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UNITS, PHYSICAL QUANTITIES AND VECTORS 11.1. IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.EXECUTE: (a) 1.00 mi = (1.00 mi) ⎛ 5280 ft ⎞ ⎛ 12 in.⎞ ⎛ 2.54 cm⎞ ⎛ 1m ⎞ ⎛ 1 km ⎞ = 1.61 km ⎝⎜ 1 mi ⎠⎟ ⎝⎜ 1 ... [收起]
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UNITS, PHYSICAL QUANTITIES AND VECTORS 1

1.1. IDENTIFY: Convert units from mi to km and from km to ft.
SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.

EXECUTE: (a) 1.00 mi = (1.00 mi) ⎛ 5280 ft ⎞ ⎛ 12 in.⎞ ⎛ 2.54 cm⎞ ⎛ 1m ⎞ ⎛ 1 km ⎞ = 1.61 km
⎝⎜ 1 mi ⎠⎟ ⎝⎜ 1 ft ⎟⎠ ⎝⎜ 1 in. ⎟⎠ ⎝⎜ 102 cm⎟⎠ ⎜⎝ 103 m⎟⎠

(b) 1.00 km = (1.00 km) ⎛ 103 m ⎞⎛ 102 cm ⎞ ⎛ 1 in. ⎞⎛ 1 ft ⎞ = 3.28 ×103 ft
⎝⎜⎜ 1 km ⎠⎟⎟⎝⎜⎜ 1 m ⎠⎟⎟ ⎝⎜ 2.54 cm ⎠⎟⎝⎜ 12 in. ⎠⎟

EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in

a km.

1.2. IDENTIFY: Convert volume units from L to in.3.

SET UP: 1 L = 1000 cm3. 1 in. = 2.54 cm

EXECUTE: 0.473 L × ⎛ 1000 cm3 ⎞ × ⎛ 1 in. ⎞3 = 28.9 in.3.
⎜⎜⎝ 1L ⎟⎟⎠ ⎜⎝ 2.54 cm ⎟⎠

EVALUATE: 1 in.3 is greater than 1 cm3, so the volume in in.3 is a smaller number than the volume in

cm3, which is 473 cm3.

1.3. IDENTIFY: We know the speed of light in m/s. t = d/v. Convert 1.00 ft to m and t from s to ns.

SET UP: The speed of light is v = 3.00 ×108 m/s. 1 ft = 0.3048 m. 1 s = 109 ns.

EXECUTE: t = 0.3048 m = 1.02 ×1029 s = 1.02 ns
3.00 ×108 m/s

EVALUATE: In 1.00 s light travels 3.00 ×108 m = 3.00 ×105 km = 1.86 ×105 mi.

1.4. IDENTIFY: Convert the units from g to kg and from cm3 to m3.

SET UP: 1 kg = 1000 g. 1 m = 1000 cm.

EXECUTE: 19.3 g × ⎛ 1 kg ⎞ ⎛ 100 cm ⎞3 = 1.93×104 kg
cm3 ⎜ ⎟ × ⎜ ⎟ m3
⎝ 1000 g ⎠ ⎝ 1m ⎠

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3.

1.5. IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L = 1000 cm3. 1 in. = 2.54 cm.
EXECUTE: (327 in.3) × (2.54 cm/in.)3 × (1 L/1000 cm3) = 5.36 L

EVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3, so the volume in cm3 is a larger number

than the volume in in.3.
1.6. IDENTIFY: Convert ft2 to m2 and then to hectares.

SET UP: 1.00 hectare = 1.00 ×104 m2. 1 ft = 0.3048 m.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-1

1-2 Chapter 1

EXECUTE: The area is (12.0 acres) ⎛ 43,600 ft2 ⎞ ⎛ 0.3048 m ⎞2 ⎛ 1.00 hectare ⎞ = 4.86 hectares.
⎝⎜⎜ 1 acre ⎠⎟⎟ ⎝⎜ 1.00 ft ⎠⎟ ⎝⎜ 1.00 ×104 m2 ⎠⎟

EVALUATE: Since 1 ft = 0.3048 m, 1 ft2 = (0.3048)2 m2.

1.7. IDENTIFY: Convert seconds to years.
1.8.
1.9. SET UP: 1 billion seconds = 1×109 s. 1 day = 24 h. 1 h = 3600 s.
1.10.
EXECUTE: 1.00 billion seconds = (1.00 × 109 s) ⎝⎜⎛ 1h ⎞ ⎛ 1 day⎞ ⎛ 1y ⎞ = 31.7 y.
1.11. 3600 s ⎟⎠ ⎝⎜ 24 h ⎠⎟ ⎝⎜ 365 days⎠⎟

EVALUATE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d, which is the average for one

extra day every four years, in leap years. The problem says instead to assume a 365-day year.
IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.

EXECUTE: (180,000 furlongs/fortnight) ⎛ 0.125 mi ⎞⎛ 1 fortnight ⎞⎛ 1 day ⎞ = 67 mi/h
⎜ 1 furlong ⎟⎜ 14 days ⎟⎠ ⎜⎝ 24 h ⎠⎟
⎝ ⎠⎝

EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a

much smaller number.

IDENTIFY: Convert miles/gallon to km/L.

SET UP: 1 mi = 1.609 km. 1 gallon = 3.788 L.

EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon ) ⎝⎜⎛ 1.609 km ⎞⎛ 1 gallon ⎞ = 23.4 km/L.
1 mi ⎠⎟⎜⎝ 3.788 L ⎠⎟

(b) The volume of gas required is 1500 km = 64.1 L. 64.1 L = 1.4 tanks.
23.4 km/L 45 L/tank

EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a

gallon, so 1 mi/gal ∼ 2 km/L, which is roughly our result.
4

IDENTIFY: Convert units.

SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg.

EXECUTE: (a) ⎛ 60 mi ⎞⎛ 1h ⎞⎛ 5280 ft ⎞ = 88 ft
⎝⎜ ⎠⎟ ⎜ 3600 ⎟⎜ 1 mi ⎟ s
h ⎝ s ⎠⎝ ⎠

(b) ⎛ 32 ft ⎞ ⎛ 30.48 cm ⎞ ⎛ 1 m ⎞ = 9.8 m
⎜⎝ s2 ⎠⎟ ⎜ ⎟ ⎜⎝ ⎟⎠ s2
⎝ 1 ft ⎠ 100 cm

(c) ⎛⎝⎜1.0 g ⎞⎛ 100 cm ⎞3 ⎛ 1 kg ⎞ = 103 kg
cm3 ⎠⎟ ⎝⎜ ⎟⎠ ⎜ ⎟ m3
1m ⎝ 1000 g ⎠

EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation

32 ft/s2 = 9.8 m/s2 is accurate to only two significant figures.

IDENTIFY: We know the density and mass; thus we can find the volume using the relation
density = mass/volume = m/V . The radius is then found from the volume equation for a sphere and the

result for the volume.

SET UP: Density = 19.5 g/cm3 and mcritical = 60.0 kg. For a sphere V = 4 π r 3.
3

EXECUTE: V = mcritical/density = ⎛ 60.0 kg ⎞⎛ 1000 g ⎞ = 3080 cm3.
⎜⎝⎜ 19.5 g/cm3 ⎠⎟⎟ ⎜ ⎟
⎝ 1.0 kg ⎠

r=3 3V =3 3 (3080 cm3 ) = 9.0 cm.
4π 4π

EVALUATE: The density is very large, so the 130-pound sphere is small in size.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Units, Physical Quantities and Vectors 1-3

1.12. IDENTIFY: Convert units.
SET UP: We know the equalities 1 mg = 10−3 g, 1 µg 10−6 g, and 1 kg = 103 g.
1.13.
1.14. EXECUTE: (a) (410 mg/day) ⎛ 10−3 g ⎞ ⎛ 1 μg ⎞ = 4.10 ×105 μg/day.
⎜ 1 mg ⎟ ⎜ 10−6 g ⎟
1.15. ⎝ ⎠ ⎝ ⎠
1.16.
1.17. (b) (12 mg/kg)(75 kg) = (900 mg) ⎛ 10−3 g ⎞ = 0.900 g.
⎜⎜⎝ 1 mg ⎠⎟⎟

(c) The mass of each tablet is (2.0 mg) ⎛ 10−3 g ⎞ = 2.0 ×10−3 g/day. The number of tablets required each
⎜ 1 mg ⎟
⎝ ⎠

day is the number of grams recommended per day divided by the number of grams per tablet:

0.0030 g/day = 1.5 tablet/day. Take 2 tablets each day.
2.0 ×10−3 g/tablet

(d) (0.000070 g/day) ⎛ 1 mg ⎞ = 0.070 mg/day.
⎝⎜⎜ 10−3 g ⎟⎟⎠

EVALUATE: Quantities in medicine and nutrition are frequently expressed in a wide variety of units.
IDENTIFY: The percent error is the error divided by the quantity.
SET UP: The distance from Berlin to Paris is given to the nearest 10 km.

EXECUTE: (a) 10 m m = 1.1×10−3,.
890 ×103

(b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total

distance to only three significant figures.

EVALUATE: In this case a very small percentage error has disastrous consequences.
IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be
no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the

location of the decimal that matters.
SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures.

EXECUTE: (a) (12 mm) × (5.98 mm) = 72 mm2 (two significant figures)

(b) 5.98 mm = 0.50 (also two significant figures)
12 mm

(c) 36 mm (to the nearest millimeter)
(d) 6 mm
(e) 2.0 (two significant figures)
EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and
(d) are known only to the nearest mm.

IDENTIFY: Use your calculator to display π ×107. Compare that number to the number of seconds in a year.
SET UP: 1 yr = 365.24 days, 1 day = 24 h, and 1 h = 3600 s.

EXECUTE: (365.24 days/1 yr ) ⎜⎛ 24 h ⎞⎛ 3600 s ⎞ = 3.15567…×107 s; π ×107 s = 3.14159…×107 s
⎝ 1 day ⎟ ⎝⎜ 1h ⎟⎠

The approximate expression is accurate to two significant figures. The percent error is 0.45%.
EVALUATE: The close agreement is a numerical accident.
IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the
number of gallons.

SET UP: Estimate 3 ×108 people, so 2 ×108 cars.
EXECUTE: (Number of cars × miles/car day)/(mi/gal) = gallons/day

(2 ×108 cars ×10000 mi/yr/car ×1 yr/365 days)/(20 mi/gal) = 3×108 gal/day

EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S.
IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express
200 months in years.

© Copyright 2012 Pearson Education, Inc. 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.

1-4 Chapter 1

SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in. = 2.54 cm. 1 y = 12 months.

EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man.

(b) 200 m = (2.00 ×104 cm) ⎛ 1 in. ⎞ = 7.9 × 103 inches. This is much greater than the height of a
⎝⎜ 2.54 cm ⎠⎟

person.
(c) 200 cm = 2.00 m = 79 inches = 6.6 ft. Some people are this tall, but not an ordinary man.
(d) 200 mm = 0.200 m = 7.9 inches. This is much too short.

