speed for that part of the

trip and the time for that part of the trip.

v2 = Δx2 → Δx2 = v2Δt2 = (65 km h )(1.96 h ) = 127.5 km = 1.3×102 km
Δt2

(a) The total distance is then Δxtotal = Δx1 + Δx2 = 130 km + 127.5 km = 257.5 km ≈ 2.6 × 102 km .

(b) The average speed is NOT the average of the two speeds. Use the definition of average speed,

Eq. 2-2.

v = Δxtotal = 257.5 km = 77 km h
Δttotal 3.33 h

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17

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

7. The distance traveled is 116 km + 1 (116 km) = 174 km, and the displacement is
2

116 km − 1 (116 km) = 58 km. The total time is 14.0 s + 4.8 s = 18.8 s.
2

(a) Average speed = distance 174 m = 9.26 m s
time elapsed = 18.8 s

(b) Average velocity = vavg = displacement 58 m = 3.1 m s
time elapsed = 18.8 s

8. (a) 50

40

30

x (m) 20

10

0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t (sec)

The spreadsheet used for this problem can be found on the Media Manager, with filename

“PSE4_ISM_CH02.XLS”, on tab “Problem 2.8a”.

(b) The average velocity is the displacement divided by the elapsed time.

v = x (3.0) − x (0.0) = ⎣⎡34 + 10 (3.0) − 2 (3.0)3 ⎤⎦ m − (34 m) = −8.0 m s

3.0s − 0.0s 3.0 s

(c) The instantaneous velocity is given by the derivative of the position function.

( )v = dx = 10 − 6t2 m s 10 − 6t2 = 0 → t = 5s = 1.3s
dt 3

This can be seen from the graph as the “highest” point on the graph.

9. Slightly different answers may be obtained since the data comes from reading the graph.
(a) The instantaneous velocity is given by the slope of the tangent line to the curve. At t = 10.0 s,

the slope is approximately v (10) ≈ 3m −0 = 0.3 m s.
10.0 s−0

(b) At t = 30.0 s, the slope of the tangent line to the curve, and thus the instantaneous velocity, is

approximately v (30) ≈ 22 m − 10 m = 1.2 m s .
35s − 25s

(c) The average velocity is given by v = x (5) − x (0) = 1.5 m − 0 = 0.30 m s.
5.0 s
5.0 s − 0 s

(d) The average velocity is given by v = x (30) − x (25) = 16 m − 9 m = 1.4 m s.
5.0 s
30.0 s − 25.0 s

(e) The average velocity is given by v = x (50) − x (40) = 10 m − 19.5 m = −0.95 m s.
10.0 s
50.0 s − 40.0 s

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18

Chapter 2 Describing Motion: Kinematics in One Dimension

10. (a) Multiply the reading rate times the bit density to find the bit reading rate.

N 1.2 m 1bit 4.3 ×106 bits s
= 1s × 0.28 ×10−6m =

(b) The number of excess bits is N − N0.

N − N0 = 4.3 ×106 bits s − 1.4 ×106 bits s = 2.9 ×106 bits s

N − N0 = 2.9 ×106 bits s = 0.67 = 67%
N 4.3 ×106 bits s

11. Both objects will have the same time of travel. If the truck travels a distance ,Δxtruck then the

distance the car travels will be Δxcar = Δxtruck + 110 m. Use Eq. 2-2 for average speed, v = Δx Δt ,

solve for time, and equate the two times.

Δt = Δxtruck = Δxcar Δxtruck = Δxtruck + 110 m
vtruck vcar 75 km 95 km h
h

Solving for Δxtruck gives Δxtruck = (110 m ) (95 (75 km h) h) = 412.5 m.

km h − 75 km

The time of travel is Δt = Δxtruck = ⎛ 412.5 m ⎞ ⎛ 60 min ⎞ = 0.33 min = 19.8 s = 2.0 × 101s .
vtruck ⎜⎝ 75000 m h ⎠⎟ ⎝⎜ 1h ⎟⎠

Also note that Δt = Δxcar = ⎛ 412.5 m + 110 m ⎞ ⎛ 60 min ⎞ = 0.33 min = 20 s.
vcar ⎝⎜ 95000 m h ⎠⎟ ⎜⎝ 1h ⎠⎟

ALTERNATE SOLUTION:

The speed of the car relative to the truck is 95 km h − 75 km h = 20 km h . In the reference frame of

the truck, the car must travel 110 m to catch it.

Δt = 0.11 km ⎛ 3600 s ⎞ = 19.8 s
20 km h ⎝⎜ 1h ⎟⎠

12. Since the locomotives have the same speed, they each travel half the distance, 4.25 km. Find the

time of travel from the average speed.

v = Δx → Δt = Δx = 4.25 km = 0.0447 h ⎛ 60 min ⎞ = 2.68 min ≈ 2.7 min
Δt v 95 km h ⎜⎝ 1h ⎟⎠

13. (a) The area between the concentric circles is equal to the length times the width of the spiral path.

π R22 − π R12 = wl →

( )l ⎡⎣(0.058 m)2 − (0.025 m)2 ⎦⎤
π R22 − R12 = π = 5.378 ×103 m ≈ 5400 m
= w 1.6 ×10−6 m

(b) 5.378 × 103 m ⎛ 1s m ⎞ ⎛ 1min ⎞ = 72 min
⎜⎝ 1.25 ⎟⎠ ⎜⎝ 60 s ⎠⎟

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19

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

14. The average speed for each segment of the trip is given by v = Δx , so Δt = Δx for each
Δt v

segment. For the first segment, Δt1 = Δx1 = 3100 km = 4.306 h. For the second segment,
v1 720 km h

Δt2 = Δx2 = 2800 km = 2.828h. .
v2 990 km h

Thus the total time is Δttot = Δt1 + Δt2 = 4.306 h + 2.828 h = 7.134 h ≈ 7.1h .

The average speed of the plane for the entire trip is v = Δxtot = 3100 km + 2800 km = 827 km h
Δttot 7.134 h

≈ 830 km h .

15. The distance traveled is 500 km (250 km outgoing, 250 km return, keep 2 significant figures). The

displacement (Δx) is 0 because the ending point is the same as the starting point.

(a) To find the average speed, we need the distance traveled (500 km) and the total time elapsed.

During the outgoing portion, v1 = Δx1 and so Δt1 = Δx1 = 250 km = 2.632 h. During the
Δt1 v1 95 km h

return portion, v2 = Δx2 , and so Δt2 = Δx2 = 250 km = 4.545 h. Thus the total time,
Δt2 v2 55 km h

including lunch, is Δttotal = Δt1 + Δtlunch + Δt2 = 8.177 h.

v = Δxtotal = 500 km = 61km h
Δttotal 8.177 h

(b) Average velocity = v = Δx Δt = 0

( )16. We are given that x (t) = 2.0 m − (3.6 m s) t + 1.1m s2 t2.

( )(a) x (1.0 s) = 2.0 m − (3.6 m s)(1.0 s) + 1.1m s2 (1.0 s)2 = −0.5 m

( )x (2.0 s) = 2.0 m − (3.6 m s)(2.0 s) + 1.1m s2 (2.0 s)2 = −0.8 m

( )x (3.0 s) = 2.0 m − (3.6 m s)(3.0 s) + 1.1m s2 (3.0 s)2 = 1.1m

(b) v = Δx = 1.1m − (−0.5m ) = 0.80 m s
Δt
2.0 s

dx (t)

dt
( )(c) s2
The instantaneous velocity is given by v (t) = = −3.6 m s+ 2.2 m t.

( )v (2.0 s) = −3.6 m s + 2.2 m s2 (2.0 s) = 0.8 m s

( )v (3.0 s) = −3.6 m s + 2.2 m s2 (3.0 s) = 3.0 m s

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20

Chapter 2 Describing Motion: Kinematics in One Dimension

17. The distance traveled is 120 m + 1 (120 m ) = 180 m, and the displacement is
2

120 m − 1 (120 m) = 60 m. The total time is 8.4 s + 1 (8.4 s) = 11.2 s.
2 3

(a) Average speed = distance 180 m = 16 m s
time elapsed = 11.2 s

(b) Average velocity = vavg = displacement 60 m = +5 m s (in original direction) (1 sig fig)
time elapsed = 11.2 s

18. For the car to pass the train, the car must travel the length of the train AND the distance the train

travels. The distance the car travels can thus be written as either dcar = vcart = (95 km h) t or

( )dcar = l train + vtraint = 1.10 km + 75 km h t. To solve for the time, equate these two expressions for

the distance the car travels.

(95 km h) t = 1.10 km + (75 km h) t → t = 1.10 km = 0.055 h = 3.3 min
20 km h

The distance the car travels during this time is d = (95 km h ) (0.055 h ) = 5.225 km ≈ 5.2 km .

If the train is traveling the opposite direction from the car, then the car must travel the length of the

train MINUS the distance the train travels. Thus the distance the car travels can be written as either

d car = (95 km h ) t or dcar = 1.10 km − (75 km h ) t. To solve for the time, equate these two

expressions for the distance the car travels.

(95 km h) t = 1.10 km − (75 km h) t → t = 1.10 km = 6.47 × 10−3 h= 23.3 s
170 km h

( )The distance the car travels during this time is d = (95 km h ) 6.47 × 10−3 h = 0.61 km .

19. The average speed of sound is given by vsound = Δx Δt , and so the time for the sound to travel from

the end of the lane back to the bowler is Δtsound = Δx = 16.5 m = 4.85 × 10−2 s. Thus the time for
vsound 340 m s

the ball to travel from the bowler to the end of the lane is given by Δtball = Δttotal − Δtsound =

2.50 s − 4.85 × 10−2 s = 2.4515s. And so the speed of the ball is as follows.

vball = Δx = 16.5 m = 6.73 m s.
Δtball 2.4515 s

20. The average acceleration is found from Eq. 2-5.

Δv 95 km h − 0 km h (95 km h ) ⎛ 1m s h ⎞
Δt 4.5 s ⎝⎜ 3.6 km ⎠⎟
a = = = = 5.9 m s2
4.5 s

21. The time can be found from the average acceleration, a = Δv Δt .

Δv 110 km h − 80 km h (30 km h ) ⎛ 1m s h ⎞
a 1.8 m s2 ⎝⎜ 3.6 km ⎟⎠
Δt = = = = 4.630 s ≈ 5s
1.8 m s2

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21

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

22. (a) The average acceleration of the sprinter is a = Δv = 9.00 m s − 0.00 m s = 7.03 m s2 .
Δt 1.28 s

( )(b)a= s2 ⎛ 1 km ⎞⎛ 3600 s ⎞2 9.11×104 km h2
7.03 m ⎜⎝ 1000 m ⎟⎠ ⎝⎜ 1h ⎟⎠ =

23. Slightly different answers may be obtained since the data comes from reading the graph.
(a) The greatest velocity is found at the highest point on the graph, which is at t ≈ 48 s .
(b) The indication of a constant velocity on a velocity–time graph is a slope of 0, which occurs
from t = 90 s to t ≈ 108 s .
(c) The indication of a constant acceleration on a velocity–time graph is a constant slope, which
occurs from t = 0 s to t ≈ 42 s , again from t ≈ 65 s to t ≈ 83 s , and again from

t = 90 s to t ≈ 108 s .
(d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest, which

occurs from t ≈ 65 s to t ≈ 83 s .

