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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.

Chapter 1 Introduction, Measurement, Estimating

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( d ) 4.76 × 10 2 = 476 ( e ) 3.62 × 10 −^5 =0.

0.25 m % uncertainty 100% 4.6% 5.48 m

= × =

6. ( a )

0.2 s % uncertainty 100% 4% 5 s

= × =

( b )

0.2 s % uncertainty 100% 0.4% 50 s

= × =

( c )

0.2 s % uncertainty 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.

3 4 6 3 3 3

3 3 5

9.2 10 s 8.3 10 s 0.008 10 s 9.2 10 s 83 10 s 8 10 s

9.2 83 8 10 s 100.2 10 s 1.00 10 s

× + × + × = × + × + ×

= + + × = × = ×

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

8. ( 2.079 × 10 m^2 ) ( 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 )

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

4 3 4 3 3 specified 3 specified 3 4 3 4 3 3 min 3 min 3 4 3 4 3 3 max 3 max 3

0.84 m 2.483m

0.80 m 2.145 m

0.88 m 2.855 m

V r

V r

V r

Δ V = 12 ( V max − V min) = 12 ( 2.855 m 3 − 2.145 m 3 )=0.355 m^3

The percent uncertainty is

3 3 specified

0.355 m 100 14.3 14 % 2.483m

V

V

= × = ≈

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

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11. ( a ) 286.6 mm 286.6 × 10 −^3 m 0.286 6 m

( b ) 85 μ V 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 ×10 volts 9 2, 500, 000, 000 volts

12. ( a ) 1 ×10 volts 6 1 megavolt =1 Mvolt

( b ) 2 × 10 −^6 meters 2 micrometers = 2 μm ( c ) 6 ×10 days (^3) 6 kilodays =6 kdays

( d ) 18 × 10 bucks^2 18 hectobucks = 18 hbucksor 1.8 kilobucks ( e ) 8 × 10 −^8 seconds 80 nanoseconds =80 ns

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/s^2.

14. ( a ) 93 million miles = ( 93 × 10 miles^6 )(1610 m 1 mile )= 1.5 × 1011 m

( b ) 1.5 × 10 11 m = 150 × 10 m^9 = 150 gigameters or 1.5 × 1011 m = 0.15 × 10 12 m =0.15 terameters

15. ( a ) ( ) ( )

2 2 2 2 1 ft = 1 ft 1 yd 3 ft = 0.111 yd , and so the conversion factor is

2 2

0.111 yd 1 ft

( b ) ( )( )

2 2 2 2 1 m = 1 m 3.28 ft 1 m = 10.8 ft , and so the conversion factor is

2 2

10.8 ft 1m

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

1 h 3600 s 1.00 km 3.8s 950 km 1 h

t = d v = ⎛⎜^ ⎞ ⎛⎟ ⎜^ ⎞⎟=

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

( b ) ( ) 10 8

1 m 1 atom 1.0 cm 1.0 10 atoms 100 cm 1.0 10 −^ m

= ×

×

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

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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 2 4 4

1 book 3500 m 7.0 10 books. 0.25 m 0.05 m

= ×

With 8 shelves of books, the total number of

books stored is as follows.

books 7.0 10 8 shelves 6 10 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 ×10 L^4

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 m^2. The mower has a cutting width of 0.5 meters. Thus the distance to be walked is as follows. area 5500 m^2 11000 m 11 km width 0.5 m

d = = = =

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.

( 6 )

visits 2 year 1 yr^ 1 dentist 2 10 people 1800 dentists 1 person 225 workdays visits 10 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

Chapter 1 Introduction, Measurement, Estimating

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To 1st sunset

To 2nd sunset

A B

Earth center

R R

d h

θ

θ

that is wearing. Also assume that there are 1.5 × 10 8 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.

( ) ( 3 ) ( 8 )^8

mass surface area thickness wear density of rubber # of tires year tire year 2 0.4 m 0.1m 0.002 m y 1200 kg m 6.0 10 tires 4 10 kg y 1 tire

= × = ×

⎜ ⎟ ⎜^ ⎟⎜ ⎟

⎝ ⎠ ⎝^ ⎠⎝ ⎠

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

( ) (^ )^ (^ )^ (^ )

(^2 2 2 2 2 2 )

2 2 2

6 2 4 4

2

2 2

2 6.4 10 m 200 m 200 m 5.1 10 m 5 10 m 80 mi

r h r d r rh h r d

rh h d d rh h

d

• = + → + + = +

• = → = +

= × + = × ≈ × ≈

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 2( t^ −^1 )on the t th^ day. On the first day, you get $0.01 2( 1 1 −^ )= $0.01. On the second day,

you get $0.01 2( 2 1−^ )= $0.02. On the third day, you get $0.01 2( 3 1 −^ )= $0.04. On the 30th^ day, you

get $0.01 2( 30 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 , h^2^  2 Rh.