(e) 200 months = (200 mon) ⎛ 1y ⎞ = 17 y. This is the age of a teenager; a middle-aged man is much
⎜⎝ 12 mon ⎟⎠

older than this.

EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give
the units in which it is being expressed.

1.18. IDENTIFY: The number of kernels can be calculated as N = Vbottle/Vkernel.
1.19.
1.20. SET UP: Based on an Internet search, Iowa corn farmers use a sieve having a hole size of 0.3125 in. ≅
8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as
1.21.
6 mm and the depth as 3 mm. We must also apply the conversion factors 1 L = 1000 cm3 and 1 cm = 10 mm.

EXECUTE: The volume of the kernel is: Vkernel = (10 mm)(6 mm)(3 mm) = 180 mm3. The bottle’s volume

is: Vbottle = (2.0 L)[(1000 cm3)/(1.0 L)][(10 mm)3 /(1.0 cm)3] = 2.0 × 106 mm3. The number of kernels is

then Nkernels = Vbottle/Vkernels ≈ (2.0 ×106 mm3)/(180 mm3) = 11,000 kernels.

EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since
these dimensions vary amongst the different available types of corn, acceptable answers could range from
6,500 to 20,000.
IDENTIFY: Estimate the number of pages and the number of words per page.
SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has
between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter
exercises and problems).

EXECUTE: An estimate for the number of words is about 106.

EVALUATE: We can expect that this estimate is accurate to within a factor of 10.

IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to

find the volume in m3 breathed in a year.

SET UP: Assume 10 breaths/min. 1 y = (365 d) ⎜⎝⎛ 24 h ⎞⎛ 60 min ⎞ = 5.3 × 105 min. 102 cm =1 m so
1d ⎠⎟ ⎝⎜ 1h ⎟⎠

106 cm3 =1 m3. The volume of a sphere is V = 4 π r 3 = 1 π d 3 , where r is the radius and d is the diameter.
3 6

Don’t forget to account for four astronauts.

EXECUTE: (a) The volume is (4)(10 breaths/min)(500 ×10−6 m3 ) ⎛ 5.3 ×105 min ⎞ = 1× 104 m3/yr.
⎜⎝⎜ 1y ⎠⎟⎟

(b) d = ⎛ 6V ⎞1/3 = ⎛ 6[1×104 m3] ⎞1/3 = 27 m
⎜⎝ π ⎟⎠ ⎜⎝⎜ π ⎟⎠⎟

EVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all

the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually
be required.
IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical
SET UP: Estimate that we blink 10 times per minute.1 y = 365 days. 1 day = 24 h, 1 h = 60 min. Use 80

years for the lifetime.

© Copyright 2012 Pearson Education, Inc. 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.

Units, Physical Quantities and Vectors 1-5

EXECUTE: The number of blinks is (10 per min) ⎛ 60 min ⎞⎛ 24 h ⎞⎛ 365 days ⎞ (80 y/lifetime) = 4 ×108
⎝⎜ 1h ⎠⎟ ⎝⎜ 1 day ⎟⎜ 1 y ⎟
⎠⎝ ⎠

EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our

calculation is surely accurate to a power of 10.

1.22. IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood
1.23.
pumped during this interval is then the volume per beat multiplied by the total beats.
1.24.
1.25. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per
1.26.
minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years.

EXECUTE: Nbeats = (75 beats/min) ⎛ 60 min ⎞⎛ 24 h ⎞⎛ 365 days ⎞⎛ 80 yr ⎞ = 3 × 109 beats/lifespan
⎜⎝ 1h ⎟⎠ ⎜ ⎟⎜ yr ⎟⎜ lifespan ⎟
⎝ 1 day ⎠⎝ ⎠⎝ ⎠

Vblood = (50 cm3/beat) ⎛ 1L ⎞⎛ 1 gal ⎠⎞⎟ ⎝⎛⎜⎜ 3×109 beats ⎞ = 4 ×107 gal/lifespan
⎝⎜ 1000 cm3 ⎠⎟ ⎝⎜ 3.788 L lifespan ⎠⎟⎟

EVALUATE: This is a very large volume.
IDENTIFY: Estimation problem
SET UP: Estimate that the pile is 18 in. × 18 in. × 5 ft 8 in.. Use the density of gold to calculate the mass
of gold in the pile and from this calculate the dollar value.

EXECUTE: The volume of gold in the pile is V = 18 in. ×18 in. × 68 in. = 22,000 in.3. Convert to cm3:

V = 22,000 in.3(1000 cm3 /61.02 in.3) = 3.6 × 105 cm3.

The density of gold is 19.3 g/cm3, so the mass of this volume of gold is

m = (19.3 g/cm3)(3.6 ×105 cm3) = 7 ×106 g.

The monetary value of one gram is \$10, so the gold has a value of (\$10/gram)(7 × 106 grams) = \$7 ×107,

or about \$100 ×106 (one hundred million dollars).
EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable.

IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3. Convert

m3 to L.

SET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is

V = 4 π r 3 = 1 π d 3. 103 cm3 = 1 L.
3 6

EXECUTE: V = 1 π (0.2 cm)3 = 4 ×10−3 cm3. The number of drops in 1.0 L is 1000 cm3 = 2 ×105
6 4 × 10−3 cm3

EVALUATE: Since V ∼ d 3, if our estimate of the diameter of a drop is off by a factor of 2 then our

estimate of the number of drops is off by a factor of 8.
IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a
school year.
SET UP: Assume a school of a thousand students, each of whom averages ten pizzas a year (perhaps an
underestimate)

EXECUTE: They eat a total of 104 pizzas.

EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each.
IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the
relative orientation of the two displacements.
SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with
the smallest magnitude is when the two displacements are antiparallel.
EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.26.
EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value
between 0.6 m and 4.2 m.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-6 Chapter 1

Figure 1.26

1.27. IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant

displacement is the single vector that poiGntsGfrom theG starting point to the stopping pGoint.
SGET UGP: GCallG the three displacements A, B, and C . The resultant displacement R is given by
R = A+ B +C.
G
EXECUTE: The vector addition diagram is given in Figure 1.27. Careful measurement gives that R is

7.8 km, 38D north of east.

EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes

of the individual displacements, 2.6 km + 4.0 km + 3.1 km.

Figure 1.27

1.28. IDENTIFY: Draw the vectorGadditioGn diagram to scale.

SET UP: The two vectors A and GB arGe spGecified in the figure that accompanies the problem.
EGXECUTE: (a) The diagram for C = A + B is given in Figure 1.28a. Measuring the length and angle of
C gives C = 9.0 m anGd anGangGle of θ = 34°.
G
(b) The diagram for D = A − B is given in Figure 1.28b. Measuring the length and angle of D gives
D = 2G2 mGand an angle of θ =G250G°.
GG
(c) − A − B = −( A + B), so −A− B has a magnitude of 9.0 m (the same as A + B ) and an angle with the
GG
+ x axis of 214° (opposite to the direction of A + B).
GG GG GG
(d) B− A = −( A − BG ), so B− A has a magnitude of 22 m and an angle with the +x axis of 70° (opposite
G
to the direction of A − B ). G
G
EVALUATE: The vector − A is equal in magnitude and opposite in direction to the vector A.

© Copyright 2012 Pearson Education, Inc. 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.

Units, Physical Quantities and Vectors 1-7

Figure 1.28

1.29. IDENTIFY: Since she returns to the starting poGintG, the vecGtor sum of the four displacements muGst be zero.

SGET UP: GCalGl the three given displacements A, B, and C, and call the fourth displacement D.
G
A + B + C + D = 0.
G
EXECUTE: The vector addition diagram is sketched in Figure 1.29. Careful measurement gives that D

is144 m, 41° soGuth of west. GGG
EVALUATE: D is equal in magnitude and opposite in direction to the sum A + B + C.

Figure 1.29

1.30. IDENTIFY: tanθ = Ay , for θ measured counterclockwise from the +x -axis.
1.31. Ax GG

SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies.

EXECUTE:

(a) tan θ = Ay = −1.00 m = −0.500. θ = tan−1(−0.500) = 360° − 26.6° = 333°.
Ax 2.00 m

(b) tan θ = Ay = 1.00 m = 0.500. θ = tan−1(0.500) = 26.6°.
Ax 2.00 m

(c) tan θ = Ay = 1.00 m = −0.500. θ = tan−1(−0.500) = 180° − 26.6° = 153°.
Ax 22.00 m

(d) tan θ = Ay = −1.00 m = 0.500. θ = tan−1(0.500) = 180° + 26.6° = 207°
Ax −2.00 m

EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct

value of θ. G G

IDENTIFY: For each vector V , use that Vx = V cosθ and Vy = V sinθ , when θ is the angle V makes

with the +x axis, measured counterclockwise from the axis.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-8 Chapter 1

G GG G
SET UP: For A, θ = 270.0°. For B, θ = 60.0°. For C, θ = 205.0°. For D, θ = 143.0°.

EXECUTE: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, By = 13.0 m. Cx = 210.9 m, Cy = −5.07 m.

Dx = −7.99 m, Dy = 6.02 m.

1.32. EVALUATE: The signs of the components correspond to the quadrant in which the vector lies.
1.33. IDENTIFY: Given the direction and one component of a vector, find the other component and the
1.34. magnitude.

SET UP: Use the tangent of the given angle and the definition of vector magnitude.

EXECUTE: (a) tan 34.0° = Ax
Ay

Ay = Ax = 16.0 m = 23.72 m
tan 34.0° tan 34.0°

Ay = −23.7 m.

(b) A = Ax2 + Ay2 = 28.6 m.

EVALUATE: The magnitude is greater than either of the components.
IDENTIFY: Given the direction and one component of a vector, find the other component and the
magnitude.
SET UP: Use the tangent of the given angle and the definition of vector magnitude.

EXECUTE: (a) tan 32.0° = Ax
Ay

Ax = (13.0 m)tan 32.0° = 8.12 m. Ax = −8.12 m.

(b) A = Ax2 + Ay2 = 15.3 m.

EVALUATE: The magnitude is greater than either of the components.

IDENTIFY: Find the vector sum of the three given displacements.

SET UP: Use coordinates for which +x is east and +y is north. The driver’s vector displacements are:
K K K
A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east.

EXECUTE: Rx = Ax + Bx + Cx = 0 + 4.0 km + (3.1 km)cos(45°) = 6.2 km; Ry = Ay + By + Cy =

2.6 km + 0 + (3.1 km)(sin 45°) = 4.8 km; R = Rx2 + Ry2 = 7.8 km; θ = tan−1[(4.8 km)/(6.2 km)] = 38°;
K
R = 7.8 km, 38° north of east. This result is confiGrmed by the sketch in Figure 1.34.
EVALUATE: Both Rx and Ry are positive and R is in the first quadrant.