24. The initial velocity of the car is the average speed of the car before it accelerates.

v = Δx = 110 m = 22 m s = v0
Δt 5.0 s

The final speed is v = 0 , and the time to stop is 4.0 s. Use Eq. 2-12a to find the acceleration.

v = v0 + at → a = v − v0 = 0 − 22 m s = −5.5 m s2
t 4.0 s

( )Thus the magnitude of the acceleration is ⎛ 1g ⎞
5.5 m s2 , or 5.5 m s2 ⎜⎝ 9.80 m s2 ⎟⎠ = 0.56 g ' s .

25. (a) v = Δx = 385 m − 25 m = 21.2 m s
Δt 20.0 s − 3.0 s

(b) a = Δv = 45.0 m s − 11.0 m s = 2.00 m s2
Δt 20.0 s − 3.0 s

26. Slightly different answers may be obtained since the data comes from reading the graph. We assume

that the short, nearly horizontal portions of the graph are the times that shifting is occurring, and

those times are not counted as being “in” a certain gear.

(a) The average acceleration in 2nd gear is given by a2 = Δv2 = 24 m s − 14 m s = 2.5 m s2 .
Δt2 8s − 4s

(b) The average acceleration in 4th gear is given by a4 = Δv4 = 44 m s − 37 m s = 0.6 m s2 .
Δt4 27 s − 16 s

(c) The average acceleration through the first four gears is given by a = Δv =
Δt

a = Δv = 44 m s − 0 m s = 1.6 m s2 .
Δt 27 s − 0 s

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22

Chapter 2 Describing Motion: Kinematics in One Dimension

27. The acceleration is the second derivative of the position function.

x = 6.8t + 8.5t 2 → v = dx = 6.8 + 17.0t → a= d2x = dv = 17.0 m s2
dt dt 2 dt

28. To estimate the velocity, find the average velocity over Table of Calculations
each time interval, and assume that the car had that velocity
at the midpoint of the time interval. To estimate the t (s) x (m) t (s) v (m/s) t (s) a (m/s2)
acceleration, find the average acceleration over each time 0.00 0.00 0.00 0.00 0.063 3.52
interval, and assume that the car had that acceleration at the 0.25 0.11 0.125 0.44 0.25 3.84
midpoint of the time interval. A sample of each calculation 0.50 0.46 0.375 1.40 0.50 4.00
is shown. 0.75 1.06 0.625 2.40 0.75 4.48
1.00 1.94 0.875 3.52 1.06 4.91
From 2.00 s to 2.50 s, for average velocity: 1.50 4.62 1.25 5.36 1.50 5.00
2.00 8.55 1.75 7.86 2.00 5.24
tmid = 2.50 s+ 2.00 s = 2.25 s 2.50 13.79 2.25 10.48 2.50 5.32
2 3.00 20.36 2.75 13.14 3.00 5.52
3.50 28.31 3.25 15.90 3.50 5.56
vavg = Δx = 13.79 m − 8.55 m = 5.24 m = 10.48 m s 4.00 37.65 3.75 18.68 4.00 5.52
Δt 2.50 s − 2.00 s 0.50 s 4.50 48.37 4.25 21.44 4.50 4.84
5.00 60.30 4.75 23.86 5.00 4.12
From 2.25 s to 2.75 s, for average acceleration: 5.50 73.26 5.25 25.92 5.50 3.76
6.00 87.16 5.75 27.80
tmid = 2.25 s+ 2.75 s = 2.50 s
2

aavg = Δv = 13.14 m s − 10.48 m s = 2.66 m s
Δt 2.75 s − 2.25 s 0.50 s

= 5.32 m s2

30 6

25 5

20 4

15 3

10 2

51

00
0 1 2 3 4 5 60 1 2 3 4 5
t (s) t (s)

The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH02.XLS,” on tab “Problem 2.28.”
v (m/s)
a (m/s2)

6

29. (a) Since the units of A times the units of t must equal meters, the units of A must be m s .
Since the units of B times the units of t2 must equal meters, the units of B must be
m s2 .

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23

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(b) The acceleration is the second derivative of the position function.

x = At + Bt2 → v = dx = A+ 2Bt → a= d2x = dv = 2B m s2
dt dt 2 dt

(c) v = A + 2Bt → v (5) = ( A + 10B) m s a = 2B m s2

(d) The velocity is the derivative of the position function.

x = At + Bt−3 → v = dx = A − 3Bt −4
dt

30. The acceleration can be found from Eq. 2-12c.

v2 = v02 + 2a ( x − x0 ) → a= v 2 − v02 = 0 − (25 m s)2 = −3.7 m s2
2 (85 m)
2 ( x − x0 )

31. By definition, the acceleration is a = v − v0 = 21 m s − 12 m s = 1.5 m s2 .
t 6.0 s

The distance of travel can be found from Eq. 2-12b.

( )x − at 2 s2 (6.0 s)2 =
x0 = v0t + 1 = (12 m s) (6.0 s) + 1 1.5 m 99 m
2 2

32. Assume that the plane starts from rest. The runway distance is found by solving Eq. 2-12c for

x − x0.

v 2 − v02 (32 m s)2 − 0
2a
2 3.0 m s2
( )v2 = v02 + 2a ( x − x0 ) → x − x0 = = = 1.7 ×102 m

33. For the baseball, v0 = 0 , x − x0 = 3.5 m, and the final speed of the baseball (during the throwing
motion) is v = 41m s. The acceleration is found from Eq. 2-12c.

v2 = v02 + 2a ( x − x0 ) → a= v2 − v02 ) = (41m s)2 − 0 = 240 m s2
2 (3.5 m)
2 ( x − x0

34. The average velocity is defined by Eq. 2-2, v = Δx = x − x0 . Compare this expression to Eq. 2-
Δt t

12d, v = (1 v + v0 ). A relation for the velocity is found by integrating the expression for the

2

acceleration, since the acceleration is the derivative of the velocity. Assume the velocity is v0 at

time t = 0.

dv vt
dt
A+ Bt → dv = ( A + Bt) dt ( A + Bt) dt v Bt 2
∫ ∫a
= = → dv = → = v0 + At + 1
2

v0 0

Find an expression for the position by integrating the velocity, assuming that x = x0 at time t = 0.

( )v v0 At Bt 2 dx v0 At Bt 2 dt
= + + 1 = dt → dx = + + 1 →
2 2

xt
∫ ∫ ( )dx =
v0 + At + 1 Bt 2 dt → x − x0 = v0t + 1 At 2 + 1 Bt 3
2 2 6

x0 0

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24

Chapter 2 Describing Motion: Kinematics in One Dimension

Compare x − x0 to 1 (v + v0 ).
t 2

v = x − x0 = v0t + 1 At 2 + 1 Bt 3 = v0 + 1 At + 1 Bt 2
t 2 t 6 2 6

( )1 v0 + v0 + At + 1 Bt 2 Bt 2
2 2
2
v + v0 = = v0 + 1 At + 1
2 4

They are different, so v ≠ 1 ( v + v0 ) .
2

35. The sprinter starts from rest. The average acceleration is found from Eq. 2-12c.

v2 = v02 + 2a ( x − x0 ) → a= v2 − v02 = (11.5 m s)2 − 0 = 4.408 m s2 ≈ 4.41 m s2
2 (15.0 m)
2 ( x − x0 )

Her elapsed time is found by solving Eq. 2-12a for time.

v = v0 + at → t= v − v0 = 11.5 m s −0 = 2.61 s
a 4.408 m s2

36. Calculate the distance that the car travels during the reaction time and the deceleration.

Δx1 = v0Δt = (18.0 m s) (0.200 s) = 3.6 m

v2 − v02 0 − (18.0 m s)2
2a
2 −3.65 m s2
( )v2 = v02 + 2aΔx2 → Δx2 = = = 44.4 m

Δx = 3.6 m + 44.4 m = 48.0 m
He will NOT be able to stop in time.

37. The words “slows down uniformly” implies that the car has a constant acceleration. The distance of

travel is found from combining Eqs. 2-2 and 2-9.

x − x0 = v0 + v t = ⎛ 18.0 m s + 0m s ⎞ (5.00 sec ) = 45.0 m
2 ⎜⎝ 2 ⎟⎠

38. The final velocity of the car is zero. The initial velocity is found from Eq. 2-12c with v = 0 and
solving for v0. Note that the acceleration is negative.

( )v2 = v02 + 2a ( x − x0 ) → v0 = v2 − 2a ( x − x0 ) = 0 − 2 −4.00 m s2 (85 m) = 26 m s

39. (a) The final velocity of the car is 0. The distance is found from Eq. 2-12c with an acceleration of

a = −0.50 m s2 and an initial velocity of 85 km h .

v 2 − v02 0− ⎢⎡⎣(85 km h ) ⎛ 1m s ⎞⎤2
2a ⎜⎝ 3.6 km h ⎟⎠⎦⎥
( )x −x0 = = = 557 m ≈ 560 m
2 −0.50 m s2

(b) The time to stop is found from Eq. 2-12a.

0− ⎡⎢⎣(85 km h ) ⎛ 1m s ⎞⎤
⎜⎝ 3.6 km ⎠⎟ ⎥
v − v0 h ⎦
a
( )t = = = 47.22 s ≈ 47 s
−0.50 m s2

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25

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(c) Take x0 = x (t = 0) = 0 m. Use Eq. 2-12b, with a = −0.50 m s2 and an initial velocity

of 85 km h . The first second is from t = 0 s to t = 1s, and the fifth second is from t = 4 s to

t = 5s.