2 2 2 2 2 2 2 2

2 2 6

4400 m 6.5 10 m 2 2 1.5 m

d R R h R Rh h d Rh h d Rh

d R h

= = = ×

A better measurement gives R = 6.38 ×10 m.^6

34. To see the Sun “disappear,” your line of sight to the top of the Sun is tangent to the Earth’s surface. Initially, you are lying down at point A, and you see the first sunset. Then you stand up, elevating your eyes by the height h. While standing, your line of sight is tangent to the Earth’s surface at point B, and so that is the direction to

the second sunset. The angle θ is the angle through

which the Sun appears to move relative to the Earth during the time to be measured. The distance d is the distance from your eyes when standing to point B.

Use the Pythagorean theorem for the following relationship.

d^2 + R^2 = R + h^2 = R^2 + 2 Rh + h^2^ → d^2 = 2 Rh + h^2

h

r r

d

Chapter 1 Introduction, Measurement, Estimating

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39. The percentage accuracy is 7 5

2 m 100% 1 10 % 2 10 m

× = × −.

×

The distance of 20,000,000 m needs to

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. chips 1 wafer 250 mm chips 100 83, 000 wafer 0.300 mm 1 cylinder cylinder

⎛ ⎞⎛ ⎞^ ⎛^ ⎞

41. ( a ) # of seconds in 1.00 y: ( )

7 1.00 y 1.00 y 3.156^ 10 s 3.16 10 s^7 1 y

×

= = ×

( b ) # of nanoseconds in 1.00 y: ( )

7 9 1.00 y 1.00 y 3.156^ 10 s^1 10 ns 3.16 10 ns^16 1 y 1 s

× ×

= = ×

( c ) # of years in 1.00 s: ( ) 7 8

1 y 1.00 s 1.00 s 3.17 10 y 3.156 10 s

= = ×^ −

×

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

soccer football

soccer football

1.09 yd 100 m 100 yd 9 yd 1m 1m 100 m 100 yd 8 m 1.09 yd

L L

L L

− = × − =

− = − × =

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 m^3. The diameter can be found from the volume of a sphere, 43 π r^3.

3 3 4 3 4 3 3

3 3 1/ 3 8 3 3 3 4 8

3 10 2 10 m m 2 10 m 6 3 10

d r d

d d

− − −

×

× = × → = = ×

×

4 2 2 4 2

1.000 10 m 3.281ft 1acre 1 hectare 1 hectare 2.471acres 1hectare 1m 4.356 10 ft

×

×

⎜ ⎟ ⎜ ⎟ ⎜^ ⎟

⎝ ⎠ ⎝ ⎠⎝^ ⎠

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 10 people^8 1 automobile^ 12, 000^ 1 gallon 1 1011 gal y 2 people 1 y 20 mi

mi auto

× ≈ ×

⎜ ⎟ ⎜ ⎟ ⎜^ ⎟

⎝ ⎠ ⎝ ⎠⎝^ ⎠

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

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46. ( a )

15 12 27

10 kg 1 proton or neutron 10 protons or neutrons 1 bacterium 10 kg

− − =

⎝ ⎠⎝^ ⎠

( b )

17 10 27

10 kg 1 proton or neutron 10 protons or neutrons 1 DNA molecule 10 kg

− − =

( c )

2 29 27

10 kg 1 proton or neutron 10 protons or neutrons 1 human 10 −^ kg

( d )

41 68 27

10 kg 1 proton or neutron 10 protons or neutrons 1 galaxy 10 −^ kg

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

3 3 4 3 3 5

1200 L day 365day 1000 cm 1km 4 10 people 4.38 10 km y 4 people 1 y 1L 10 cm

× ⎛^ ⎞ ⎛^ ⎞^ ⎛^ ⎞⎛^ ⎞ = ×^ −

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜^ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝^ ⎠

The depth of water is found by dividing the volume by the area. 3 3 5 5 2

4.38 10 km y km 10 cm 8.76 10 8.76 cm y 9 cm y 50 km y 1 km

V

d A

−

= = × = ⎛^ × − ⎞^ ⎛^ ⎞= ≈

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 10 people^6 1.5 loads week^ 52 weeks^ 0.1kg^ 2.34 109 kg^2 109 kg 1person 1 y 1load y y

× × × × = × ≈ ×

50. The volume of a sphere is given by V = 43 π r^3 ,and so the radius is

V

r

= ⎛⎜^ ⎞⎟

For a 1-ton rock,

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

3 1 T 2000 lb^ 1ft 10.8 ft^3 1 T 186 lb

V = =

⎛ ⎞^ ⎛^ ⎞

1/ 3 3 1/ 3 3 3 10.8 ft 2 2 2 2.74 ft 3 ft 4 4

V

d r

⎛ ⎞ ⎡^ ⎤

⎜ ⎟ ⎢^ ⎥

8 bits 1sec 1min 783.216 10 bytes 74.592 min 75 min 1byte 1.4 10 bits 60 sec

× × × × = ≈

×

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

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 60^ o tan 60 o^ 120 yd tan 60o 210 yd

3ft 0.305 m 210 yd 190 m 1yd 1ft

d = → d = = =

l

l

8 s 1 y 100% 3 10 % 1 y 3.156 10 s

× = ×^ −

×

⎜ ⎟ ⎜^ ⎟

⎝ ⎠⎝^ ⎠

59. ( a )

o o^10 o 9

10 m 1 nm 1.0 A 1.0 A 0.10 nm 1A^10 m

− = (^) − =

⎛ ⎞^ ⎛^ ⎞ ⎛ ⎞

⎜ ⎟ ⎜⎜^ ⎟⎟⎜ ⎟

( b )

o o 10 5 o 15

10 m 1 fm 1.0 A 1.0 A 1.0 10 fm 1A^10 m

− = (^) − = ×

⎛ ⎞^ ⎛^ ⎞⎛^ ⎞

⎜ ⎟ ⎜⎜^ ⎟⎟⎜ ⎟

( c ) ( )

o 10 o 10

1A

1.0 m 1.0 m 1.0 10 A 10 − m

= = ×

( d ) ( )

15 o 25 o 10

9.46 10 m 1A 1.0 ly 1.0 ly 9.5 10 A 1 ly 10 − m

×

= = ×

⎛ ⎞^ ⎛^ ⎞

⎜ ⎟ ⎜^ ⎟

⎝ ⎠ ⎜^ ⎟

60. The volume of a sphere is found by V =^43 π r^3.

4 3 4 6 3 19 3 V Moon (^) = 3 π R Moon= 3 π 1.74 × 10 m = 2.21 × 10 m

4 3 3 6 3 Earth 3 Earth Earth 4 3 6 Moon 3 Moon Moon

6.38 10 m

1.74 10 m

V R R

V R R

×

×

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

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

⎛ ⎞^ ⎛^ ⎞

sin 8 103 100 100 3% sin 0.

θ θ

Δ × −

⎛ ⎞^ ⎛^ ⎞

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

⎛ ⎞^ ⎛^ ⎞

sin 2 103 100 100 0.2% sin 0.

Δ × −

⎛ ⎞^ ⎛^ ⎞

A consequence of this result is that when using a protractor, and you have a fixed uncertainty in the angle ( ±0.5 oin 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.

l

d

60 o

Chapter 1 Introduction, Measurement, Estimating

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

o ( 3 )

o

1 2 6.38^ 10 km 0.621 mi 1 minute 1.15 mi 60 minute 360 1 km

π ×

⎛ ⎞ ⎛^ ⎞⎛ ⎞

⎜ ⎟ ⎜⎜^ ⎟ ⎜⎟^ ⎟

⎝ ⎠ ⎝ ⎠⎝^ ⎠

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.