Figure 1.34 G G G Cx Ax + Bx and Cy Ay + By. Cx and Cy to find the magnitude
1.35. IDENTIFY: If C= A + B, then = = Use
G
and direction of C.

SET UP: From Figure E1.28 in the textbook, Ax = 0, Ay = −8.00 m and Bx = +Bsin 30.0° = 7.50 m,

By = +B cos30.0° = 13.0 m.

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Units, Physical Quantities and Vectors 1-9

GGG
EXECUTE: (a) C = A + B so Cx = Ax + Bx = 7.50 m and Cy = Ay + By = +5.00 m. C = 9.01 m.

tanθ = Cy = 5.00 m and θ = 33.7°.
G CGx G7.50G m GG

(b) BG + AG = AG + B, so B + A has magnitude 9.01 m and direction specified by 33.7°.
(c) D = A − B so Dx = Ax − Bx = −7.50 m and Dy = Ay − By = 221.0 m. D = 22.3 m.

tanφ = Dy = 221.0 m and φ = 70.3°. G 3rd quadrant and the angle θ counterclockwise from the
Dx 27.50 m D is in the

+ x aGxis Gis 180°G + 7G0.3° = 2G50.3G°.
(d) B − A = −( A − B), so B − A has magnitude 22.3 m and direction specified by θ = 70.3°.

EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.28.

1.36. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given
1.37.
vectors. G G

SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies.

EXECUTE: (a) (−8.60 cm)2 + (5.20 cm)2 = 10.0 cm, arctan ⎛ 5.20 ⎞ = 148.8° (which is 180° − 31.2° ).
⎝⎜ −8.60 ⎠⎟

(b) (−9.7 m)2 + (−2.45 m)2 = 10.0 m, arctan ⎛ −2.45 ⎞ = 14° + 180° = 194°.
⎝⎜ −9.7 ⎟⎠

(c) (7.75 km)2 + (−2.70 km)2 = 8.21 km, arctan ⎛ −2.7 ⎞ = 340.8° (which is 360° −19.2° ).
⎜⎝ 7.75 ⎠⎟

EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ

agree with our sketches.

IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are

asked to find their sum.

SET UP:

A = 3.25 km
B = 2.90 km
C = 1.50 km

Figure 1.37a
GG G

Select a coordinate system where + x is east and + y is north. LeGt A, B and CGbe tGhe tGhreeG
displacements of the professor. Then the resultant displacement R is given by R = A + B + C. By the
method of components, Rx = Ax + Bx + Cx and Ry = Ay + By + Cy . Find the x and y components of each
vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant
can be found from its x and y components that we have calculated. As always it is essential to draw a
sketch.

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1-10 Chapter 1
EXECUTE:

Ax = 0, Ay = +3.25 km
Bx = −2.90 km, By = 0
Cx = 0, Cy = −1.50 km
Rx = Ax + Bx + Cx
Rx = 0 − 2.90 km + 0 = −2.90 km
Ry = Ay + By + Cy
Ry = 3.25 km + 0 −1.50 km = 1.75 km

Figure 1.37b

R = Rx2 + Ry2 = (−2.90 km)2 + (1.75 km)2

R = 3.39 km

tanθ = Ry = 1.75 km = −0.603
Rx −2.90 km

θ = 148.9°

Figure 1.37c

The angle θ measured counterclockwise from the +x-axis. In terms of compass directions, the resultant

displacement is 31.1° N of W. G

EVALUATE: Rx < 0 and Ry > 0, so R is in 2nd quadrant. This agrees with the vector addition diagram.

1.38. IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of

components. GG G

SET UP: The two vectors A and B and their resultant C are shown in Figure 1.38. Let + y be in the

direction of the resultant. A = B.

EXECUTE: Cy = Ay + By. 372 N = 2Acos 43.0° and A = 254 N.

EVALUATE: The sum of the magnitudes of the two forces exceeds the magnitude of the resultant force
because only a component of each force is upward.

Figure 1.38

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Units, Physical Quantities and Vectors 1-11

1.39. GGG G
IDENTIFY: Vector addition problem. A − BG= A + (G−B).
SET UP: Find the x- and y-components of AG and BG. Then the x- and y-components of the vector sum are
calculated from the x- and y-components of A and B.
EXECUTE:

Ax = Acos(60.0°)
Ax = (2.80 cm)cos(60.0°) = +1.40 cm
Ay = Asin (60.0°)
Ay = (2.80 cm)sin (60.0°) = +2.425 cm
Bx = B cos(−60.0°)
Bx = (1.90 cm)cos(−60.0°) = +0.95 cm
By = Bsin (−60.0°)
By = (1.90 cm)sin (−60.0°) = −1.645 cm
Note that the signs of the components correspond
to the directions of the component vectors.

Figure 1.39a

G GG
(a) Now let R = A + B.
Rx = Ax + Bx = +1.40 cm + 0.95 cm = +2.35 cm.
Ry = Ay + By = +2.425 cm −1.645 cm = +0.78 cm.

R = Rx2 + Ry2 = (2.35 cm)2 + (0.78 cm)2

R = 2.48 cm

tanθ = Ry = +0.78 cm = +0.3319
Rx +2.35 cm

θ = 18.4°

Figure 1.39b

G GG
EVALUATE: The vector addition diagram for R = A + B is

G
R is in the 1st quadrant, with |Ry | < |Rx | ,
in agreement with our calculation.

Figure 1.39c
G GG

(b) EXECUTE: Now let R = A − B.
Rx = Ax − Bx = +1.40 cm − 0.95 cm = +0.45 cm.
Ry = Ay − By = +2.425 cm +1.645 cm = +4.070 cm.

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1-12 Chapter 1

R = Rx2 + Ry2 = (0.45 cm)2 + (4.070 cm)2

R = 4.09 cm

tanθ = Ry = 4.070 cm = +9.044
Rx 0.45 cm

θ = 83.7°

Figure 1.39d
GG G

EVALUATE: The vector addition diagram for R = A + (−B) is
G
R is in the 1st quadrant, with |Rx| < |Ry|,

in agreement with our calculation.

Figure 1.39e
(c) EXECUTE:

GG GG
B − A = −( AG− B)
G G G
B − A and A − B are equal in magnitude and

opposite in direction.

R = 4.09 cm and θ = 83.7° +180° = 264°

Figure 1.39f
GG G

EVALUATE: The vector addition diagram for R = B + (− A) is

G
R is in the 3rd quadrant, with |Rx | < |Ry |,
in agreement with our calculation.

Figure 1.39g

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Units, Physical Quantities and Vectors 1-13

1.40. ISADGEET=NUATPxIFi:ˆY+:5A.y0TˆjBhG. e=g5e.n0e(r4aiˆl expression for a vector written in terms of components and unit vectors is
1.41. − 30 G
1.42. − 6 ˆj) = G j
20i

EXECUTE: (a) Ax = 5.0, Ay = −6.3 (b) Ax = 11.2, Ay = −9.91 (c) Ax = −15.0, Ay = 22.4

(d) Ax = 20, Ay = −30

EVALUATE: The components are signed scalars.
IDENTIFY: Find the components of each vector and then use Eq. (1.14).

SET UP: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, By = 13.0 m. Cx = 210.9 m, Cy = −5.07 m.

Dx = −7.99 m, Dy = 6.02 m. G
G G C
EXECUTE: A = (−8.00 m) ˆj; B = (7.50 m)iˆ + (13.0 m) ˆj; = (−10.9 m)iˆ + (−5.07 m) ˆj;
G
D = (−7.99 m)iˆ + (6.02 m) ˆj.

EVALUATE: All these vectors lie in the xy-plane and have no z-component.
IDENTIFY: Find A and B. Find the vector difference using components.

SET UP: Deduce the x- and y-components and use Eq. (1.8).
G
EXECUTE: (a) A = 4.00iˆ + 7.00 ˆj; =A8x .=06+. 4BG.0=0;5.A0y0=iˆ +7.00.
A = Ax2 + Ay2 = − 2.00 ˆj;
(4.00)2 + (7.00)2 Bx = +5.00; By = −2.00;

B= Bx2 + By2 = (5.00)2 + (−2.00)2 = 5.39. G
G
EVALUATE: Note that the magnitudes of A and B are each larger than either of their components.
G G
EXECUTE: (b) A − B = 4.00iˆ + 7.00 ˆj − (5.00iˆ − 2.00 ˆj) = (4.00 − 5.00)iˆ + (7.00 + 2.00) ˆj.
G G
A − B = −1.00iˆ + 9.00 ˆj

(c) Let G G − G = −1.00iˆ + 9.00 ˆj. Then Rx = −1.00, Ry = 9.00.
R= A B

R = Rx2 + Ry2

R = (−1.00)2 + (9.00)2 = 9.06.

tanθ = Ry = 9.00 = −9.00
Rx −1.00

θ = −83.6° + 180° = 96.3°.

Figure 1.42

1.43. G
EVALUATE: Rx < 0 and Ry > 0, so R is in the 2nd quadrant.

IDENTIFY: Use trig to find the components of each vector. Use Eq. (1.11) to find the components of the
vector sum. Eq. (1.14) expresses a vector in terms of its components.

SET UP: Use the coordinates in the figure that accompanies the problem.
G
EXECUTE: (a) A = (3.60 m) cos 70.0°iˆ + (3.60 m) sin 70.0° ˆj = (1.23 m)iˆ + (3.38 m) ˆj
G
B = −(2.40 m) cos 30.0°iˆ − (2.40 m) sin 30.0° ˆj = (−2.08 m)iˆ + (−1.20 m) ˆj

(b) G = (3.00) G − (4.00) G = (3.00)(1.23 m)iˆ + (3.00)(3.38 m) ˆj − (4.00)(−2.08 m)iˆ − (4.00)(−1.20 m) ˆj
C A B

= (12.01 m)iˆ + (14.94) ˆj

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1-14 Chapter 1

(c) From Equations (1.7) and (1.8),

C= (12.01 m)2 + (14.94 m)2 = 19.17 m, arctan ⎛ 14.94 m⎞ = 51.2°
⎝⎜ 12.01 m ⎟⎠

EVALUATE: Cx and Cy are both positive, so θ is in the first quadrant.