⎛ 1m s ⎞
⎝⎜ 3.6 km ⎠⎟
( )x (0) = 0 (1s)2 = 23.36 m
; x (1) = 0 + (85 km h ) h (1s ) + 1 −0.50 m s2 →
2

x (1) − x (0) = 23 m

⎛ 1m s ⎞
⎜⎝ 3.6 km ⎠⎟
( )x (4) = 0 + (85 km ) ( ) s2 (4 s)2 = 90.44 m
h h 4 s + 1 −0.50 m
2

⎛ 1m s ⎞
⎝⎜ 3.6 km ⎠⎟
( )x (5) = 0 + (85 km h ) (5 s ) s2 (5 s)2 = 111.81m
h + 1 −0.50 m
2

x (5) − x (4) = 111.81m − 90.44 m = 21.37m ≈ 21m

40. The final velocity of the driver is zero. The acceleration is found from Eq. 2-12c with v = 0 and
solving for a .

v2 − v02 0 − ⎣⎡⎢(105 km h ) ⎛ 1m s ⎞⎤2
⎜⎝ 3.6 km h ⎠⎟⎦⎥
a= 2 ( x − x0 ) = = −531.7 m s2 ≈ −5.3×102 m s2
2 (0.80 m)

−531.7 m s2
9.80 m s2 g
( )Converting to “g’s”: a= = −54 g's

41. The origin is the location of the car at the beginning of the reaction time. The initial speed of the car

is (95 km h ) ⎛ 1m s h ⎞ = 26.39 m s. The location where the brakes are applied is found from
⎜⎝ 3.6 km ⎟⎠

the equation for motion at constant velocity: x0 = v0tR = (26.39 m s) (1.0 s) = 26.39 m. This is now

the starting location for the application of the brakes. In each case, the final speed is 0.
(a) Solve Eq. 2-12c for the final location.

v 2 − v02 0 − (26.39 m s)2
2a
2 −5.0 m s2
( )v2 = v02 + 2a ( x − x0 ) x m+ 96 m
→ = x0 + = 26.39 =

(b) Solve Eq. 2-12c for the final location with the second acceleration.

x0 v 2 − v02 0 − (26.39 m s)2
2a 2
−7.0 m s2
( )x = m
+ = 26.39 + = 76 m

42. Calculate the acceleration from the velocity–time data using Eq. 2-12a, and then use Eq. 2-12b to

calculate the displacement at t = 2.0 s and t = 6.0 s. The initial velocity is v0 = 65 m s .

a = v − v0 = 162 m s − 65 m s = 9.7 m s2 x = x0 + v0t + 1 at 2 →
t 10.0 s 2

( ) ( )x (6.0s) − x (2.0s) = ⎣⎡
x0 + v0 (6.0 s) + 1 a (6.0 s)2 − x0 + v0 ( 2.0 s ) + 1 a (2.0 s)2 ⎦⎤
2 2

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26

Chapter 2 Describing Motion: Kinematics in One Dimension

( )( )= s)2
v0 (6.0 s − 2.0 s) + 1 a ⎡⎣(6.0 − (2.0 s)2 ⎦⎤ = (65 m s) (4.0 s) + 1 9.7 m s2 32 s2
2 2

= 415 m ≈ 4.2 ×102 m

43. Use the information for the first 180 m to find the acceleration, and the information for the full

motion to find the final velocity. For the first segment, the train has v0 = 0 m s , v1 = 23 m s , and a

displacement of x1 − x0 = 180 m. Find the acceleration from Eq. 2-12c.

( )v12 = v02 + 2a x1 − x0 → a= v12 − v02 = (23 m s)2 − 0 = 1.469 m s2
2 (180 m)
2 ( x1 − x0 )

Find the speed of the train after it has traveled the total distance (total displacement of

x2 − x0 = 255 m) using Eq. 2-12c.

( )( )v22 = v02 + 2a x2 − x0 → v2 = v02 + 2a ( x2 − x0 ) = 2 1.469 m s2 ( 255 m) = 27 m s

44. Define the origin to be the location where the speeder passes the police car. Start a timer at the
instant that the speeder passes the police car, and find another time that both cars have the same
displacement from the origin.

For the speeder, traveling with a constant speed, the displacement is given by the following.

Δxs = vst = (135 km h ) ⎛ 1m s h ⎞ ( t ) = ( 37.5 t ) m
⎜⎝ 3.6 km ⎠⎟

For the police car, the displacement is given by two components. The first part is the distance

traveled at the initially constant speed during the 1 second of reaction time.

Δx p1 = vp1 (1.00 s) = (95 km h ) ⎛ 1m s h ⎞ (1.00 s) = 26.39 m
⎜⎝ 3.6 km ⎠⎟

The second part of the police car displacement is that during the accelerated motion, which lasts for

(t − 1.00) s. So this second part of the police car displacement, using Eq. 2-12b, is given as follows.

( )Δxp2
= vp1 (t − 1.00) + 1 ap (t − 1.00)2 = ⎣⎡(26.39 m s)(t − 1.00) + 1 2.00 m s2 (t − 1.00)2 ⎦⎤ m
2 2

( )So the total police car displacement is Δx p = Δx p1 + Δx p2 = 26.39 + 26.39 (t − 1.00) + (t − 1.00)2 m.

Now set the two displacements equal, and solve for the time.

26.39 + 26.39 (t − 1.00) + (t − 1.00)2 = 37.5 t → t2 − 13.11t + 1.00 = 0

t = 13.11 ± (13.11)2 − 4.00 = 7.67 × 10−2 s , 13.0 s

2

The answer that is approximately 0 s corresponds to the fact that both vehicles had the same

displacement of zero when the time was 0. The reason it is not exactly zero is rounding of previous

values. The answer of 13.0 s is the time for the police car to overtake the speeder.

As a check on the answer, the speeder travels Δxs = (37.5 m s) (13.0 s) = 488 m, and the police car
travels Δxp = ⎡⎣26.39 + 26.39 (12.0) + (12.0)2 ⎦⎤ m = 487 m. . The difference is due to rounding.

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27

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

45. Define the origin to be the location where the speeder passes the police car. Start a timer at the
instant that the speeder passes the police car. Both cars have the same displacement 8.00 s after the
initial passing by the speeder.

For the speeder, traveling with a constant speed, the displacement is given by Δxs = vst = (8.00vs ) m.

For the police car, the displacement is given by two components. The first part is the distance
traveled at the initially constant speed during the 1.00 s of reaction time.

Δx p1 = vp1 (1.00 s) = (95 km h ) ⎛ 1m s h ⎞ (1.00 s) = 26.39 m
⎜⎝ 3.6 km ⎠⎟

The second part of the police car displacement is that during the accelerated motion, which lasts for

7.00 s. So this second part of the police car displacement, using Eq. 2-12b, is given by the following.

( )Δxp2 )2 (7.00 s)2 =233.73 m
= vp1 (7.00 s) + 1 a p ( 7.00 s = (26.39 m s) (7.00 s) + 1 2.00 m s2
2 2

Thus the total police car displacement is Δx p = Δx p1 + Δx p2 = (26.39 + 233.73) m = 260.12 m.

Now set the two displacements equal, and solve for the speeder’s velocity.

(8.00vs ) m = 260.12 m → vs = (32.5 m s) ⎛ 3.6 km h ⎞ = 117 km h
⎜ 1m s ⎟
⎝ ⎠

46. During the final part of the race, the runner must have a displacement of 1100 m in a time of 180 s

(3.0 min). Assume that the starting speed for the final part is the same as the average speed thus far.

v = Δx = 8900 m s = 5.494 m s = v0
Δt
(27 × 60)

The runner will accomplish this by accelerating from speed v0 to speed v for t seconds, covering a

distance d1, and then running at a constant speed of v for (180 − t ) seconds, covering a distance d2.

We have these relationships from Eq. 2-12a and Eq. 2-12b.

v = vo + at d1 = vot + 1 at 2 d2 = v (180 − t ) = (v0 + at ) (180 − t )
2

1100 m = d1 + d2 = vot + 1 at 2 + (v0 + at ) (180 − t) → 1100 m = 180v0 + 180at − 1 at 2 →
2 2

( ) ( )1100 m = (180 s) (5.494 m s) + (180 s) s2 t s2 t2
0.2 m − 1 0.2 m
2

0.1t2 − 36t + 111 = 0 t = 357 s , 3.11 s

Since we must have t < 180 s , the solution is t = 3.1s .

47. For the runners to cross the finish line side-by-side means they must both reach the finish line in the
same amount of time from their current positions. Take Mary’s current location as the origin. Use
Eq. 2-12b.

For Sally: 22 = 5 + 5t + 1 ( −.5) t2 → t2 − 20t + 68 = 0 →
2

t = 20 ± 202 − 4(68) = 4.343s, 15.66 s
2

The first time is the time she first crosses the finish line, and so is the time to be used for the

problem. Now find Mary’s acceleration so that she crosses the finish line in that same amount of

time.

For Mary: 22 = 0 + 4t + 1 at 2 → a= 22 − 4t = 22 − 4(4.343) = 0.49 m s2
2 t1 2
1 ( 4.343)2
2 2

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28

Chapter 2 Describing Motion: Kinematics in One Dimension

48. Choose downward to be the positive direction, and take y0 = 0 at the top of the cliff. The initial

velocity is v0 = 0, and the acceleration is a = 9.80 m s2 . The displacement is found from Eq. 2-

12b, with x replaced by y.

( )y at 2 (3.75 s)2
= y0 + v0t + 1 → y − 0 = 0 + 1 9.80 m s2 → y = 68.9 m
2 2

49. Choose downward to be the positive direction. The initial velocity is v0 = 0, the final velocity is

v = (55 km h ) ⎛ 1m s h ⎞ = 15.28 m s, and the acceleration is a = 9.80 m s2 . The time can be
⎝⎜ 3.6 km ⎟⎠

found by solving Eq. 2-12a for the time.

v = v0 + at → t = v − v0 = 15.28 m s − 0 = 1.6 s
a 9.80 m s2

50. Choose downward to be the positive direction, and take y0 = 0 to be at the top of the Empire State

Building. The initial velocity is v0 = 0, and the acceleration is a = 9.80 m s2 .

(a) The elapsed time can be found from Eq. 2-12b, with x replaced by y.

y − y0 = v0t + 1 at 2 → t= 2y 2(380 m) = 8.806 s ≈ 8.8 s .
2 a=
9.80 m s2

(b) The final velocity can be found from Eq. 2-12a.

( )v = v0 + at = 0 + 9.80 m s2 (8.806 s) = 86 m s

51. Choose upward to be the positive direction, and take y0 = 0 to be at the height where the ball was

hit. For the upward path, v0 = 20 m s , v = 0 at the top of the path, and a = −9.80 m s2 .
(a) The displacement can be found from Eq. 2-12c, with x replaced by y .

v 2 − v02 0 − (20 m s)2
2a
2 −9.80 m s2
( )v2 = v02 + 2a ( y − y0 ) → y = y0 + = 0+ = 20 m

(b) The time of flight can be found from Eq. 2-12b, with x replaced by y , using a displacement of 0
for the displacement of the ball returning to the height from which it was hit.

y = y0 + v0t + 1 at 2 = 0 → t (v0 + 1 at ) = 0 → t=0,t= 2v0 = 2 (20 m s) = 4s
2 2 −a
9.80 m s2

The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and
the result of t = 4 s is the time to return to the original displacement. Thus the answer is t = 4 s.

