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CHAPTER 1: INTRODUCTION: THE NATURE
OF SCIENCE AND PHYSICS
1.2 PHYSICAL QUANTITIES AND UNITS
1. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in
meters per second? (b) How many miles per hour is this?
Solution (a) 100hkm × 1000
1 km
m
×
1h
3600 s
= 27.77 m/s = 27.8 m/s
(b) 100hkm × 1.609
1 mi
km
= 62 mi/h
2. A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?
Solution (a) 33sm × 1000
1 km
m
×
3600 s
1h
= 118.8 km/h = 1.2 × 102 km/h
(b) At 120 km/h, the car is travelling faster than the speed limit.
3. Show that 1.0 𝑚/𝑠 = 3.6 𝑘𝑚/ℎ. Hint: Show the explicit steps involved in
converting 1.0 𝑚/𝑠 = 3.6 𝑘𝑚/ℎ.
Solution 1.0 m 1.0 m 3600 s 1 km
= × × = 3.6 km/h
s s hr 1000 m
4. American football is played on a 100-yd-long field, excluding the end zones. How
long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
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Solution 3 ft 1m
100 yd × × = 91.44 m = 91.4 m
1 yd 3.281 ft
5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its
dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
Solution 1 ft
115 m × = 377.3 ft = 377 ft long
0.3048 m
12 in.
377.3 ft × = 4528 in. = 4.53 × 103 in. long
1.0 ft
1 ft
85 m × = 278.9 ft = 2.8 × 102 ft wide
0.3048 m
12 in.
278.9 ft × = 3346 in. = 3.3 × 103 in. wide
1.0 ft
6. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1
meter equals 39.37 in.)
Solution 6ft, 1.0 in. = (6 ft × 12 in.) + 1.0 in. = 73.0 in.; 73.0 in.× 1 m = 1.85 m
ft 39.37 in.
7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its
height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
Solution 1 km
29,028 ft × = 8.847 km
3281 ft
8. The speed of sound is measured to be 342 𝑚/𝑠 on a certain day. What is this in
km/h?
Solution 342 m 1 km 3600 s
× × = 1.23 × 103 km/h
s 1000 m 1h
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9. Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose
that one such plate has an average speed of 4.0 cm/year. (a) What distance does it
move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
Solution (a) (4.0 cm × 1 s) × 1 m × 1 y × 1 d × 1 h = 1.3 × 10−9 m
y 100 cm 365.25 d 24.0 h 3600 s
6
(b) 4.0ycm × 100
1m
cm
×
1 km
1000 m
×
10 y
1 My
= 40 km/My
10. (a) Refer to Table 1.3 to determine the average distance between the Earth and the
Sun. Then calculate the average speed of the Earth in its orbit in kilometers per
second. (b) What is this in meters per second?
Solution (a) 𝑣 = 𝑑 = 2𝜋𝑟 = 2𝜋( 108 km) × 1 d × 1 h = 20 km/s
𝑡 𝑡 365.25 d 24 h 3600 s
(b) 𝑣 = 20 skm × 1000
1 km
m
= 2.0 × 104 m/s
1.3 ACCURACY, PRECISION, AND SIGNIFICANT FIGURES
11. Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty.
What is the uncertainty in your mass (in kilograms)?
Solution 3%
𝛿m= × 65 kg = 2 kg
100%
12. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What
is its percent uncertainty?
Solution 𝛿𝑙 0.50 cm 1m
% 𝑢𝑛𝑐 𝑙 = × 100% = × × 100% = 2.5 × 10−2 %
𝑙 20 m 100 cm
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CHAPTER 1: INTRODUCTION: THE NATURE
OF SCIENCE AND PHYSICS
1.2 PHYSICAL QUANTITIES AND UNITS
1. The speed limit on some interstate highways is roughly 100 km/h. (a) What is this in
meters per second? (b) How many miles per hour is this?
Solution (a) 100hkm × 1000
1 km
m
×
1h
3600 s
= 27.77 m/s = 27.8 m/s
(b) 100hkm × 1.609
1 mi
km
= 62 mi/h
2. A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?
Solution (a) 33sm × 1000
1 km
m
×
3600 s
1h
= 118.8 km/h = 1.2 × 102 km/h
(b) At 120 km/h, the car is travelling faster than the speed limit.
3. Show that 1.0 𝑚/𝑠 = 3.6 𝑘𝑚/ℎ. Hint: Show the explicit steps involved in
converting 1.0 𝑚/𝑠 = 3.6 𝑘𝑚/ℎ.
Solution 1.0 m 1.0 m 3600 s 1 km
= × × = 3.6 km/h
s s hr 1000 m
4. American football is played on a 100-yd-long field, excluding the end zones. How
long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
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Solution 3 ft 1m
100 yd × × = 91.44 m = 91.4 m
1 yd 3.281 ft
5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m wide. What are its
dimensions in feet and inches? (Assume that 1 meter equals 3.281 feet.)
Solution 1 ft
115 m × = 377.3 ft = 377 ft long
0.3048 m
12 in.
377.3 ft × = 4528 in. = 4.53 × 103 in. long
1.0 ft
1 ft
85 m × = 278.9 ft = 2.8 × 102 ft wide
0.3048 m
12 in.
278.9 ft × = 3346 in. = 3.3 × 103 in. wide
1.0 ft
6. What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1
meter equals 39.37 in.)
Solution 6ft, 1.0 in. = (6 ft × 12 in.) + 1.0 in. = 73.0 in.; 73.0 in.× 1 m = 1.85 m
ft 39.37 in.
7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its
height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)
Solution 1 km
29,028 ft × = 8.847 km
3281 ft
8. The speed of sound is measured to be 342 𝑚/𝑠 on a certain day. What is this in
km/h?
Solution 342 m 1 km 3600 s
× × = 1.23 × 103 km/h
s 1000 m 1h
Page 3 OpenStax College Physics 2e Instructor Solutions
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9. Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose
that one such plate has an average speed of 4.0 cm/year. (a) What distance does it
move in 1.0 s at this speed? (b) What is its speed in kilometers per million years?
Solution (a) (4.0 cm × 1 s) × 1 m × 1 y × 1 d × 1 h = 1.3 × 10−9 m
y 100 cm 365.25 d 24.0 h 3600 s
6
(b) 4.0ycm × 100
1m
cm
×
1 km
1000 m
×
10 y
1 My
= 40 km/My
10. (a) Refer to Table 1.3 to determine the average distance between the Earth and the
Sun. Then calculate the average speed of the Earth in its orbit in kilometers per
second. (b) What is this in meters per second?
Solution (a) 𝑣 = 𝑑 = 2𝜋𝑟 = 2𝜋( 108 km) × 1 d × 1 h = 20 km/s
𝑡 𝑡 365.25 d 24 h 3600 s
(b) 𝑣 = 20 skm × 1000
1 km
m
= 2.0 × 104 m/s
1.3 ACCURACY, PRECISION, AND SIGNIFICANT FIGURES
11. Suppose that your bathroom scale reads your mass as 65 kg with a 3% uncertainty.
What is the uncertainty in your mass (in kilograms)?
Solution 3%
𝛿m= × 65 kg = 2 kg
100%
12. A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What
is its percent uncertainty?
Solution 𝛿𝑙 0.50 cm 1m
% 𝑢𝑛𝑐 𝑙 = × 100% = × × 100% = 2.5 × 10−2 %
𝑙 20 m 100 cm
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