1.44. IDENTIFY: A unit vector has magnitude equal to 1.
1.45. SET UP: The magnitude of a vector is given in terms of its components by Eq. (1.12).
1.46.
1.47. EXECUTE: (a) |iˆ + ˆj + kˆ| = 12 +12 +12 = 3 ≠ 1 so it is not a unit vector. G
G |A| > 1,
1.48. (b) |A| = Ax2G+ Ay2 + Az2 . If any component is greater than +1 or less than −1, so it cannot be a

unit vector. A can have negative components since the minus sign goes away when the component is

squared.
G
(c) |A| = 1 gives a2(3.0)2 + a2 (4.0)2 = 1 and a2 25 = 1. a = ± 1 = ±0.20.
5.0

EVALUATE: GThGe magnitude of a vector is greater than the magnitude of any of its components.
IDENTIFY: A ⋅GB = ABGcosφ
B, φ = 150.0°. GG φ = 145.0°. GG φ = 65.0°.
SET UP: For A and = (8.00 m)(15.0 For B and C, 2104 m2 For A and C,
GG m)cos150.0° =
EXECUTE: (a) A ⋅ B
GG
(b) B⋅C = (15.0 m)(12.0 m)cos145.0° = −148 m2
GG
(c) A⋅C = (8.00 m)(12.0 m)cos65.0° = 40.6 m2

EVALUATE: When φ < 90° the scGalaGr product is positive and when φ > 90° the scalar product is negative.
IDENTIFY: Target variables are A ⋅ B and the angle φ between the two vectors.
G G
SET UP: We are given A and B in unit vector form and can take the scalar product using Eq. (1.19).

The angle φ can then be found from Eq. (1.18).
G G
EXECUTE: (a) A = 4.00iˆ + 7.00 ˆj, B = 5.00iˆ − 2.00 ˆj; A = 8.06, B = 5.39.
GG
A⋅B = (4.00iˆ + 7.00 ˆj) ⋅ (5.00iˆ − 2.00 ˆj) = (4.00)(5.00) + (7.00)(−2.00) = 20.0 − 14.0 = +6.00.
GG
A⋅B 6.00
(b) cosφ = AB = (8.06)(5.39) = 0.1382; φ= 82.1°.
G G
G
EVALUATE: The component of B along A is in the same direction as A, so the scalar product is

positive and the angle φ is less than 90°.

IDENTIFY: For all of tφhe=searpcaciorss⎛⎝⎜⎜oAfGAv⋅BeBGct⎠⎞⎟⎟o=rsa,rtchceoasn⎝⎛⎜gAlexBisxA+fBoAuynBdyf⎟⎞⎠ro. m combining Eqs. (1.18) and (1.21),
to give the angle φ as

SET UP: Eq. (1.14) shows how to obtain the components for a vector written in terms of unit vectors.

EXECUTE: GG 40, B = 13, and so φ = arccos ⎛ −22 ⎞ = 165°.
(a) A ⋅ B = −22, A = ⎜⎝ 40 13 ⎠⎟

GG 34, B = 136, φ = arccos ⎛ 60 ⎞ = 28°.
(b) A⋅B = 60, A = ⎝⎜ 34 136 ⎠⎟

GG
(c) A⋅B = 0 and φ = 90°.
GG GG GG
EVALUATE: If A ⋅ B > 0, 0 ≤ φ < 90°. If A ⋅ B < 0, 90° < φ ≤ 180°. If A ⋅ B = 0, φ = 90° and the two

vectors are perpendicular. GG

IDENTIFY: Target variabGle is theG vector A × B expressed in terms of unit vectors.
SEEXTECUUPT: E:WeAGar=e4g.i0v0eiˆn+ A and BG in unit vector form and can take the vector product using Eq. (1.24).
7.00 ˆj, B = 5.00iˆ − 2.00 ˆj.

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Units, Physical Quantities and Vectors 1-15

G G = (4.00iˆ + 7.00 ˆj) × (5.00iˆ − 2.00 ˆj) = 20.0iˆ × iˆ − 8.00iˆ × ˆj + 35.0 ˆj × iˆ −14.0 ˆj × ˆj. But
A× B G G
A B
iˆG× iˆ G= ˆj × ˆj = 0 and iˆ × ˆj = kˆ, ˆj × iˆ = −kˆ, so × = −8.00kˆ + 35.0(−kˆ) = −43.0kˆ. The magnitude of

A × B is 43.0. GG

EVALUATE: Sketch the vectors A and B in a coordinate system wherGe thGe xy-plane is in the plane of the
paper and the z-axis is directed out toward you. By the right-hand rule A × B is directed into the plane of
the paper, in the −z-direction. This agrees with the above calculation that used unit vectors.

Figure 1.48

1.49. GG
1.50. IDENTIFY: A × D has magnitude ADsinφ. Its direction is given by the right-hand rule.

1.51. SET UP: φ = 180G° − 5G3° = 127° m) sin127° = 63.9 m2. The right-hand rule says GG is in the
1.52. EXECUTE: (a) |A × D| = (8.00 m)(10.0 A× D

−z-diGrectGion (into the page). GG
(b) D× A has the same magnitude as A× D and is in the opposite direction.
G G
EVALUATE: The component of D perpendicular to A is D⊥ = D sin 53.0° = 7.99 m.
GG agrees with our previous result.
|A× D| = AD⊥ = 63.9 m2, which

IDENTIFY: The right-hand rule gives the direction and Eq. (1.22) gives the magnitude.

SET UP: φ = 120.0°. GG

EXECUTE: (a) The direction of A× B is into the page (the −z-direction ). The magnitude of the vector

product is AB sinφ = (2.80 cm)(1.90 cm)sin120° = 4.61 cm2. GG
B×A
(b) Rather than repeat the calculations, Eq. (1.23) may be used to see that has magnitude 4.61 cm2

and is in the + z-direction (out of the page).
EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is

Cz = Ax By − Ay Bx = (2.80 cm)cos60.0°(−1.90 cm)sin 60°

− (2.80 cm)sin 60.0°(1.90 cm)cos60.0° = 24.61 cm2.

This gives the same result.

IDENTIFY: Apply Eqs. (1.18) and (1.22).

SET UP: The angle between the vGecGtors is 20° + 90° + 30° = 140°. −6.62 m2 .
EXECUTE: (a) Eq. (1.18) gives A ⋅ B = (3.60 m)(2.40 m)cos140° =

(b) From Eq. (1.22), the magnitude of the cross product is (3.60 m)(2.40 m)sin140° = 5.55 m2 and the

direction, from the right-hand rule, is out of the page (the + z-direction ). GG

EVALUATE: We could also use Eqs. (1.21) and (1.27), with the components of A and B.

IDENTIFY: Use Eq. (1.27) for the components of the vector product.

SET UP: Use coordinates with the + x-axis to the right, + y-axis toward the top of the page, and + z-axis

out of the page. Ax = 0, Ay = 0 and Az = −3.50 cm. The page is 20 cm by 35 cm, so Bx = −20 cm and

By = 35 cm. G × G = 122 cm 2 , G × G = 70 cm2 , G × G = 0.
EXECUTE: (A B)x (A B)y (A B)z

EVALUATE: From the components we calculated the magnitude of the vector product is 141 cm2.

B = 40.3 cm and φ = 90°, so ABsinφ = 141 cm2, which agrees.

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1-16 Chapter 1
1.53.
GG GG
1.54. IDENTIFY: G A and BG are giGven in unGit vGector foGrm. FindG A, B anGd the vector difference A − B.
1.55. SET UP: A = 22.00i + 3.00 j + 4.00k, B = 3.00i + 1.00 j − 3.00k

1.56. Use Eq. (1.8) to find the magnitudes of the vectors.

EXECUTE: (a) A = Ax2 + Ay2 + Az2 = (−2.00)2 + (3.00)2 + (4.00)2 = 5.38

B = Bx2 + By2 + Bz2 = (3.00)2 + (1.00)2 + (−3.00)2 = 4.36
G G
(b) A − B = (−2.00iˆ + 3.00 ˆj + 4.00kˆ) − (3.00iˆ + 1.00 ˆj − 3.00kˆ)
G G
A − B =G(−2.G00 −G3.00)iˆ + (3.00 − 1.00) ˆj + (4.00 − (−3.00))kˆ = 25.00iˆ + 2.00 ˆj + 7.00kˆ.

(c) Let C = A − B, so Cx = −5.00, Cy = +2.00, Cz = +7.00

C= Cx2 + C 2 + Cz2 = (−5.00)2 + (2.00)2 + (7.00)2 = 8.83
y
GG GG GG GG
B − A = −( A − B), so A − B and B − A have the same magnitude but opposite directions.

EVALUATE: A, B and C are each larger than any of their components.
IDENTIFY: Area is length times width. Do unit conversions.

SET UP: 1 mi = 5280 ft. 1 ft3 = 7.477 gal.

EXECUTE: (a) The area of one acre is 1 mi × 1 mi = 1 mi2, so there are 640 acres to a square mile.
8 80 640

(b) (1 acre) × ⎛ 1 mi2 ⎞ × ⎛ 5280 ft ⎞2 = 43,560 ft 2
⎜⎜⎝ 640 acre ⎠⎟⎟ ⎜⎝ 1 mi ⎟⎠

(all of the above conversions are exact).

(c) (1 acre-foot) = (43,560 ft 3 ) × ⎛ 7.477 gal⎞ = 3.26 × 105 gal, which is rounded to three significant figures.
⎜⎝ 1 ft3 ⎠⎟

EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-
foot is much larger than a gallon.
IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and
from this the radius.

SET UP: The earth has mass mE = 5.97 ×1024 kg and radius rE = 6.38×106 m. The volume of a sphere is

V = 4 π r 3. ρ = 1.76 g/cm3 = 1760 km/m3.
3

EXECUTE: (a) The planet has mass m = 5.5mE = 3.28 ×1025 kg. V = m = 3.28 ×1025 kg = 1.86 ×1022 m3.
ρ 1760 kg/m3

r = ⎛ 3V ⎞1/3 = ⎛ 3[1.86 ×1022 m3] ⎞1/3 = 1.64 ×107 m = 1.64 ×104 km
⎜⎝ 4π ⎠⎟ ⎜⎝⎜ 4π ⎟⎠⎟

(b) r = 2.57rE

EVALUATE: Volume V is proportional to mass and radius r is proportional to V1/3, so r is proportional to

m1/3. If the planet and earth had the same density its radius would be (5.5)1/3rE = 1.8rE. The radius of the

planet is greater than this, so its density must be less than that of the earth.

IDENTIFY and SET UP: Unit conversion.

EXECUTE: (a) f = 1.420 ×109 cycles/s, so 1 s = 7.04 ×10−10 s for one cycle.
1.420 ×109

(b) 7.04 3600 s/h = 5.11×1012 cycles/h
×10−10 s/cycle

(c) Calculate the number of seconds in 4600 million years = 4.6 ×109 y and divide by the time for 1 cycle:

(4.6 ×109 y)(3.156 ×107 s/y) = 2.1×1026 cycles
7.04 ×10−10 s/cycle

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Units, Physical Quantities and Vectors 1-17

(d) The clock is off by 1 s in 100,000 y = 1 ×105 y, so in 4.60 ×109 y it is off by

(1 s) ⎛ 4.60 ×109 ⎞ = 4.6 × 104 s (about 13 h).
⎜⎝⎜ 1×105 ⎟⎟⎠

EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the

1.57. IDENTIFY: Using the density of the oxygen and volume of a breath, we want the mass of oxygen (the
1.58. target variable in part (a)) breathed in per day and the dimensions of the tank in which it is stored.
1.59.
SET UP: The mass is the density times the volume. Estimate 12 breaths per minute. We know 1 day = 24 h,
1 h = 60 min and 1000 L = 1 m3. The volume of a cube having faces of length l is V = l3.