52. Choose upward to be the positive direction, and take y0 = 0 to be the height from which the ball

was thrown. The acceleration is a = −9.80 m s2 . The displacement upon catching the ball is 0,

assuming it was caught at the same height from which it was thrown. The starting speed can be
found from Eq. 2-12b, with x replaced by y.

y = y0 + v0t + 1 at 2 = 0 →
2

y − y0 − 1 at 2
( )v0 2
= = − 1 at = − 1 −9.80 m s2 (3.2 s) = 15.68 m s ≈ 16 m s
t 2 2

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29

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The height can be calculated from Eq. 2-12c, with a final velocity of v = 0 at the top of the path.

v 2 − v02 0 − (15.68 m s)2
2a
2 −9.80 m s2
( )v2 = v02 + 2a ( y − y0 ) → y = y0 + = 0+ = 12.54 m ≈ 13 m

53. Choose downward to be the positive direction, and take y0 = 0 to be at the maximum height of the
kangaroo. Consider just the downward motion of the kangaroo. Then the displacement is

y = 1.65 m, the acceleration is a = 9.80 m s2 , and the initial velocity is v0 = 0 m s . Use Eq. 2-

12b to calculate the time for the kangaroo to fall back to the ground. The total time is then twice the
falling time.

y = y0 + v0t + 1 at 2 = 0 → y = 1 at 2 → tfall = 2y →
2 2 a

2y 2 (1.65 m)
a
9.80 m s2
( )ttotal = 2 = 2 = 1.16 s

54. Choose upward to be the positive direction, and take y0 = 0 to be at the floor level, where the jump

starts. For the upward path, y = 1.2 m , v = 0 at the top of the path, and a = −9.80 m s2 .
(a) The initial speed can be found from Eq. 2-12c, with x replaced by y .

v2 = v02 + 2a ( y − y0 ) →

( )v0 = v2 − 2a ( y − y0 ) = −2ay = −2 −9.80 m s2 (1.2 m ) = 4.8497 m s ≈ 4.8 m s

(b) The time of flight can be found from Eq. 2-12b, with x replaced by y , using a displacement of 0
for the displacement of the jumper returning to the original height.

( )y= y0 + v0t + 1 at 2 = 0 → t v0 + 1 at =0 →
2 2

t =0,t = 2v0 = 2 (4.897 m s) = 0.99 s
−a
9.80 m s2

The result of t = 0 s is the time for the original displacement of zero (when the jumper started to

jump), and the result of t = 0.99 s is the time to return to the original displacement. Thus the

answer is t = 0.99 seconds.

55. Choose downward to be the positive direction, and take y0 = 0 to be the height where the object

was released. The initial velocity is v0 = −5.10 m s , the acceleration is a = 9.80 m s2 , and the

displacement of the package will be y = 105 m. The time to reach the ground can be found from

Eq. 2-12b, with x replaced by y.

y = y0 + v0t + 1 at 2 → t2 + 2v0 t − 2y = 0 → t2 + 2 (−5.10 m s) t − 2 (105 m) = 0 →
2 a a
9.80 m s2 9.80 m s2

t = 5.18s , − 4.14 s

The correct time is the positive answer, t = 5.18s .

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30

Chapter 2 Describing Motion: Kinematics in One Dimension

56. Choose downward to be the positive direction, and take y0 = 0 to be the height from which the

object is released. The initial velocity is v0 = 0, and the acceleration is a = g. Then we can

calculate the position as a function of time from Eq. 2-12b, with x replaced by y, as y (t) = 1 gt 2 . At
2

the end of each second, the position would be as follows.

y(0) = 0 ; y (1) = 1 g ; y (2) = 1 g (2)2 = 4 y (1) ; y (3) = 1 g (3)2 = 9 y (1)
2 2 2

The distance traveled during each second can be found by subtracting two adjacent position values

from the above list.

d (1) = y (1) − y (0) = y (1) ; d (2) = y (2) − y (1) = 3y (1) ; d (3) = y (3) − y (2) = 5 y (1)

We could do this in general.

y(n) = 1 gn2 y(n + 1) = 1 g (n + 1)2
2 2

( )d + 1)2
(n + 1) = y(n + 1) − y(n) = 1 g (n − 1 gn2 = 1 g ( n + 1)2 − n2
2 2 2

( )=
1 g n2 + 2n +1− n2 = 1 g (2n + 1)
2 2

The value of (2n +1) is always odd, in the sequence 1, 3, 5, 7, ….

57. Choose upward to be the positive direction, and y0 = 0 to be the level from which the ball was

thrown. The initial velocity is v0, , the instantaneous velocity is v = 14 m s , the acceleration is

a = −9.80 m s2 , and the location of the window is y = 23 m.
(a) Using Eq. 2-12c and substituting y for x, we have

v2 = v02 + 2a ( y − y0 ) →

( )v0 = ± v2 − 2a ( y − y0 ) = ± (14 m s)2 − 2 −9.8 m s2 ( 23 m ) = 25.43 m s ≈ 25 m s

Choose the positive value because the initial direction is upward.
(b) At the top of its path, the velocity will be 0, and so we can use the initial velocity as found

above, along with Eq. 2-12c.

v2 − v02 0 − (25.43 m s)2
2a
2 −9.80 m s2
( )v2 = v02 + 2a ( y − y0 ) → y= y0 + = 0+ = 33 m

(c) We want the time elapsed from throwing (speed v0 = 25.43 m s ) to reaching the window (speed

v = 14 m s ). Using Eq. 2-12a, we have the following.

v = v0 + at → t = v − v0 = 14 m s − 25.43 m s = 1.166 s ≈ 1.2 s
a −9.80 m s2

(d) We want the time elapsed from the window (speed v0 = 14 m s ) to reaching the street (speed

v = −25.43 m s ). Using Eq. 2-12a, we have the following.

v = v0 + at → t = v − v0 = −25.43 m s − 14 m s = 4.0 s
a −9.80 m s2

This is the elapsed time after passing the window. The total time of flight of the baseball from

passing the window to reaching the street is 4.0s + 1.2 s = 5.2 s .

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31

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

58. (a) Choose upward to be the positive direction, and y0 = 0 at the ground. The rocket has v0 = 0,

a = 3.2 m s2 , and y = 950 m when it runs out of fuel. Find the velocity of the rocket when it

runs out of fuel from Eq 2-12c, with x replaced by y.

( )v2 v02
950 m
= + 2a y − y0 →

( )v950 m = ± v02 + 2a ( y − y0 ) = ± 0 + 2 3.2 m s2 (950 m) = 77.97 m s ≈ 78 m s

The positive root is chosen since the rocket is moving upwards when it runs out of fuel.

(b) The time to reach the 950 m location can be found from Eq. 2-12a.

v950 m = v0 + at950 m → t950 m = v950 m − v0 = 77.97 m s− 0 = 24.37 s ≈ 24 s
a 3.2 m s2

(c) For this part of the problem, the rocket will have an initial velocity v0 = 77.97 m s , an

acceleration of a = −9.80 m s2 , and a final velocity of v = 0 at its maximum altitude. The

altitude reached from the out-of-fuel point can be found from Eq. 2-12c.

v2 = v2 + 2a ( y − 950 m) →
950 m

950 m 0 − v2 = 950 m + − (77.97 m s)2 = 950 m + 310 m = 1260 m
950
2 −9.80 m s2
( )ymax = + m

2a

(d) The time for the “coasting” portion of the flight can be found from Eq. 2-12a.

v = v950 m + atcoast → tcoast = v − v0 = 0 − 77.97 m s = 7.96 s
a −9.80 m s2

Thus the total time to reach the maximum altitude is t = 24.37 s + 7.96 s = 32.33s ≈ 32 s .

(e) For the falling motion of the rocket, v0 = 0 m s , a = −9.80 m s2 , and the displacement is

−1260 m (it falls from a height of 1260 m to the ground). Find the velocity upon reaching the
Earth from Eq. 2-12c.

v2 = v02 + 2a ( y − y0 ) →

( )v = ± v02 + 2a ( y − y0 ) = ± 0 + 2 −9.80 m s2 (−1260 m ) = −157 m s ≈ −160 m s

The negative root was chosen because the rocket is moving downward, which is the negative

direction.

(f) The time for the rocket to fall back to the Earth is found from Eq. 2-12a.

v = v0 + at → tfall = v − v0 = −157 m s − 0 = 16.0 s
a −9.80 m s2

Thus the total time for the entire flight is t = 32.33 s + 16.0 s = 48.33s ≈ 48 s . .

59. (a) Choose y = 0 to be the ground level, and positive to be upward. Then y = 0 m,

y0 = 15 m, a = −g, and t = 0.83s describe the motion of the balloon. Use Eq. 2-12b.

y = y0 + v0t + 1 at 2 →
2

( )v0 (0.83s)2 = −14 m s
y− y0 − 1 at 2 0 − 15 m − 1 −9.80 m s2
t 2 2
= = (0.83s)

So the speed is 14 m s .

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32

Chapter 2 Describing Motion: Kinematics in One Dimension

(b) Consider the change in velocity from being released to being at Roger’s room, using Eq. 2-12c.

v2 − v02 − (−14 m s)2
2a
2 −9.8 m s2
( )v2 = v02 + 2aΔy → Δy = = = 10 m

Thus the balloons are coming from 2 floors above Roger, and so the fifth floor .

60. Choose upward to be the positive direction, and y0 = 0 to be the height from which the stone is

thrown. We have v0 = 24.0 m s , a = −9.80 m s2 , and y − y0 = 13.0 m.

(a) The velocity can be found from Eq, 2-12c, with x replaced by y.

v2 = v02 + 2a ( y − y0 ) = 0 →

( )v = ± v02 + 2ay = ± ( 24.0 m s)2 + 2 −9.80 m s2 (13.0 m ) = ±17.9 m s

Thus the speed is v = 17.9 m s .

(b) The time to reach that height can be found from Eq. 2-12b.

y = y0 + v0t + 1 at 2 → t2 + 2 (24.0 m s) t + 2 (−13.0 m) = 0 →
2
−9.80 m s2 −9.80 m s2

t 2 − 4.898 t + 2.653 = 0 → t = 4.28 s , 0.620 s

(c) There are two times at which the object reaches that height – once on the way up (t = 0.620 s) ,
and once on the way down (t = 4.28s).