2 2 2 3 2 3

V = π r h = π 0.12 m 1.7 m = 7.69 × 10 −^ m ≈ 8 × 10 −m

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. 1bus 1driver 50, 000 passengers 1000 drivers 50 passengers 1bus

× × ≈

( 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. 1bus 1driver 2500 passengers 50 drivers 50 passengers 1bus

× × ≈

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

[ ][ ] [ ] [ ]

[ ][ ] [ ] [ ]

[ ]

L

units of 4.1 m L units of 4. m

L units of 0.018 y L units of 0. y

units of 2.69 L

mass 8 g density 2.82 g cm 3g cm volume 2.8325cm

67. ( a )

2 2 3 2 Earth Earth Earth 2 2 3 2 Moon Moon Moon

SA 4 6.38^ 10 km

SA (^4) 1.74 10 km

R R

R R

π π

×

×

( b )

4 3 3 3 3 Earth 3 Earth Earth 4 3 3 3 3 Moon 3 Moon Moon

V 6.38^ 10 km

V (^) 1.74 10 km

R R

R R

π π

×

×

23 23 9 2 2 6 2 2 Earth

atoms 6.02 10 atoms 6.02 10 atoms atoms

1.18 10

m 4 π R 4 π 6.38 10 m m

× ×

= = = ×

×

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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.

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

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 2 x. The total time required to travel this distance is x /70 plus x /90. Then 2 2(90)(70) 79 km/h. 70 90 90 70

d x v t x x

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.

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

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7. The distance traveled is 116 km + 12 ( 116 km) = 174 km,and the displacement is

116 km − 12 ( 116 km )= 58 km.The total time is 14.0 s + 4.8 s = 18.8 s.

( a ) Average speed =

distance 174 m 9.26 m s time elapsed 18.8 s

( b ) Average velocity = (^) avg

displacement 58 m 3.1m s time elapsed 18.8 s

v = = =

8. ( a )

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.

( ) ( ) (^ )^ (^ )^ (^ )

3 3.0 0.0 34 10 3.0^ 2 3.0^ m^ 34 m 8.0 m s 3.0s 0.0s 3.0s

x x v

− +^ −^ −

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

10 6 2 m s 10 6 2 0 5 s 1.3s 3

dx v t t t dt

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 ( )

3 m 0 10 0.3 m s 10.0 s 0

v.

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

approximately ( )

22 m 10 m 30 1.2 m s 35s 25s

v.

( c ) The average velocity is given by

( ) 5 ( ) 0 1.5 m 0

0.30 m s 5.0 s 0 s 5.0 s

x x v

( d ) The average velocity is given by

( 30 )^ ( 25 )^ 16 m 9 m

1.4 m s 30.0 s 25.0 s 5.0 s

x x v

( e ) The average velocity is given by

( 50 ) ( 40 ) 10 m^ 19.5 m^

0.95 m s 50.0 s 40.0 s 10.0 s

x x v

0

10

20

30

40

50

0.0 0.5 1.0 1.5 2.0 2.5 3. t (sec)

x^

(m)

Chapter 2 Describing Motion: Kinematics in One Dimension

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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

6 6

1.2 m 1bit 4.3 10 bits s 1s 0.28 10 m

N = × − = ×

×

( b ) The number of excess bits is N − N 0.

6 6 6 N − N 0 = 4.3 × 10 bits s − 1.4 × 10 bits s = 2.9 ×10 bits s 6 0 6

2.9 10 bits s 0.67 67% 4.3 10 bits s

N N

N

− ×

×

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

distance the car travels will be Δ x car = Δ x truck + 110 m. Use Eq. 2-2 for average speed, v = Δ x Δ t ,

solve for time, and equate the two times. truck car truck truck truck car

110 m 75 km h 95 km h

x x x x t v v

Solving for Δ x truck gives ( )

truck ( )

75 km h 110 m 412.5 m. 95 km h 75 km h

Δ x = = −

The time of travel is truck^1 truck

412.5 m 60 min 0.33 min 19.8s 2.0 10 s 75000 m h 1h

x t v

Δ = = = = = ×

Also note that car car

412.5 m 110 m 60 min 0.33 min 20 s. 95000 m h 1h

x t v

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. 0.11 km 3600 s 19.8 s 20 km h 1 h

Δ = t ⎛⎜^ ⎞⎟=

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. 4.25 km 60 min 0.0447 h 2.68 min 2.7 min 95 km h 1 h

x x v t t v

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

( ) (^ )^ (^ )

2 2 2 1 2 2 2 2 2 1 3 6

0.058 m 0.025 m 5.378 10 m 5400 m 1.6 10 m

R R w

R R w

π π

π π −

= = = × ≈

×

l

l

( b ) 3

1s 1min 5.378 10 m 72 min 1.25 m 60 s

× =