EXECUTE: (a) (12 breaths/min ) ⎛ 60 min ⎞⎛ 24 h ⎞ = 17,280 breaths/day. The volume of air breathed in
⎜⎝ 1h ⎟⎠⎜⎝ 1 day ⎟

one day is ( 1 L/breath )(17, 280 breaths/day) = 8640 L = 8.64 m3. The mass of air breathed in one day is the
2

density of air times the volume of air breathed: m = (1.29 kg/m3)(8.64 m3) = 11.1 kg. As 20% of this

quantity is oxygen, the mass of oxygen breathed in 1 day is (0.20)(11.1 kg) = 2.2 kg = 2200 g.

(b) V = 8.64 m3 and V = l3, so l = V 1/3 = 2.1 m.

EVALUATE: A person could not survive one day in a closed tank of this size because the exhaled air is
breathed back into the tank and thus reduces the percent of oxygen in the air in the tank. That is, a person
cannot extract all of the oxygen from the air in an enclosed space.

IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area.

SET UP: The length could be as large as 7.61 cm and the width could be as large as 1.91 cm.

EXECUTE: The area is 14.44 ± 0.095 cm2. The fractional uncertainty in the area is 0.095 cm2 = 0.66%,
14.44 cm2

and the fractional uncertainties in the length and width are 0.01 cm = 0.13% and 0.01 cm = 0.53%. The
7.61 cm 1.9 cm

sum of these fractional uncertainties is 0.13% + 0.53% = 0.66%, in agreement with the fractional

uncertainty in the area.

EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in

any of the individual numbers.

IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities.

SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target
variables and from these we get the uncertainty.

EXECUTE: (a) The volume of a disk of diameter d and thickness t is V = π (d/2)2t.

The average volume is V = π (8.50 cm/2)2(0.50 cm) = 2.837 cm3. But t is given to only two significant

figures so the answer should be expressed to two significant figures: V = 2.8 cm3.
We can find the uncertainty in the volume as follows. The volume could be as large as
V = π (8.52 cm/2)2 (0.055 cm) = 3.1 cm3, which is 0.3 cm3 larger than the average value. The volume

could be as small as V = π (8.48 cm/2)2(0.045 cm) = 2.5 cm3, which is 0.3 cm3 smaller than the average

value. The uncertainty is ±0.3 cm3, and we express the volume as V = 2.8 ± 0.3 cm3.

(b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm = 170. By taking the
largest possible value of the diameter and the smallest possible thickness we get the largest possible value
for this ratio: 8.52 cm/0.045 cm = 190. The smallest possible value of the ratio is 8.48/0.055 = 150. Thus
the uncertainty is ±20 and we write the ratio as 170 ± 20.

EVALUATE: The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much
less, so the percentage uncertainty in the volume and in the ratio should be about 10%.

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1-18 Chapter 1

1.60. IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume.

1.61. SET UP: The volume of a sphere of radius r is V = 4 π r 3. The volume of a cylinder of radius r and length
1.62. 3
1.63.
l is V = π r2l. The density of water is 1000 kg/m3.
1.64.
EXECUTE: (a) Estimate the volume as that of a sphere of diameter 10 cm: V = 5.2 ×10−4 m3.
m = (0.98)(1000 kg/m3)(5.2 ×10−4 m3) = 0.5 kg.

(b) Approximate as a sphere of radius r = 0.25μm (probably an overestimate): V = 6.5×10−20 m3.

m = (0.98)(1000 kg/m3)(6.5 ×10−20 m3) = 6 ×10−17 kg = 6 ×10−14 g.

(c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: V = π r2l = 2.8 ×10−7 m3.

m = (0.98)(1000 kg/m3)(2.8 ×10−7 m3) = 3 ×10−4 kg = 0.3 g.

EVALUATE: The mass is directly proportional to the volume.
IDENTIFY: The number of atoms is your mass divided by the mass of one atom.
SET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the

mass of one H2O molecule: 18.015 u ×1.661×10−27 kg/u = 2.992 ×10−26 kg/molecule.

EXECUTE: (70 kg)/(2.992 × 10−26 kg/molecule) = 2.34 × 1027 molecules. Each H2O molecule has

3 atoms, so there are about 6 ×1027 atoms.

EVALUATE: Assuming carbon to be the most common atom gives 3 ×1027 molecules, which is a result of
the same order of magnitude.
IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill.
SET UP: Estimate the thickness of a dollar bill by measuring a short stack, say ten, and dividing the
measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the

distance from the earth to the moon is 3.8 ×108 m.

EXECUTE: Nbills = ⎛ 3.8 ×108 m ⎞⎛ 103 mm ⎞ = 3.8 × 1012 bills ≈ 4 ×1012 bills
⎜⎜⎝ 0.1 mm/bill ⎟⎟⎠⎜⎝⎜ 1 m ⎟⎠⎟

EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is

significantly less—roughly 1 billion dollars.

IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided

by the surface area of a single dollar bill.

SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or

3,380,000 mi2. This estimate is within 10 percent of the actual area, 3,794,083 mi2. The population is

roughly 3.0 ×108 while the area of a dollar bill, as measured with a ruler, is approximately 6 1 in. by
8

2 5 in.
8

EXECUTE: AU.S. = (3,380,000 mi2 )[(5280 ft)/(1 mi)]2[(12 in.)/(1 ft)]2 = 1.4 × 1016 in.2
Abill = (6.125 in.)(2.625 in.) = 16.1 in.2

Total cost = Nbills = AU.S. /Abill = (1.4 ×1016 in.2)/(16.1 in.2 /bill) = 9 × 1014 bills
Cost per person = (9 ×1014 dollars)/(3.0 ×108 persons) = 3×106dollars/person

EVALUATE: The actual cost would be somewhat larger, because the land isn’t flat.
IDENTIFY: Estimate the volume of sand in all the beaches on the earth. The diameter of a grain of sand
determines its volume. From the volume of one grain and the total volume of sand we can calculate the
number of grains.

SET UP: The volume of a sphere of diameter d is V = 1 π d 3. Consulting an atlas, we estimate that the
6

continents have about 1.45 ×105 km of coastline. Add another 25% of this for rivers and lakes, giving
1.82 ×105 km of coastline. Assume that a beach extends 50 m beyond the water and that the sand is 2 m
deep. 1 billion = 1×109.

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Units, Physical Quantities and Vectors 1-19

EXECUTE: (a) The volume of sand is (1.82 ×108 m)(50 m)(2 m) = 2 ×1010 m3. The volume of a grain is

V = 1 π (0.2 × 10−3 m)3 = 4 ×10−12 m3. The number of grains is 2 × 1010 m3 = 5 × 1021. The number of
6 4 ×10−12 m3

grains of sand is about 1022.

(b) The number of stars is (100 ×109)(100 ×109) = 1022. The two estimates result in comparable numbers

1.65. for these two quantities.
EVALUATE: Both numbers are crude estimates but are probably accurate to a few powers of 10.
IDENTIFY: We know the magnitude and direction of the sum of the two vector pulls and the direction of
one pull. We also know that one pull has twice the magnitude of the other. There are two unknowns, the

magnitude of the smaller pull and its direction. Ax + Bx = Cx and Ay + By = Cy give two equations for

these two unknowns. G G GGG

SET UP: Let the smaller pull be A and the larger pull be B. B = 2A. C = A + B has magnitude 460.0 N

and is northward. Let +x be east and +y be north. Bx = −Bsin 25.0° and By = B cos 25.0°. CGx = 0,
G
Cy = 460.0 N. A must have an eastward component to cancel the westward component of B. There are
GG
then two possibilities, as sketched in Figures 1.65 a and b. A can have a northward component or A can

have a southward component.

EXECUTE: In either Figure 1.65 a or b, Ax + Bx = Cx and B = 2 A gives (2A)sin 25.0° = Asinφ and

φ = 57.7°. In Figure 1.65a, Ay + By = Cy gives 2Acos 25.0° + Acos57.7° = 460.0 N and A = 196 N. In

Figure 1.65b, 2Acos 25.0° − Acos57.7° = 460.0 N and A = 360 N. One solution is for the smaller pull to

be 57.7° east of north. In this case, the smaller pull is 196 N and the larger pull is 392 N. The other

solution is for the smaller pull to be 57.7° east of south. In this case the smaller pull is 360 N and the

larger pull is 720 N. G

EVALUATE: For the first solution, with A east of north, each worker has to exert less force to produce the

given resultant force and this is the sensible direction for the worker to pull.

Figure 1.65

G G GGGG G GGG
1.66. IDENTIFY: Let D be the fourth force. Find D such that A + B + C + D =G 0, so D = −( A + B + C ).

SET UP: Use components and solve for the components Dx and Dy of D.

EXECUTE: Ax = + Acos30.0° = +86.6 N, Ay = + Asin 30.0° = +50.00 N.

Bx = −Bsin 30.0° = −40.00 N, By = +B cos30.0° = +69.28 N.

Cx = −C cos53.0° = −24.07 N, Cy = −C sin 53.0° = −31.90 N.

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1-20 Chapter 1

Then Dx = −22.53 N, Dy = −87.34 N and D = Dx2 + Dy2 = 90.2 N. tanα = |Dy/Dx| = 87.34/22.53.

α = 75.54°. φ = 180° + α = 256°, counterclockwise from the + x-axis. G

EVALUATE: As shown in Figure 1.66, since Dx and Dy are both negative, D must lie in the third

Figure 1.66

GGG GGG G
1.67. IDENTIFY: A + B = C (or B + A = C ). The target variable is vector A. G

SET UP: UseG components and Eq. (1.10) to solve for the components of A. Find the magnitude and
direction of A from its components.

EXECUTE: (a)

Cx = Ax + Bx, so Ax = Cx − Bx
Cy = Ay + By , so Ay = Cy − By
Cx = C cos 22.0° = (6.40 cm)cos 22.0°
Cx = +5.934 cm
Cy = C sin 22.0° = (6.40 cm)sin 22.0°

Cy = +2.397 cm

Bx = B cos(360° − 63.0°) = (6.40 cm)cos 297.0°
Bx = +2.906 cm
By = Bsin 297.0° = (6.40 cm)sin 297.0°
By = −5.702 cm

Figure 1.67a

(b) Ax = Cx − Bx = +5.934 cm − 2.906 cm = +3.03 cm
Ay = Cy − By = +2.397 cm − (−5.702) cm = +8.10 cm

A = Ax2 + Ay2

A = (3.03 cm)2 + (8.10 cm)2 = 8.65 cm

tanθ = Ay = 8.10 cm = 2.67
Ax 3.03 cm

θ = 69.5°

Figure 1.67b

GG
EVALUATE: The A we calculated agrees qualitatively with vector A in the vector addition diagram in

part (a).