61. Choose downward to be the positive direction, and y0 = 0 to be the height from which the stone is

dropped. Call the location of the top of the window yw , and the time for the stone to fall from

release to the top of the window is tw. Since the stone is dropped from rest, using Eq. 2-12b with y

substituting for x, we have yw = y0 + v0t + 1 at 2 = 0+ 0+ 1 gtw2 . The location of the bottom of the
2 2

window is yw + 2.2 m, and the time for the stone to fall from release to the bottom of the window is

tw + 0.33s. Since the stone is dropped from rest, using Eq. 2-12b, we have the following:

yw + 2.2 m = y0 + v0 + 1 at 2 = 0+0+ 1 g (tw + 0.33s)2 . Substitute the first expression for yw into
2 2

the second expression.

gt( )12 2.2 m g 2 → tw = 0.515 s
w
2
+ = 1 tw + 0.33 s
2

Use this time in the first equation to get the height above the top of the window from which the stone

fell.

( )yw gtw2 s2 (0.515 s)2 = 1.3 m
= 1 = 1 9.80 m
2 2

62. Choose upward to be the positive direction, and y0 = 0 to be the location of the nozzle. The initial

velocity is v0, the acceleration is a = −9.80 m s2 , the final location is y = −1.5 m, and the time of

flight is t = 2.0 s. Using Eq. 2-12b and substituting y for x gives the following.

( )y (2.0 s)2 = 9.1m s
= y0 + v0t + 1 at 2 → v0 = y − 1 at 2 = −1.5 m − 1 −9.80 m s2
2 2 2 2.0 s

t

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33

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

63. Choose up to be the positive direction, so a = − g. Let the ground be the y = 0 location. As an

intermediate result, the velocity at the bottom of the window can be found from the data given.
Assume the rocket is at the bottom of the window at t = 0, and use Eq. 2-12b.

12
2 pass
y = y + v t + attop of →
bottom of bottom of pass

window window window window window

( )( )10.0 m = 8.0 m + vbottom of ( 0.15 s )2
0.15 s + 1 −9.80 m s2 → vbottom of = 14.07 m s
2

window window

Now use the velocity at the bottom of the window with Eq. 2-12c to find the launch velocity,

assuming the launch velocity was achieved at the ground level.

( )v = v + 2a2
bottom of
2 y − y0 →
launch

window

( )v =launch
( )v2 (14.07 m s)2 + 2 9.80 m s2 (8.0 m) = 18.84 m s
bottom of
− 2a y − y0 =

window

≈ 18.8 m s

The maximum height can also be found from Eq. 2-12c, using the launch velocity and a velocity of 0

at the maximum height.

( )v2
maximum
= v2 + 2a ymax − y0 →
launch

height

v − v2 2 s2
maximum launch s2
(( ) )ymax
= y0 + height = − 18.84 m = 18.1 m
2 −9.80 m
2a

64. Choose up to be the positive direction. Let the bottom of the cliff be the y = 0 location. The

( )equation of motion for the dropped ball is t2.
yball = y0 + v0t + 1 at 2 = 50.0 m+ 1 −9.80 m s2 The
2 2

( )equation of motion for the thrown stone is
( )ystone v0t at 2 t −9.80 m s2 t2.
= y0 + + 1 = 24.0 m s + 1
2 2

Set the two equations equal and solve for the time of the collision. Then use that time to find the

location of either object.

( ) ( )( )y = yball stone
→ 50.0 m+ 1 −9.80 m s2 t2 = 24.0 m s t + 1 −9.80 m s2 t2 →
2 2

50.0 m = (24.0 m s) t → t = 50.0 m = 2.083 s
24.0 m s

( )yball at 2 s2 (2.083s)2 =
= y0 + v0t + 1 = 50.0 m + 1 −9.80 m 28.7 m
2 2

65. For the falling rock, choose downward to be the positive direction, and y0 = 0 to be the height from

which the stone is dropped. The initial velocity is v0 = 0 m s , the acceleration is a = g, the

displacement is y = H , and the time of fall is t1. Using Eq. 2-12b with y substituting for x, we have

H = y0 + v0t + t1 2 = 0+0+ 1 gt12 . For the sound wave, use the constant speed equation that
2
2

vs = Δx = H, which can be rearranged to give t1 =T − H , where T = 3.4 s is the total time
Δt vs
T − t1

elapsed from dropping the rock to hearing the sound. Insert this expression for t1 into the equation

for H from the stone, and solve for H.

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34

Chapter 2 Describing Motion: Kinematics in One Dimension

H = 1 g ⎛ T − H ⎞2 → g H2 − ⎛ gT + 1⎟⎠⎞ H + 1 gT 2 =0 →
2 ⎜ vs ⎟ 2vs2 ⎜ vs 2
⎝ ⎠ ⎝

4.239 × 10−5 H 2 − 1.098H + 56.64 = 0 → H = 51.7 m, 2.59 × 104m

If the larger answer is used in t1 = T − H , a negative time of fall results, and so the physically
vs

correct answer is H = 52 m .

66. (a) Choose up to be the positive direction. Let the throwing height of both objects be the y = 0

location, and so y0 = 0 for both objects. The acceleration of both objects is a = − g. The

equation of motion for the rock, using Eq. 2-12b, is yrock = y0 + v0 trock + 1 at 2 = v0 trock − 1 gt2 ,
2 2

where t is the time elapsed from the throwing of the rock. The equation of motion for the ball,

being thrown 1.00 s later, is y ball = y0 + v0 ball (t − 1.00 s) + 1 a (t − 1.00 s)2 =
2

v0 ball (t − 1.00 s) − 1 g (t − 1.00 s)2 . Set the two equations equal (meaning the two objects are at
2

the same place) and solve for the time of the collision.

gt 2 t − 1.00 s 2
( ) ( )y = yrock ball g
→ v0 trock − 1 = v0 ball t − 1.00 s − 1 →
2 2

( ) ( )(12.0 m s2 t2 s2 (t − 1.00 s)2
s)t − 1 9.80 m = (18.0 m s)(t − 1.00 s) − 1 9.80 m →
2 2

(15.8 m s) t = (22.9 m) → t = 1.45s

(b) Use the time for the collision to find the position of either object.

( )( )( )yrock gt 2 s2 (1.45 s)2 =
= v0 trock − 1 = 12.0 m s 1.45 s − 1 9.80 m 7.10 m
2 2

(c) Now the ball is thrown first, and so yball = v0 tball − 1 gt 2 and
2

yrock = v0 rock (t − 1.00s) − 1 g (t − 1.00 s)2 . Again set the two equations equal to find the time of
2

collision.

gt 2 t − 1.00 s 2
( ) ( )y = yball rock
→ v0 tball − 1 = v0 rock t − 1.00 s − 1 g →
2 2

( ) ( )(18.0 m s)t s2 t2 s2 (t − 1.00 s)2
− 1 9.80 m = (12.0 m s)(t − 1.00 s) − 1 9.80 m →
2 2

(3.80 m s) t = 16.9 m → t = 4.45s

But this answer can be deceptive. Where do the objects collide?

( )( )( )yball (4.45 s)2 = −16.9 m
= v0 tball − 1 gt 2 = 18.0 m s 4.45 s − 1 9.80 m s2
2 2

Thus, assuming they were thrown from ground level, they collide below ground level, which

cannot happen. Thus they never collide .

67. The displacement is found from the integral of the velocity, over the given time interval.

∫ ∫ ( )t2 t=3.1s t =3.1s ⎣⎡25 (3.1) + 9 (3.1)2 ⎤⎦ ⎣⎡25 (1.5) + 9 (1.5)2 ⎤⎦
t =1.5s
Δx = vdt = (25 + 18t ) dt =
25t + 9t 2 = −

t1 t =1.5s

= 106 m

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35

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

68. (a) The speed is the integral of the acceleration.

vt

dv = A
∫ ∫a = dv
→ dv = adt → dv = A tdt → t dt →
dt v0 0

( )v − v0
= 2 At 3/ 2 → v = v0 + 2 At 3/ 2 → v = 7.5 m s+ 2 2.0 m s5/2 t3/2
3 3 3

(b) The displacement is the integral of the velocity.

( )v dx
= dt → dx = vdt → dx = v0 + 2 At3/ 2 dt →
3

xt

∫ ∫ ( ) ( )( )dx =
v0 + 2 At3/ 2 dt → x = v0t + 2 2 At5/ 2 = 7.5 m s t + 4 2.0 m s5/2 t5/2
3 3 5 15

0m 0

( )(c) a (t = 5.0 s) = 2.0 m s5/2 5.0 s = 4.5 m s2

( )v (t s5/2 (5.0 s)3/2 = 22.41m s ≈
= 5.0 s) = 7.5 m s+ 2 2.0 m 22 m s
3

( )x (t (5.0 s)5/2 = 67.31m ≈
= 5.0 s) = (7.5 m s)(5.0 s) + 4 2.0 m s5/2 67 m
15

69. (a) The velocity is found by integrating the acceleration with respect to time. Note that with the

substitution given in the hint, the initial value of u is u0 = g − kv0 = g.

a = dv → dv = adt → dv = ( g − kv) dt → dv = dt
dt g − kv

Now make the substitution that u ≡ g − kv.

u ≡ g − kv → dv = − du g dv = dt → − du 1 = dt → du = −kdt
k − kv k u u

∫ ∫udu = t dt → ln u u = −kt → ln u = −kt → u = ge−kt = g − kv →
u g g
g −k

0

( )v g
= 1 − e−kt
k

( )(b) g g.
As t goes to infinity, the value of the velocity is vterm = lim k 1 − e−kt = k We also note that

t→∞

if the acceleration is zero (which happens at terminal velocity), then a = g − kv = 0 →

vterm = g .
k

70. (a) The train's constant speed is vtrain = 5.0 m s , and the location of the empty box car as a

function of time is given by xtrain = vtraint = (5.0 m s) t. The fugitive has v0 = 0 m s and

a = 1.2 m s2 until his final speed is 6.0 m s. The elapsed time during the acceleration is

tacc = v − v0 6.0 m s = 5.0 s. Let the origin be the location of the fugitive when he starts to
a = 1.2 m s2

run. The first possibility to consider is, “Can the fugitive catch the empty box car before he
reaches his maximum speed?” During the fugitive's acceleration, his location as a function of

( )time is given by Eq. 2-12b,
xfugitive = x0 + v0t + 1 at 2 = 0+0+ 1 1.2 m s2 t2. For him to catch
2 2

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36

Chapter 2 Describing Motion: Kinematics in One Dimension

( )the train, we must have xtrain = xfugitive s2 t2.
→ (5.0 m s)t = 1 1.2 m The solutions of this
2

are t = 0 s, 8.3s. Thus the fugitive cannot catch the car during his 5.0 s of acceleration.

Now the equation of motion of the fugitive changes. After the 5.0 s of acceleration, he runs
with a constant speed of 6.0 m s. Thus his location is now given (for times t > 5s ) by the

following.

( )x =fugitive (5.0 s)2 + (6.0 m s) (t − 5.0 s) = (6.0 m s) t − 15.0 m
1 1.2 m s2
2

So now, for the fugitive to catch the train, we again set the locations equal.

x = xtrain fugitive → (5.0 m s) t = (6.0 m s) t − 15.0 m → t = 15.0 s

(b) The distance traveled to reach the box car is given by the following.