1.68. IDENTIFY: Find the vector sum of theGtwo disGplacements. G G GG

SGET UP: Call the two displacements A and B, where A = 170 km and B = 230 km. A + B = R. A and
B are as shown in Figure 1.68.

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Units, Physical Quantities and Vectors 1-21

EXECUTE: Rx = Ax + Bx = (170 km)sin 68° + (230 km)cos 48° = 311.5 km.

Ry = Ay + By = (170 km)cos68° − (230 km)sin 48° = −107.2 km.

R= Rx2 + Ry2 = (311.5 km)2 + (−107.2 km)2 = 330 km. tanθR = | Ry | = 107.2 km = 0.344.
Rx 311.5 km

θR = 19° south of east. G

EVALUATE: Our calculation using components agrees with R shown in the vector addition diagram,

Figure 1.68.

Figure 1.68

1.69. IDENTIFY: Vector addition. Target variable is the 4th displacement.

SET UP: Use a coordinate system where east is in the + x-direction and north is in the + y-direction.
GG G G
Let A, B, and C be the three displacements thatGare gGivenG andG letGD be the fourth unmeasured
displacementG. Then the resultant displacement is R = A + B + C + D. And since she ends up back where
she sGtarteGd, RG = 0G. G G G G
0 = A + B + C + D, so D = −( A + B + C )

Dx = −( Ax + Bx + Cx ) and Dy = −( Ay + By + Cy )

EXECUTE:

Ax = −180 m, Ay = 0
Bx = B cos315° = (210 m)cos315° = +148.5 m
By = Bsin 315° = (210 m)sin 315° = −148.5 m
Cx = C cos60° = (280 m)cos60° = +140 m
Cy = C sin 60° = (280 m)sin 60° = +242.5 m

Figure 1.69a

Dx = −( Ax + Bx + Cx ) = −(−180 m +148.5 m +140 m) = −108.5 m
Dy = −( Ay + By + Cy ) = −(0 −148.5 m + 242.5 m) = −94.0 m

D = Dx2 + Dy2

D = (−108.5 m)2 + (−94.0 m)2 = 144 m

tanθ = Dy = −94.0 m = 0.8664
Dx −108.5 m

θ G= 180° + 40.9° = 220.9°
( D is in the third quadrant since both

Dx and Dy are negative.)

Figure 1.69b

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1-22 Chapter 1

GG
The direction of D can also be specified in terms of φ = θ − 180° = 40.9°; D is 41° south of west.
EVALUATE: The vector addition diagram, approximately to scale, is

G
Vector D in this diagram agrees qualitatively
with our calculation using components.

Figure 1.69c

1.70. IDENTIFY: Add the vectors using the method of components.

SET UP: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, By = 13.0 m. Cx = −10.9 m, Cy = −5.07 m.

EXECUTE: (a) Rx = Ax + Bx + Cx = −3.4 m. Ry = Ay + By + Cy = −0.07 m. R = 3.4 m. tanθ = −0.07 m .
−3.4 m

θ = 1.2° below the −x-axis.

(b) Sx = Cx − Ax − Bx = −18.4 m. Sy = Cy − Ay − By = −10.1 m. S = 21.0 m. tanθ = Sy = −10.1 m .
Sx −18.4 m

θ = 28.8° below the −x-axis. GG

EVALUATE: The magnitude and direction we calculated for R and S agree with our vector diagrams.

Figure 1.70

1.71. IDENTIFY: Find the vector sum of the two forces.
1.72. SET UP: Use components to add the two forces. Take the +x-direction to be forward and the
+ y-direction to be upward.

EXECUTE: The second force has components F2x = F2 cos32.4° = 433 N and F2y = F2 sin 32.4° = 275 N.

The first force has components F1x = 480 N and F1y = 0. Fx = F1x + F2x = 913 N and

Fy = F1y + F2y = 275 N. The resultant force is 954 N in the direction 16.8° above the forward direction.

EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less

than the sum of their magnitudes.

IDENTIFY: Solve for one of the vectors in the vector sum. Use components.

SET UP: Use coordinates for which + x is east and + y is north. The vector displacements are:
K K K
A = 2.00 km, 0°of east; B = 3.50 m, 45° south of east; and R = 5.80 m, 0° east

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Units, Physical Quantities and Vectors 1-23

EXECUTE: Cx = Rx − Ax − Bx = 5.80 km − (2.00 km) − (3.50 km)(cos 45°) = 1.33 km; Cy = Ry − Ay − By

= 0 km − 0 km − (−3.50 km)(sin 45°) = 2.47 km; C = (1.33 km)2 + (2.47 km)2 = 2.81 km;
θ = tan−1[(2.47 km)/(1.33 km)] = 61.7° north of east. The vector addition diagram in Figure 1.72 shows
good qualitative agreement with these values.
EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive.

Figure 1.72

1.73. IDENTIFY: We know the resultant of two forces of known equal magnitudes and want to find that
1.74. magnitude (the target variable).
1.75.
SET UP: Use coordinates having a horizontal +x axis and an upward + y axis. Then Ax + Bx = Rx and

Rx = 5.60 N.

SOLVE: Ax + Bx = Rx and Acos32° + B sin 32° = Rx. Since A = B,

2Acos32° = Rx, so A = Rx = 3.30 N.
(2)(cos 32°)

EVALUATE: The magnitude of the x component of each pull is 2.80 N, so the magnitude of each pull

(3.30 N) is greater than its x component, as it should be. G GGG GG

IDENTIFY: The four displacements return her to her starting point, so D = −( A + B + C), where A, B
G G
and C are in the three given displacements and D is the displacement for her return.

START UP: Let + x be east and + y be north.

EXECUTE: (a) Dx = −[(147 km)sin85° + (106 km)sin167° + (166 km)sin 235°] = −34.3 km.
Dy = −[(147 km)cos85° + (106 km)cos167° + (166 km)cos 235°] = +185.7 km.

D = (−34.3 km)2 + (185.7 km)2 = 189 km.

(b) The direction relative to north is φ = arctan ⎛ 34.3 km ⎞ = 10.5°. Since Dx <0 and Dy > 0, the
G ⎜ 185.7 km ⎟
⎝ ⎠

direction of D is 10.5° west of north.

EVALUATE: The four displacements add to zero.

IDENTIFY: The sum of the vector forces on the beam suGm to zero, so their x components and their y
components sum to zero. Solve for the components of F.

SET UP: The forces on the beam are sketched in Figure 1.75a. Choose coordiGnates as shown in the sketch.
The 100-N pull makes an angle of 30.0° + 40.0° = 70.0° with the horizontal. F and the 100-N pull have

been replaced by their x and y components.

EXECUTE: (a) The sum of the x-components is equal to zero gives Fx + (100 N)cos70.0° = 0 and

Fx = −34.2 N. The sum of the y-components is equal to zero gives Fy + (100 N)sin 70.0° −124 N = 0 and
G
Fy = +30.0 N. F and its components are sketched in Figure 1.75b. F = Fx2 + Fy2 = 45.5 N.

tanφ = |Fy | = 30.0 N and φ = 41.3°. G is directed at 41.3° above the −x -axis in Figure 1.75a.
|Fx | 34.2 N F

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1-24 Chapter 1

G
(bG) The vector addition diagram is given in Figure 1.75c. F determined from the diagram agrees with
F calculated in part (a) using components.
G
EVALUATE: The vertical component of the 100 N pull is less than the 124 N weight so F must have an

upward component if all three forces balance.

Figure 1.75

GG G
1.76. IDENTIFY: Let the three given displacements be A, B and C, where A = 40 steps, B = 80 steps and
G GGG G G
C = 50 steps. R = A + B + C. The displacement C that will return him to his hut is −R.

SET UP: Let the east direction be the + x-direction and the north direction be the + y-direction.

EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.76.
(b) Rx = (40)cos 45° − (80)cos60° = −11.7 and Ry = (40)sin 45° + (80)sin 60° − 50 = 47.6.

The magnitude and direction of the resultant are (−11.7)2 + (47.6)2 = 49, acrtan ⎛ 47.6⎞ = 76°, north of
⎜⎝ 11.7 ⎠⎟
G
west. We know that R is in the second quadrant because Rx < 0, Ry > 0. To return to the hut, the explorer

must take 49 steps in a direction 76° south of eGast, which is 14° east of south.
EVALUATE: It is useful to show Rx , Ry and R on a sketch, so we can specify what angle we are

computing.

Figure 1.76

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Units, Physical Quantities and Vectors 1-25

1.77. G
IDENTIFY and SET UP: The vector A that connects points (x1, y1) and (x2, y2) has components

Ax = x2 − x1 and Ay = y2 − y1.

EXECUTE: (a) Angle of first line is θ = tan −1 ⎛ 200 − 20 ⎞ = 42°. Angle of second line is 42° + 30° = 72°.
⎜⎝ 210 − 10 ⎟⎠

Therefore X = 10 + 250cos72° = 87, Y = 20 + 250sin 72° = 258 for a final point of (87,258).

(b) The computer screen now looks something like Figure 1.77. The length of the bottom line is

(210 − 87)2 + (200 − 258)2 = 136 and its direction is tan −1 ⎛ 258 − 200 ⎞ = 25° below straight left.
⎜⎝ 210 − 87 ⎟⎠

EVALUATE: Figure 1.77 is a vector addition diagram. The vector first line plus the vector arrow gives the
vector for the second line.

Figure 1.77

1.78. IDENTIFY: Vector addition. One vector and the sum are given; find the second vector (magnitude and

direction). G

SETGUP: Let + x be east and + y be north. Let A be the displacement 285 km at 40.0° north of west and
leGt BG be Gthe unknowG n displacement.
AG + BG = RG where R = 115 km, east
B = R− A

Bx = Rx − Ax, By = Ry − Ay

EXECUTE: Ax = − Acos 40.0° = 2218.3 km, Ay = + Asin 40.0° = +183.2 km

Rx = 115 km, Ry = 0

Then Bx = 333.3 km, By = 2183.2 km. B = Bx2 + By2 = 380 km;

tanα = |By/Bx| = (183.2 km)/(333.3 km)
α = 28.8°, south of east

Figure 1.78 G G

EVALUATE: TGhe southward component of B cancels the northward component of A. The Geastward
component of B must be 115 km larger than the magnitude of the westward component of A.

1.79. IDENTIFY: Vector addition. One force and the vector sum are given; find the second force.

SET UP: Use components. Let + y be upward.

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1-26 Chapter 1

G
B is the force the biceps exerts.