(x tfugitive = 15.0 s) = (6.0 m s) (15.0 s) − 15.0 m = 75 m

71. Choose the upward direction to be positive, and y0 = 0 to be the level from which the object was

thrown. The initial velocity is v0 and the velocity at the top of the path is v = 0 m s . The height at
the top of the path can be found from Eq. 2-12c with x replaced by y.

v2 = v02 + 2a ( y − y0 ) → y− y0 = −v02
2a

From this we see that the displacement is inversely proportional to the acceleration, and so if the

acceleration is reduced by a factor of 6 by going to the Moon, and the initial velocity is unchanged,

the displacement increases by a factor of 6 .

72. (a) For the free-falling part of the motion, choose downward to be the positive direction, and

y0 = 0 to be the height from which the person jumped. The initial velocity is v0 = 0,

acceleration is a = 9.80 m s2 , and the location of the net is y = 15.0 m. Find the speed upon

reaching the net from Eq. 2-12c with x replaced by y.

( ) ( )v2 = v02 + 2a y − y( ) → v = ± 0 + 2a ( y − 0) = ± 2 9.80 m s2 (15.0 m ) = 17.1m s

The positive root is selected since the person is moving downward.
For the net-stretching part of the motion, choose downward to be the positive direction, and
y0 = 15.0 m to be the height at which the person first contacts the net. The initial velocity is

v0 = 17.1 m s , the final velocity is v = 0, and the location at the stretched position is

y = 16.0 m. . Find the acceleration from Eq. 2-12c with x replaced by y.

v2 = v02 + 2a ( y − y0 ) → a= v2 − v02 = 02 − (17.1m s)2 = −150 m s2
2 (1.0 m)
2( y − y0 )

(b) For the acceleration to be smaller, in the above equation we see that the displacement should

be larger. This means that the net should be “loosened” .

73. The initial velocity of the car is v0 = (100 km h ) ⎛ 1m s h ⎞ = 27.8 m s. Choose x0 = 0 to be the
⎝⎜ 3.6 km ⎠⎟

location at which the deceleration begins. We have v = 0 m s and a = −30g = −294 m s2 . Find

the displacement from Eq. 2-12c.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

v2 − v02 0 − (27.8 m s)2
2a
−2.94 ×102 m s2
( )v2 = v02 + 2a ( x − x0 ) x x0
→ = + =0+ 2 = 1.31m ≈ 1.3m

74. Choose downward to be the positive direction, and y0 = 0 to be at the start of the pelican’s dive.

The pelican has an initial velocity is v0 = 0, an acceleration of a = g, and a final location of
y = 16.0 m. Find the total time of the pelican’s dive from Eq. 2-12b, with x replaced by y.

y = y0 + v0t + 1 at 2 → y = 0 + 0 + 1 at 2 → tdive = 2y 2 (16.0 m) = 1.81 s.
2 2 a=
9.80 m s2

The fish can take evasive action if he sees the pelican at a time of 1.81 s – 0.20 s = 1.61 s into the
dive. Find the location of the pelican at that time from Eq. 2-12b.

( )y (1.61 s)2 = 12.7 m
= y0 + v0t + 1 at = 0+ 0+ 1 9.80 m s2
2 2

Thus the fish must spot the pelican at a minimum height from the surface of the water of

16.0 m − 12.7 m = 3.3 m .

75. (a) Choose downward to be the positive direction, and y0 = 0 to be the level from which the

car was dropped. The initial velocity is v0 = 0, the final location is y = H , and the
acceleration is a = g. Find the final velocity from Eq. 2-12c, replacing x with y.

v2 = v02 + 2a ( y − y0 ) → v = ± v02 + 2a ( y − y0 ) = ± 2gH .

The speed is the magnitude of the velocity, v = 2 gH .

(b) Solving the above equation for the height, we have that H = v2 . Thus for a collision of

2g

v = (50 km h) ⎛ 1m s h ⎞ = 13.89 m s, the corresponding height is as follows.
⎝⎜ 3.6 km ⎟⎠

v2 (13.89 m s)2 10 m
= 2g
2 9.80 m s2
( )H = = 9.84 m ≈

(c) For a collision of v = (100 km h ) ⎛ 1m s h ⎞ = 27.78 m s , the corresponding height is as
⎜⎝ 3.6 km ⎠⎟

follow.

v2 (27.78 m s)2
2g
2 9.80 m s2
( )H= = = 39.37 m ≈ 40 m

76. Choose downward to be the positive direction, and y0 = 0 to be at the roof from which the stones

are dropped. The first stone has an initial velocity of v0 = 0 and an acceleration of a = g. Eqs. 2-
12a and 2-12b (with x replaced by y) give the velocity and location, respectively, of the first stone as

a function of time.

v = v0 + at → v1 = gt1 y = y0 + v0t + 1 at 2 → y1 = 1 gt12
2 2

The second stone has the same initial conditions, but its elapsed time t − 1.50 s, and so has velocity

and location equations as follows.

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38

Chapter 2 Describing Motion: Kinematics in One Dimension

v2 = g (t1 − 1.50 s) y2 = 1 g (t1 − 1.50 s)2
2

The second stone reaches a speed of v2 = 12.0 m s at a time given by the following.

t1 = 1.50 s + v2 = 1.50 s + 12.0 m s = 2.72 s
g 9.80 m s2

( )The location of the first stone at that time is (2.72 s)2 = 36.4 m. .
y1 = 1 gt12 = 1 9.80 m s2
2 2

The location of the second stone at that time is y2 = 1 g ( t1 − 1.50 s)2 =
2

( )1 (2.72 − 1.50 s)2 = 7.35 m. Thus the distance between the two stones is

2
9.80 m s2

y1 − y2 = 36.4 m − 7.35 m = 29.0 m .

77. The initial velocity is v0 = (15 km h ) ⎛ 1m s ⎞ = 4.17 m s. The final velocity is
⎝⎜ 3.6 km h ⎟⎠

v0 = (75 km h ) ⎛ 1m s h ⎞ = 20.83 m s. The displacement is x − x0 = 4.0 km = 4000 m. Find the
⎝⎜ 3.6 km ⎟⎠

average acceleration from Eq. 2-12c.

v2 = v02 + 2a ( x − x0 ) → a= v2 − v02 = (20.83 m s)2 − (4.17 m s)2 = 5.2 ×10−2 m s2
2 (4000 m)
2 ( x − x0 )

78. The speed limit is 50 km h ⎛ 1m s h ⎞ = 13.89 m s.
⎝⎜ 3.6 km ⎟⎠

(a) For your motion, you would need to travel (10 + 15 + 50 + 15 + 70 + 15) m = 175 m to get the

front of the car all the way through the third intersection. The time to travel the 175 m is found

using the distance and the constant speed.

Δx = v Δ t → Δt = Δx = 175 m s = 12.60 s
v 13.89 m

Yes , you can make it through all three lights without stopping.

(b) The second car needs to travel 165 m before the third light turns red. This car accelerates from

v0 = 0 m s to a maximum of v = 13.89 m s with a = 2.0 m s2 . Use Eq. 2-12a to determine

the duration of that acceleration.

v = v0 + at → tacc = v − v0 = 13.89 m s − 0 m s = 6.94 s
a 2.0 m s2

The distance traveled during that time is found from Eq. 2-12b.

6.94 s 2 = 48.2 m
( )( ) ( )x − x0
acc = v0tacc + at1 2 = 0+ 1 2.0 m s2
2
2 acc

Since 6.94 s have elapsed, there are 13 – 6.94 = 6.06 s remaining to clear the intersection. The

car travels another 6.06 s at a speed of 13.89 m/s, covering a distance of Δxconstant = vavgt =

speed

(13.89 m s) (6.06 s) = 84.2 m. Thus the total distance is 48.2 m + 84.2 m = 132.4 m. No ,

the car cannot make it through all three lights without stopping.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The car has to travel another 32.6 m to clear the third intersection, and is traveling at a speed of

13.89 m/s. Thus the care would enter the intersection a time t = Δx 32.6 m = 2.3 s after
v = 13.89 m s

the light turns red.

79. First consider the “uphill lie,” in which the ball is being putted down the hill. Choose x0 = 0 to be
the ball’s original location, and the direction of the ball’s travel as the positive direction. The final

velocity of the ball is v = 0 m s , the acceleration of the ball is a = −1.8 m s2 , and the displacement

of the ball will be x − x0 = 6.0 m for the first case and x − x0 = 8.0 m for the second case. Find the
initial velocity of the ball from Eq. 2-12c.

( )v2 ⎪⎧
v2 = v02 + 2a ( x − x0 ) → v0 = 2a (x x0 ) ⎨ 0 − 2 −1.8 m s2 (6.0 m) = 4.6 m s

− − = ( )⎪⎩ 0 − 2 −1.8 m s2 (8.0 m) = 5.4 m s

The range of acceptable velocities for the uphill lie is 4.6 m s to 5.4 m s , a spread of 0.8 m/s.

Now consider the “downhill lie,” in which the ball is being putted up the hill. Use a very similar set-
up for the problem, with the basic difference being that the acceleration of the ball is now

a = −2.8 m s2 . Find the initial velocity of the ball from Eq. 2-12c.

( )v2 ⎧
⎪ 0 − 2 −2.8 m s2 (6.0 m) = 5.8 m s
v2 = v02 + 2a ( x − x0 ) → v0 = 2a (x x0 ) ⎨
− − = ( )⎪⎩ 0 − 2 −2.8 m s2 (8.0 m) = 6.7 m s

The range of acceptable velocities for the downhill lie is 5.8 m s to 6.7 m s , a spread of 0.9 m/s.

Because the range of acceptable velocities is smaller for putting down the hill, more control in
putting is necessary, and so putting the ball downhill (the “uphill lie”) is more difficult.

80. To find the distance, we divide the motion of the robot into three segments. First, the initial
acceleration from rest; second, motion at constant speed; and third, deceleration back to rest.

(5.0 s)2 = 2.5 m
( ) ( )d1
= v0t + 1 a1t12 = 0+ 1 0.20 m s2 v1 = a1t1 = 0.20 m s2 (5.0 s) = 1.0 m s
2 2

d2 = v1t2 = (1.0 m s) (68s) = 68 m v2 = v1 = 1.0 m s

( )d3 a1t12 (2.5s)2 = 1.25 m
= v2t3 + 1 = (1.0 m s) (2.5s) + 1 −0.40 m s2
2 2

d = d1 + d2 + d3 = 2.5 m + 68 m + 1.25 m = 71.75 m ≈ 72 m

81. Choose downward to be the positive direction, and y0 = 0 to be at the top of the cliff. The initial

velocity is v0 = −12.5 m s , the acceleration is a = 9.80 m s2 , and the final location is y = 75.0 m.

(a) Using Eq. 2-12b and substituting y for x, we have the following.

( )y
= y0 + v0t + 1 at 2 → 4.9 m s2 t2 − (12.5 m s) t − 75.0 m = 0 → t = −2.839 s , 5.390 s
2

The positive answer is the physical answer: t = 5.39 s .

( )(b) Using Eq. 2-12a, we have v = v0 + at = −12.5 m s + 9.80 m s2 (5.390 s) = 40.3 m s .

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40

Chapter 2 Describing Motion: Kinematics in One Dimension

(c) The total distance traveled will be the distance up plus the distance down. The distance down
will be 75.0 m more than the distance up. To find the distance up, use the fact that the speed at
the top of the path will be 0. Using Eq. 2-12c we have the following.

v 2 − v02 0 − (−12.5 m s)2
2a
2 9.80 m s2
( )v2 = v02 + 2a ( y − y0 ) → y = y0 + =0+ = −7.97 m

Thus the distance up is 7.97 m, the distance down is 82.97 m, and the total distance traveled is

90.9 m .

82. (a) In the interval from A to B, it is moving in the negative direction , because its displacement is
negative.

(b) In the interval from A to B, it is speeding up , because the magnitude of its slope is increasing
(changing from less steep to more steep).

(c) In the interval from A to B, the acceleration is negative , because the graph is concave down,
indicating that the slope is getting more negative, and thus the acceleration is negative.

(d) In the interval from D to E, it is moving in the positive direction , because the displacement is
positive.

(e) In the interval from D to E, it is speeding up , because the magnitude of its slope is increasing
(changing from less steep to more steep).

(f) In the interval from D to E, the acceleration is positive , because the graph is concave upward,
indicating the slope is getting more positive, and thus the acceleration is positive.

(g) In the interval from C to D, the object is not moving in either direction .

The velocity and acceleration are both 0.

83. This problem can be analyzed as a series of three one-dimensional motions: the acceleration phase,
the constant speed phase, and the deceleration phase. The maximum speed of the train is as follows.

(95 km h ) ⎛ 1m s h ⎞ = 26.39 m s
⎜⎝ 3.6 km ⎟⎠

In the acceleration phase, the initial velocity is v0 = 0 m s , the acceleration is a = 1.1 m s2 , and
the final velocity is v = 26.39 m s. Find the elapsed time for the acceleration phase from Eq. 2-12a.

v = v0 + at → tacc = v − v0 = 26.39 m s−0 = 23.99 s
a 1.1 m s2

Find the displacement during the acceleration phase from Eq. 2-12b.

( )( )x − x0 (23.99 s)2 = 316.5 m
= v0t + 1 at 2 = 0+ 1 1.1m s2
2 2
acc

In the deceleration phase, the initial velocity is v0 = 26.39 m s, the acceleration is a = −2.0 m s2 ,
and the final velocity is v = 0 m s . Find the elapsed time for the deceleration phase from Eq. 2-12a.

v = v0 + at → tdec = v − v0 = 0 − 26.39 m s = 13.20 s
a −2.0 m s2

Find the distance traveled during the deceleration phase from Eq. 2-12b.

( )( x − at 2 (13.20s)2 = 174.1m
)x0 dec = v0t + 1 = (26.39 m s) (13.20s) + 1 −2.0 m s2
2 2

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The total elapsed time and distance traveled for the acceleration / deceleration phases are:
tacc + tdec = 23.99 s + 13.20 s = 37.19 s

( x − )x0 acc + ( x − )x0 dec = 316.5 m + 174.1 m = 491 m

(a) If the stations are spaced 1.80 km = 1800 m apart, then there is a total of 9000 m = 5 inter-
1800 m

station segments. A train making the entire trip would thus have a total of 5 inter-station

segments and 4 stops of 22 s each at the intermediate stations. Since 491 m is traveled during

acceleration and deceleration, 1800 m − 491m = 1309 m of each segment is traveled at an

average speed of v = 26.39 m s. The time for that 1309 m is given by Δx = v Δt →

Δtconstant = Δx = 1309 m = 49.60 s. Thus a total inter-station segment will take 37.19 s +
v 26.39 m s
speed

49.60 s = 86.79 s. With 5 inter-station segments of 86.79 s each, and 4 stops of 22 s each, the

total time is given by t0.8 km = 5(86.79 s) + 4 (22 s) = 522 s = 8.7 min .

(b) If the stations are spaced 3.0 km = 3000 m apart, then there is a total of 9000 m = 3 inter-
3000 m

station segments. A train making the entire trip would thus have a total of 3 inter-station

segments and 2 stops of 22 s each at the intermediate stations. Since 491 m is traveled during

acceleration and deceleration, 3000 m − 491m = 2509 m of each segment is traveled at an

average speed of v = 26.39 m s . The time for that 2509 m is given by d = vt →

t = d = 2509 m = 95.07 s. Thus a total inter-station segment will take 37.19 s + 95.07 s =
v 26.39 m s

132.3 s. With 3 inter-station segments of 132.3 s each, and 2 stops of 22 s each, the total time is

t3.0 km = 3(132.3s) + 2 (22 s) = 441s = 7.3min .

84. For the motion in the air, choose downward to be the positive direction, and y0 = 0 to be at the

height of the diving board. The diver has v0 = 0 (assuming the diver does not jump upward or

downward), a = g = 9.80 m s2 , and y = 4.0 m when reaching the surface of the water. Find the
diver’s speed at the water’s surface from Eq. 2-12c, with x replaced by y.

v2 = v02 + 2a ( y − y0 ) x →

( )v = ± v02 + 2a ( y − y0 ) = 0 + 2 9.80 m s2 (4.0 m ) = 8.85 m s

For the motion in the water, again choose down to be positive, but redefine y0 = 0 to be at the

surface of the water. For this motion, v0 = 8.85 m s , v = 0 , and y − y0 = 2.0 m . Find the
acceleration from Eq. 2-12c, with x replaced by y.

v2 = v02 + 2a ( y − y0 ) → a= v2 − v02 = 0 − (8.85 m s)2 = −19.6 m s2 ≈ −20 m s2
2 (2.0 m)
2( y − y0 ) x

The negative sign indicates that the acceleration is directed upwards.

85. Choose upward to be the positive direction, and the origin to be at the level where the ball was
thrown. The velocity at the top of the ball’s path will be v = 0, and the ball will have an

acceleration of a = − g. If the maximum height that the ball reaches is y = H , then the relationship

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42

Chapter 2 Describing Motion: Kinematics in One Dimension

between the initial velocity and the maximum height can be found from Eq. 2-12c, with x replaced

by y.

v2 = v02 + 2a ( y − y0 ) → 0 = v02 + 2 (− g ) H → H = v02 2g

(( )) (( ))It is given that H Bill v0 Bill 2 2g v0 Bill 2 = 1.52 2.3 .
v0 Bill = 1.5v0 Joe , so H Joe = v0 Joe 2 2g = v0 Joe 2 = 2.25 ≈

86. The v vs. t graph is found by taking the slope of the x vs. t graph.
Both graphs are shown here.

87. The car’s initial speed is vo = (45 km h) ⎛ 1m s h ⎞ = 12.5 m s.
⎝⎜ 3.6 km ⎟⎠

( )Case I: trying to stop. The constraint is, with the braking deceleration of the car a = −5.8 m s2 ,

can the car stop in a 28 m displacement? The 2.0 seconds has no relation to this part of the problem.

Using Eq. 2-12c, the distance traveled during braking is as follows.

v 2 − v02 0 − (12.5 m s)2
2a
2 −5.8 m s2
x0 ) =( )( x − = = 13.5 m → She can stop the car in time.

Case II: crossing the intersection. The constraint is, with the given acceleration of the car

⎣⎡⎢a = ⎝⎛⎜ 65 km h − 45 km h ⎠⎟⎞ ⎛ 1m s ⎞ = 0.9259 m s2 ⎤ , can she get through the intersection
6.0 s ⎝⎜ 3.6 km ⎠⎟ ⎥
h ⎦

(travel 43 meters) in the 2.0 seconds before the light turns red? Using Eq. 2-12b, the distance

traveled during the 2.0 sec is as follows.

( )( x − at 2 s2 (2.0 s)2 = 26.9 m
x0 ) = v0t + 1 = (12.5 m s ) ( 2.0 s) + 1 0.927 m
2 2

She should stop.

88. The critical condition is that the total distance covered by the passing car and the approaching car
must be less than 400 m so that they do not collide. The passing car has a total displacement
composed of several individual parts. These are: i) the 10 m of clear room at the rear of the truck, ii)
the 20 m length of the truck, iii) the 10 m of clear room at the front of the truck, and iv) the distance

the truck travels. Since the truck travels at a speed of v = 25 m s , the truck will have a

displacement of Δxtruck = (25 m s) t. Thus the total displacement of the car during passing is

Δxpassing = 40 m + (25 m s) t.
car

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

To express the motion of the car, we choose the origin to be at the location of the passing car when

the decision to pass is made. For the passing car, we have an initial velocity of v0 = 25 m s and an

acceleration of a = 1.0 m s2 . Find Δxpassing from Eq. 2-12b.

car

( ) ( )Δxpassing t2
= xc − x0 = v0t + 1 at = 25 m s t + 1 1.0 m s2
2 2

car

Set the two expressions for Δxpassing equal to each other in order to find the time required to pass.

car

( ) ( )( ) ( )40 m + s2 t2 s2 t2
25 m s tpass = 25 m s t +1 1.0 m pass → 40 m = 1 1.0 m pass →
pass 2 2

tpass = 80s2 = 8.94 s

Calculate the displacements of the two cars during this time.

Δxpassing = 40 m + (25 m s) (8.94 s) = 264 m
car

( ) ( )Δx = v t =approaching
approaching 25 m s 8.94 s = 224 m

car car

Thus the two cars together have covered a total distance of 488 m, which is more than allowed.

The car should not pass.

89. Choose downward to be the positive direction, and y0 = 0 to be at the height of the bridge. Agent

Bond has an initial velocity of v0 = 0, an acceleration of a = g, and will have a displacement of
y = 13 m − 1.5 m = 11.5 m. Find the time of fall from Eq. 2-12b with x replaced by y.

y = y0 + v0t + 1 at 2 → t= 2y 2 (11.5 m) = 1.532 s
2 a=
9.80 m s2

If the truck is approaching with v = 25 m s , then he needs to jump when the truck is a distance

away given by d = vt = (25 m s) (1.532 s) = 38.3 m. Convert this distance into “poles.”

d = (38.3 m ) (1 pole 25 m ) = 1.53 poles

So he should jump when the truck is about 1.5 poles away from the bridge.