Figure 1.79a

G GGG
E is the force the elbow exerts. E + B = R, where R = 132.5 N and is upward.
Ex = Rx − Bx, Ey = Ry − By
EXECUTE: Bx = −Bsin 43° = −158.2 N, By = +B cos 43° = +169.7 N, Rx = 0, Ry = +132.5 N
Then Ex = +158.2 N, Ey = −37.2 N.
E = Ex2 + Ey2 = 160 N;

tanα = |Ey/Ex| = 37.2/158.2
α = 13°, below horizontal

1.80. Figure 1.79b
1.81.
GG
EVALUATE: The x-component Gof E cancels the x-component of B. The resultant upward force is less
than the upward component of B, so Ey must be downward.
IDENTIFY: Find the vector sum of the four displacements.
SET UP: Take the beginning of the journey as the origin, with north being the y-direction, east the
x-direction, and the z-axis vertical. The first displacement is then (−30 m) kˆ, the second is (−15 m) ˆj, the
third is (200 m) iˆ, and the fourth is (100 m) ˆj.
EXECUTE: (a) Adding the four displacements gives
(−30 m) kˆ + (−15 m) ˆj + (200 m) iˆ + (100 m) ˆj = (200 m) iˆ + (85 m) ˆj − (30 m) kˆ.
(b) The total distance traveled is the sum of the distances of the individual segments:
30 m + 15 m + 200 m + 100 m = 345 m. The magnitude of the total displacement is:

D = Dx2 + Dy2 + Dz2 = (200 m)2 + (85 m)2 + (−30 m)2 = 219 m.

EVALUATE: The magnitude of the displacement is much less than the distance traveled along the path.
IDENTIFY: The sum of the force dGisplGacemG ents mGust be zeroG. Use components.
SET UP: Call the displacements A, B, C and GD, wG here DG is the final unknown displacement for the
reGturnG froGm thGe treasure to the oak tree. Vectors A, B, and C are sketched in Figure 1.81a.
A + B + C + D = 0 says Ax + Bx + Cx + Dx = 0 and Ay + By + Cy + Dy = 0. A = 825 m, B = 1250 m, and
C = 1000 m. Let + x be eastward and + y be north.

EXECUTE: (a) Ax + Bx + Cx + Dx = 0 gives
Dx = −( Ax + Bx + Cx ) = −(0 − [1250 m]sin 30.0° + [1000 m]cos 40.0°) = −141 m. Ay + By + Cy + Dy = 0

gives Dy = −( Ay + By + Cy ) = −(−825 m + [1250 m]cos30.0° + [1000 m]sin 40.0°) = −900 m. The fourth

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Units, Physical Quantities and Vectors 1-27

G Dx2 + Dy2 = 911 m.
displacement D and its components are sketched in Figure 1.81b. D =

tanφ = | Dx | = 141 m and φ = 8.9°. You should head 8.9° west of south and must walk 911 m.
|Dy | 900 m G

(b) The vector diGagram is sketched in Figure 1.81c. The final displacement D from this diagram agrees
with the vector D calculaGted in part (a) using componentGs. G
G
EVALUATE: Note that D is the negative of the sum of A, B, and C.

Figure 1.81

1.82. IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and want to
1.83.
find the magnitude Gof one of the other vectors. G
SGET UP: Calling A the vector from you to the first postG, B Gthe Gvector from you to the second post, and
C the vector frGom the first to the second post, we have A + C + B. Solving using components and the
magnitude of C gives Ax + Cx = Bx and Ay + Cy = By.

EXECUTE: Bx = 0, Ax = 41.53 m and Cx = Bx − Ax = −41.53 m.

C = 80.0 m, so Cy = ± C2 − Cx2 = ±68.38 m.

The post is 37.1 m from you.

EVALUATE: By = −37.1 m (negative) since post is south of you (in the negative y direction).

IDENTIFY: We are given the resultant of three vectors, two of which we know, and want to find the

magnitude and direGction of the third vector. G G GGGG
SET UP: Calling C the unknown vector and A and B the known vectors, we have A + B + C = R. The

components are Ax + Bx + Cx = Rx and Ay + By + Cy = Ry.

EXECUTE: The components of the known vectors are Ax = 12.0 m, Ay = 0,

Bx = −Bsin 50.0° = −21.45 m, By = B cos50.0° = +18.00 m, Rx = 0, and Ry = −10.0 m. Therefore the
G
components of C are Cx = Rx − Ax − Bx = 0 −12.0 m − (−21.45 m) = 9.45 m and

Cy = Ry − Ay − By = −10.0 m − 0 −18.0 m = −28.0 m.

Using these components to find the magnitude and direction of G gives C = 29.6 m and tanθ = 9.45 and
C 28.0

θ = 18.6° east of south

EVALUATE: A graphical sketch shows that this answer is reasonable.

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1-28 Chapter 1

1.84. IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and want to

1.85. find the magnitude Gof one of the other vectors. G
1.86. SET UP: Calling A the vector oGf Ricardo’s displacement from the tree, B thGe veGctor Gof Jane’s
displacement from the tree, and C the vector from Ricardo to Jane, we have A + C = B. Solving using
1.87.
1.88. components we have Ax + Cx = Bx and Ay + C y = By .
G G
EXECUTE: (a) The components of A and B are Ax = −(26.0 m)sin 60.0° = −22.52 m,

Ay = (26.0 m)cos60.0° = +13.0 m, Bx = −(16.0 m)cos30.0° = −13.86 m,

By = −(16.0 m)sin 30.0° = −8.00 m, Cx = Bx − Ax = −13.86 m − (−22.52 m) = +8.66 m,

Cy = By − Ay = −8.00 m − (13.0 m) = −21.0 m

Finding the magnitude from the components gives C = 22.7 m.

(b) Finding the direction from the components gives tanθ = 8.66 and θ = 22.4°, east of south.
21.0

EVALUATE: A graphical sketch confirms that this answer is reasonable.

IDENTIFY: Think of the displacements of the three people as vectors. We know two of them and want to

find their resultant.G G G
SET UP: Calling A the vector GfromG JohGn to Paul, B the vector from Paul to George, and C the vector
from John to George, we have A + B = C , which gives Ax + Bx = Cx and Ay + By = Cy.

EXECUTE: The known components are Ax = −14.0 m, Ay = 0, Bx = B cos37° = 28.75 m, and

By = −Bsin 37° = −21.67 m. Therefore Cx = −14.0 m + 28.75 m = 14.75 m, Cy = 0 − 21.67 m = −21.67 m.

These components give C = 26.2 m and tanθ = 14.75 , which gives θ = 34.2° east of south.
21.67

EVALUATE: A graphical sketch confirms that this aGnswer is reasonable. G
IDENTIFY: If the vector from yoGur tent to Joe’s is A and from your tent to Karl’s is B, then the vector
G
from Joe’s tent to Karl’s is B − A.

SET UP: Take your tent’s position as the origin. Let + x be east and + y be north.

EXECUTE: The position vector for Joe’s tent is
([21.0 m]cos 23°) iˆ − ([21.0 m]sin 23°) ˆj = (19.33 m) iˆ − (8.205 m) ˆj.

The position vector for Karl’s tent is ([32.0 m]cos 37°)iˆ + ([32.0 m]sin 37°) ˆj = (25.56 m)iˆ + (19.26 m) ˆj.

The difference between the two positions is
(19.33 m − 25.56 m)iˆ + (−8.205 m −19.25 m) ˆj = −(6.23 m)iˆ − (27.46 m) ˆj. The magnitude of this vector is

the distance between the two tents: D = (−6.23 m)2 + (−27.46 m)2 = 28.2 m

EVALUATE: If both tents were due east of yours, the distance between them would be
32.0 m − 21.0 m = 11.0 m. If Joe’s was due north of yours and Karl’s was due south of yours, then the
distance between them would be 32.0 m + 21.0 m = 53.0 m. The actual distance between them lies

between these limiting values.

IDENTIFY: We know the scalar product and the magnitude of the vector product of two vectors and want

to know the angle between them.G G GG
SET UP: The scalar product is A ⋅ B = AB cosθ and the vector product is A × B = ABsinθ .

EXECUTE: GG AB cosθ = −6.00 and GG = AB sin θ = +9.00. Taking the ratio gives tanθ = 9.00 ,
A⋅B = A× B −6.00

so θ = 124°.

EVALUATE: Since the scalar product is negative, the angle must be between 90° and 180°.

IDENTIFY: Calculate the scalar product and use Eq. (1.18) to determine φ.

SET UP: The unit vectors are perpendicular to each other.

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Units, Physical Quantities and Vectors 1-29

EXECUTE: The direction vectors each have magnitude 3, and their scalar product is
(1)(1) + (1)(−1) + (1)(−1) = −1, so from Eq. (1.18) the angle between the bonds is

arccos ⎛ −1 3 ⎞ = arccos ⎛ − 1 ⎞ = 109°.
⎝⎜ 3 ⎠⎟ ⎜⎝ 3 ⎠⎟

EVALUATE: The angle between the two vectors in the bond directions is greater than 90°.

1.89. IDENTIFY: We know the magnitude of two vectors and their scalar product and want to find the
1.90.
1.91. magnitude of their vector producGt. G GG
SET UP: The scalar product is A ⋅ B = AB cosθ and the vector product is A × B = ABsinθ .
1.92.
EXECUTE: GG AB cosθ = 90.0 m2, which gives cosθ = 90.0 m2 = 90.0 m2 m) = 0.4688, so
A⋅B = AB (12.0 m)(16.0

GG
θ = 62.05°. Therefore A × B = ABsinθ = (12.0 m)(16.0 m)sin 62.05° = 170 m2.

EVALUATE: The magnitude of the vector product is greater than the scalar product because the angle

between the vectoGrs isGgreaGter than 45º. GG
IDENTIFY: Let C = A + BG ,anVGd⋅cVGal=cuVla2t.eAGth⋅eBGsc=alAaBr pcorosdφu.ct C ⋅C.
SET UP: For any vector V

EGXECGUTE: (a) Use the linearity of the dot product to show that the square of the magnitude of the sum
A + B is

( G G G G = G G G G + G G GG = G G GG + 2 G G = A2 + B2 + G G
A+ B) ⋅(A+ B) A⋅ A+ A⋅ B B⋅ A+ B⋅B A⋅ A+ B⋅B A⋅ B 2A⋅ B

= A2 + B2 + 2 AB cosφ

(b) Using the result of part (a), with A = B, the condition is that A2 = A2 + A2 + 2A2cosφ, which solves

for 1 = 2 + 2cosφ, cos φ = − 1 , and φ = 120°.
2

EVALUATE: The expression C2 = A2 + B2 + 2AB cosφ is called the law of cosines.

IDENTIFY: Find the anGgleGbetween specified pairs of vectors.
SEBGEXT=ECUiˆU+PT:Eˆj :+Uksˆ(ea()caolAGosnφ=g=kˆlinA(Aea⋅BlaBodn)g line ab)

A ⋅=BG1,=Bkˆ=⋅ (iˆ1+2 + 12 + 12 = 3
G ˆj + kˆ) =1
A GG

So cosφ = A⋅B = 1/ 3; φ = 54.7°
(b) (along line ad)
G G = iˆ + AB
B= A ˆj + kˆ

ˆj + kˆ (along line ac)

A =G 12 + 12 + 12 = 3; B = 12 + 12 = 2
G = (iˆ + ˆj + kˆ) ⋅ (iˆ + ˆj) = 1+ 1 = 2
A⋅B
GG
A⋅B 2 2;
So cosφ = AB = 3 2 = 6 φ = 35.3°

EVALUATE: Each angle is computed to be less than 90°, in agreement with what is deduced from

Figure P1.91 in the textbook.

IDENTIFY: We know the magnitude of two vectors and the magnitude of their vector product, and we

want to find the possible values of their scalar product.

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1-30 Chapter 1

G G GG
SET UP: The vector product is A × B = ABsinθ and the scalar product is A ⋅ B = AB cosθ.

EXECUTE: GG = AB sinθ = 12.0 m2, so sinθ = 12.0 m2 m) = 0.6667, which gives two possible
A× B (6.00 m)(3.00

vGaluGes: θ = 41.81° or θ = 138.19°. Therefore the two possible values of the scalar product are
A ⋅ B = AB cosθ = 13.4 m2 or −13.4 m2.

EVALUATE: The two possibilities have equal magnitude but opposite sign because the two possible angles
are supplementary to each other. The sines of these angles are the same but the cosines differ by a factor
of −1. See Figure 1.92.

Figure 1.92

1.93. IDENTIFY: We know the scalar product of two vectors, both their directions, and the magnitude of one of
1.94. them, and wGe wG ant to find the magnitude of the other vector.
SET UP: A ⋅ B = AB cosθ. Since we know the direction of each vector, we can find the angle between
1.95.
them. GG

EXECUGTEG: The angle between the vectors is θ = 79.0°. Since A ⋅ B = AB cosθ, we have
A⋅B 48.0 m2
B= Acosθ = (9.00 m)cos79.0° = 28.0 m.

G G
EVALUATE: Vector B has theGsamGe units as vector A.
GG
IDENTIFY: The cross product A × B is perpendicular to bGothG A and B.
SET UP: Use Eq. (1.27) to calculate the components of A × B.

EXECUTE: The cross product is

(−13.00) iˆ + (6.00) ˆj + (−11.00) kˆ = 13 ⎣⎡⎢−(1.00) iˆ + ⎛ 6.00 ⎞ ˆj − 11.00 kˆ ⎦⎥⎤. The magnitude of the vector in
⎜⎝ 13.00⎟⎠ 13.00

square brackets is 1.93, and so a unit vector in this direction is

⎡ −(1.00) iˆ + (6.00/13.00) ˆj − (11.00/13.00) kˆ ⎤⎥.
⎢ 1.93 ⎦

The negative of this vector,

⎡ (1.00) iˆ − (6.00/13.00) ˆj + (11.00/13.00) kˆ ⎤ ,
⎢ 1.93 ⎥
⎣ ⎦
GG
is also a unit vector perpendicular to A and B.

EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular

to both vectors is perpendicular to this plane. G GG

IDENTIFGY Gand SETGUPG: The target variables are the components of C. We are given A and B. We also
know A⋅C and B⋅C, and this gives us two equations in the two unknowns Cx and Cy.
G G GG
EXECUTE: A and C are perpendicular, so A ⋅ C = 0. AxCx + AyCy = 0, which gives 5.0Cx − 6.5Cy = 0.
GG
B ⋅ C = 15.0, so −3.5Cx + 7.0Cy = 15.0

We have two equations in two unknowns Cx and Cy. Solving gives Cx = 8.0 and Cy = 6.1.

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Units, Physical Quantities and Vectors 1-31

G
EGVAGLUATE: WGe cGan check that our result does give us a vector C that satisfies the two equations
A ⋅ C = 0 and B ⋅ C = 15.0.

1.96. IDENTIFY: Calculate the magnitude of the vector product and then use Eq. (1.22).
1.97.
SET UP: The magnitude of a vector is related to its components by Eq. (1.12).
1.98. GG
GG | A× B| (−5.00)2 + (2.00)2
EXECUTE: |A× B| = AB sin θ . sinθ = AB = (3.00)(3.00) = 0.5984 and

θ = sin−1(0.5984) = 36.8°. GG

EVALUATE: We haven’t foGundG A aGnd B,G jusGt theG angle between them.
(a) IDENTIFY: Prove that A ⋅ (B × C ) = ( A × B) ⋅ C.

SET UP: Express the scalar and vector products in terms of components.

EXECUTE: GG G GG GG GG

A ⋅ (B × C ) = Ax (B × C )x + Ay (B × C ) y + Az (B × C )z

GG G
A ⋅ (B × C ) = Ax (ByCz − BzCy ) + Ay (BzCx − BxCz ) + Az (BxCy − ByCx )

GGG GG GG GG
(A × B)⋅C = (A × B)xCx + (A × B)yCy + (A × B)zCz

G GG
( A × B) ⋅ C = ( AyBz − Az By )Cx + ( Az Bx − AxBz )Cy + ( AxBy − AyBx )Cz
GG G GGG
Comparison of the expressions for A⋅(B × C) and (A × B)⋅C shows they contain the same terms, so
GGG GGG
A⋅(B × C) = (A × B)⋅C. G G G
given the magnitude and direction of GG G
(b) IDENTIFY: Calculate (A × B)⋅C, GG A, B and C.

SGET UGP: Use EGq. (1.22) to find the magnitude and direction of A × B. Then we know the components of
A × B and of C and can use an expression like Eq. (1.21) to find the scalar product in terms of

components.

EXECUTE: A = 5.00; θ A = 26.0°; B = 4.00, θB = 63.0°
GG
|A × B| = ABsinφ.
G G
The angle φ between A and B is equal to φ = θB −θA = 63.0° − 2G6.0°G= 37.0°. So
GG
| AG× B| = (G5.00)(4.00) sin 37.0° = 12.04, and by the right hand-rule A×B is in the + z-direction. Thus
G
( A × B) ⋅ C = (1G2.04G)(6.00) = 72.2
G
EVGALUGATEG : A × B is a vector, so taking its scalar product with C is a legitimate vector operation.
( A × B) ⋅ C is a scalar product between two vectors so the result is a scalar.

IDENTIFY: Use the maximum and minimum values of the dimensions to find the maximum and minimum
areas and volumes.
SET UP: For a rectangle of width W and length L the area is LW. For a rectangular solid with dimensions
L, W and H the volume is LWH.
EXECUTE: (a) The maximum and minimum areas are (L + l)(W + w) = LW + lW + Lw,

(L − l)(W − w) = LW − lW − Lw, where the common terms wl have been omitted. The area and its

uncertainty are then WL ± (lW + Lw), so the uncertainty in the area is a = lW + Lw.

(b) The fractional uncertainty in the area is a lW + Wl = l + w , the sum of the fractional uncertainties
A= WL L W

in the length and width.

(c) The similar calculation to find the uncertainty v in the volume will involve neglecting the terms lwH,

lWh and Lwh as well as lwh; the uncertainty in the volume is v = lWH + LwH + LWh, and the fractional

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1-32 Chapter 1

uncertainty in the volume is v = lWH + LwH + LWh = l + w + h , the sum of the fractional
V LWH L W H

1.99. uncertainties in the length, width and height.
1.100. EVALUATE: The calculation assumes the uncertainties are small, so that terms involving products of two
or more uncertainties can be neglected.
IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to
SET UP: Add the x-components and the y-components.
EXECUTE: The receiver’s position is
[(+1.0 + 9.0 − 6.0 + 12.0)yd]iˆ + [(−5.0 + 11.0 + 4.0 + 18.0) yd] ˆj = (16.0 yd)iˆ + (28.0 yd) ˆj.

The vector from the quarterback to the receiver is the receiver’s position minus the quarterback’s position,

or (16.0 yd)iˆ + (35.0 yd) ˆj, a vector with magnitude (16.0 yd)2 + (35.0 yd)2 = 38.5 yd. The angle is

arctan ⎛ 16.0 ⎞ = 24.6° to the right of downfield.
⎝⎜ 35.0⎟⎠

EVALUATE: The vector from the quarterback to receiver has positive x-component and positive

y-component.

IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the

distance between two objects is the magnitude of this vector. Use the scalar product to find the angle

between two vectors. coordinates (xA, yA) and object B has coordinates (xB , yB ), the vector rGAB from A
SET UP: If object A has

to B has x-component xB − xA and y-component yB − yA.
EXECUTE: (a) The diagram is sketched in Figure 1.100.

(b) (i) In AU, (0.3182)2 + (0.9329)2 = 0.9857.

(ii) In AU, (1.3087)2 + (−0.4423)2 + (−0.0414)2 = 1.3820.

(iii) In AU (0.3182 − 1.3087)2 + (0.9329 − (−0.4423))2 + (0.0414)2 = 1.695.

(c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot
product. Combining Eqs. (1.18) and (1.21),

φ = arccos ⎛⎜ (−0.3182)(1.3087 − 0.3182) + (−0.9329)(−0.4423 − 0.9329) + (0) ⎞ = 54.6°.
⎝ (0.9857)(1.695) ⎟

(d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90°.
EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that
on this date Mars was farther from the earth than it is from the Sun.

Figure 1.100

1.101. IDENTIFY: Draw the vector addition diagram for the position vectors. G

SET UP: Use coordinates in which thGe Sun to Merak line lies along the x-axis. Let A be the posGition
vector of Alkaid relative to the Sun, M is the position vector of Merak relative to the Sun, and R is the

position vector for Alkaid relative to Merak. A = 138 ly and M = 77 ly.

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Units, Physical Quantities and Vectors 1-33

G GG
EXECUTE: The relative positions are shown in Figure 1.101. M + R = A. Ax = M x + Rx so

Rx = Ax − M x = (138 ly)cos 25.6° − 77 ly = 47.5 ly. Ry = Ay − M y = (138 ly)sin 25.6° − 0 = 59.6 ly.

R = 76.2 ly is the distance between Alkaid and Merak.

(b) The angle is angle φ in Figure 1.101. cosθ = Rx = 47.5 ly and θ = 51.4°. Then φ = 180° − θ = 129°.
R 76.2 ly

EVALUATE: The concepts of vector addition and components make these calculations very simple.

Figure 1.101

1.102. IDENTIFY: Define G = Aiˆ + Bˆj + Ckˆ. Show that rG ⋅ G = 0 if Ax + By + Cz = 0.
S S

SET UP: Use EGq. (1.21) to calculate the scalar product.
EXECUTE: rG ⋅S = (xiˆ + yˆj +
zkˆ)⋅( Aiˆ + Bthˆje+nCrGkˆ⋅ )SG==A0xa+ndByal+l Cz rG G
If the points satisfy Ax + By + Cz = 0, points are perpendicular to S. The vector and

plane are sketched in Figure 1.102.
EVALUATE: If two vectors are perpendicular their scalar product is zero.

Figure 1.102

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