90. Take the origin to be the location where the speeder passes the police car. The speeder’s constant

speed is vspeeder = (130 km h ) ⎛ 1 ms ⎞ = 36.1m s, and the location of the speeder as a function
⎝⎜ 3.6 km h ⎠⎟

( )of time is given by xspeeder = v tspeeder speeder = 36.1 m s t .speeder The police car has an initial velocity of

v0 = 0 m s and a constant acceleration of a .police The location of the police car as a function of time

is given by Eq. 2-12b: xpolice = v0t + 1 at 2 = a t .1 2
2
2 police police

(a) The position vs. time graphs would qualitatively look

like the graph shown here. x

(b) The time to overtake the speeder occurs when the speeder Speeder Police car
has gone a distance of 750 m. The time is found using the t t1
speeder’s equation from above.

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44

Chapter 2 Describing Motion: Kinematics in One Dimension

( )750 m = t 750 m 21s
36.1 m s t → speeder = 36.1m s = 20.8 s ≈
speeder

(c) The police car’s acceleration can be calculated knowing that the police car also had gone a
distance of 750 m in a time of 22.5 s.

750 m = 1 ap (20.8 s)2 → ap = 2 (750 m) = 3.47 m s2 ≈ 3.5 m s2
2 (20.8 s)2

(d) The speed of the police car at the overtaking point can be found from Eq. 2-12a.

( )v = v0 + at = 0 + 3.47 m s2 (20.8 s) = 72.2 m s ≈ 72 m s

Note that this is exactly twice the speed of the speeder.

91. The speed of the conveyor belt is given by d = v Δt → v d 1.1 m 0.44 m min . The rate
= Δt = 2.5 min =

of burger production, assuming the spacing given is center to center, can be found as follows.

⎛ 1 burger ⎞ ⎛ 0.44 m ⎞ = 2.9 burgers
⎝⎜ 0.15 m ⎟⎠ ⎜⎝ 1 min ⎠⎟ min

92. Choose downward to be the positive direction, and the origin to be at the top of the building. The

barometer has y0 = 0, v0 = 0, and a = g = 9.8 m s2 . Use Eq. 2-12b to find the height of the

building, with x replaced by y.

( )y t2
= y0 + v0t + 1 at 2 = 0+ 0 + 1 9.8 m s2
2 2

( ) ( )y =t=2.0 ( ) ( )y =t=2.3
1 9.8 m s2 2.0 s 2 = 20 m 1 9.8 m s2 2.3 s 2 = 26 m
2 2

The difference in the estimates is 6 m. If we assume the height of the building is the average of the

two measurements, then the % difference in the two values is 6m × 100 = 26% .
23 m

93. (a) The two bicycles will have the same velocity at A
any time when the instantaneous slopes of their

x vs. t graphs are the same. That occurs near the x B
time t1 as marked on the graph.
(b) Bicycle A has the larger acceleration, because

its graph is concave upward, indicating a positive

acceleration. Bicycle B has no acceleration because t

its graph has a constant slope. t1
(c) The bicycles are passing each other at the times

when the two graphs cross, because they both have the same position at that time. The graph

with the steepest slope is the faster bicycle, and so is the one that is passing at that instant. So at

the first crossing, bicycle B is passing bicycle A. At the second crossing, bicycle A is passing

bicycle B.

(d) Bicycle B has the highest instantaneous velocity at all times until the time t1, where both graphs
have the same slope. For all times after t1, bicycle A has the highest instantaneous velocity.
The largest instantaneous velocity is for bicycle A at the latest time shown on the graph.

(e) The bicycles appear to have the same average velocity. If the starting point of the graph for a

particular bicycle is connected to the ending point with a straight line, the slope of that line is

the average velocity. Both appear to have the same slope for that “average” line.

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

45

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

94. In this problem, note that a < 0 and x > 0. Take your starting position as 0. Then your position is

given by Eq. 2-12b, x1 = vM t + 1 at 2 , and the other car’s position is given by x2 = x + vAt. Set the
2

two positions equal to each other and solve for the time of collision. If this time is negative or

imaginary, then there will be no collision.

at 2 at 2
( )x1 = x2
→ vM t + 1 = x + vAt → 1 + vM − vA t−x=0
2 2

t = (vA − vM ) ± ( vM )− vA 2 − 4 1 a ( − x )
2

2 1 a
2

No collision: ( vM )− vA 2 −4 1 a(−x) < 0 → x > (vM )− vA 2
2
−2a

95. The velocities were changed from km/h to m/s by multiplying the conversion factor that 1 km/hr =
1/3.6 m/s.

(a) The average acceleration for each interval is calculated by a = Δv Δt , and taken to be the

acceleration at the midpoint of the time interval. In the spreadsheet, an + 1 = vn +1 − vn . The
2 tn +1 − tn

accelerations are shown in the table below.

(b) The position at the end of each interval is calculated by ( ) ( )xn+1
= xn + 1 vn + vn+1 tn+1 − tn .
2

This can also be represented as x = x0 + v Δt. These are shown in the table below.

t (s) v (km/h) v (m/s) t (s) a (m/s2) t (s) x (m)

0.0 0.0 0.0 0.25 3.33 0.0 0.00
0.5 6.0 1.7 0.75 4.00 0.5 0.42
1.0 13.2 3.7 1.25 5.06 1.0 1.75
1.5 22.3 6.2 1.75 5.50 1.5 4.22
2.0 32.2 8.9 2.25 6.00 2.0 8.00
2.5 43.0 11.9 2.75 5.83 2.5 13.22
3.0 53.5 14.9 3.25 5.06 3.0 19.92
3.5 62.6 17.4 3.75 4.44 3.5 27.99
4.0 70.6 19.6 4.25 4.33 4.0 37.24
4.5 78.4 21.8 4.75 3.72 4.5 47.58
5.0 85.1 23.6 5.0 58.94

(c) The graphs are shown below. The spreadsheet used for this problem can be found on the Media
Manager, with filename “PSE4_ISM_CH02.XLS,” on tab “Problem 2.95c.”

6.0 60

5.0 50

4.0 40
Acc (m/s2)
Distance (m)
3.0 30

2.0 20

1.0 10

0.0 0

0123450 12 34 5

Time (s) Time (s)

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

46

Chapter 2 Describing Motion: Kinematics in One Dimension

96. For this problem, a spreadsheet was designed. The columns of the spreadsheet are time,

acceleration, velocity, and displacement. The time starts at 0 and with each interval is incremented

by 1.00 s. The acceleration at each time is from the data t (s) a (m/s2) v (m/s) x (m)
given in the problem. The velocity at each time is found 0.0 1.25 0.0 0
by multiplying the average of the accelerations at the 1.0 1.58 1.4 1
current time and the previous time, by the time interval, 2.0 1.96 3.2 3
and then adding that to the previous velocity. Thus 3.0 2.40 5.4 7

( ) ( )vn+1 The displacement from
= vn + 1 an + an+1 tn+1 − tn . 4.0 2.66 7.9 14
2
5.0 2.70 10.6 23
the starting position at each time interval is calculated by a
6.0 2.74 13.3 35
constant acceleration model, where the acceleration is as
7.0 2.72 16.0 50
given above. Thus the positions is calculated as follows.
8.0 2.60 18.7 67
tn+1 − tn 2
( ) [ ( )]( )xn+1 = xn + vn
tn+1 − tn + 1 1 an + an+1 9.0 2.30 21.1 87
2 2

The table of values is reproduced here. 10.0 2.04 23.3 109

(a) v (17.00) = 30.3 m s 11.0 1.76 25.2 133
12.0 1.41 26.8 159
(b) x (17.00) = 305 m
13.0 1.09 28.0 187

The spreadsheet used for this problem can be found on the 14.0 0.86 29.0 215
Media Manager, with filename “PSE4_ISM_CH02.XLS,” 15.0 0.51 29.7 245
on tab “Problem 2.96.” 16.0 0.28 30.1 275
17.0 0.10 30.3 305

97. (a) For each segment of the path, 8.6
8.4
the time is given by the distance divided 8.2
8.0
by the speed.

t = tland + tpool = d land + d pool time to get to child (s)
vland vpool

x D2 + (d − x)2 7.8
= vR +
vS 7.6

(b) The graph is shown here. The minimum 7.4

time occurs at a distance along the pool of 7.2
0 1 2 3 4 5 6 7 8 9 10
about x = 6.8 m .
distance along pool (m)

An analytic differentiation to solve for the minimum point gives x = 6.76 m.

The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH02.XLS,” on tab “Problem 2.97b.”

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

47

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors

Responses to Questions

1. No. Velocity is a vector quantity, with a magnitude and direction. If two vectors have different
directions, they cannot be equal.

2. No. The car may be traveling at a constant speed of 60 km/h and going around a curve, in which case
it would be accelerating.

3. Automobile races that begin and end at the same place; a round-trip by car from New York to San
Francisco and back; a balloon flight around the world.

4. The length of the displacement vector is the straight-line distance between the beginning point and
the ending point of the trip and therefore the shortest distance between the two points. If the path is a
straight line, then the length of the displacement vector is the same as the length of the path. If the
path is curved or consists of different straight line segments, then the distance from beginning to end
will be less than the path length. Therefore, the displacement vector can never be longer than the
length of the path traveled, but it can be shorter.

5. The player and the ball have the same displacement.

G
6. V is the magGnitude oGf the vector V ; it is not necessarily larger than the magnitudes V1 and V2. For

instance, if V1 and V2 have the same magnitude as each other and are in opposite directions, then V
is zero.

7. The maximum magnitude of the sum is 7.5 km, in the case where the vectors are parallel. The
minimum magnitude of the sum is 0.5 km, in the case where the vectors are antiparallel.

8. No. The only way that two vectors can add up to give the zero vector is if they have the same
magnitude and point in exactly opposite directions. However, three vectors of unequal magnitudes
can add up to the zero vector. As a one-dimensional example, a vector 10 units long in the positive x
direction added to two vectors of 4 and 6 units each in the negative x direction will result in the zero
vector. In two dimensions, consider any three vectors that when added form a triangle.

9. (a) Yes. In three dimensions, the magnitude of a vector is the square root of the sum of the squares
of the components. If two of the components are zero, the magnitude of the vector is equal to
the magnitude of the remaining component.

(b) No.

10. Yes. A particle traveling around a curve while maintaining a constant speed is accelerating because
its direction is changing. A particle with a constant velocity cannot be accelerating, since the velocity
is not changing in magnitude or direction.

11. The odometer and the speedometer of the car both measure scalar quantities (distance and speed,
respectively).

12. Launch the rock with a horizontal velocity from a known height over level ground. Use the equations
for projectile motion in the y-direction to find the time the rock is in the air. (Note that the initial
velocity has a zero y-component.) Use this time and the horizontal distance the rock travels in the

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

48