1
Introduction and Vectors
CHAPTER OUTLINE
1.1 Standards of Length, Mass, and Time
1.2 Dimensional Analysis
1.3 Conversion of Units
1.4 Order-of-Magnitude Calculations
1.5 Significant Figures
1.6 Coordinate Systems
1.7 Vectors and Scalars
1.8 Some Properties of Vectors
1.9 Components of a Vector and Unit Vectors
1.10 Modeling, Alternative Representations, and Problem-Solving Strategy
* An asterisk indicates an item new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
*OQ1.1 The answer is yes for (a), (c), and (e). You cannot add or subtract a
number of apples and a number of jokes. The answer is no for (b) and
(d). Consider the gauge of a sausage, 4 kg/2 m, or the volume of a
cube, (2 m)3. Thus we have (a) yes (b) no (c) yes (d) no (e) yes.
*OQ1.2 41 € ≈ 41 € (1 L/1.3 €)(1 qt/1 L)(1 gal/4 qt) ≈ (10/1.3) gal ≈ 8 gallons,
answer (c).
*OQ1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041
kg (e) 0.27 kg. Then the ranking is c = e > d > a > b.
*OQ1.4 Answer (c). The vector has no y component given. It is therefore 0.
*OQ1.5 The population is about 6 billion = 6 × 109. Assuming about 100 lb per
person = about 50 kg per person (1 kg has the weight of about 2.2 lb),
1
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 2 Introduction and Vectors
the total mass is about (6 × 109)(50 kg) = 3 × 1011 kg, answer (d).
*OQ1.6 The number of decimal places in a sum of numbers should be the same
as the smallest number of decimal places in the numbers summed.
21.4 s
15 s
17.17 s
4.003 s
57.573 s = 58 s, answer (d).
*OQ1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The
meterstick measurement and (c) can both be 4.24 cm. Only (d) does not
overlap. Thus (a), (b), and (c) all agree with the meterstick
measurement.
*OQ1.8 Mass is measured in kg; acceleration is measured in m/s2. Force =
mass × acceleration, so the units of force are answer (a) kg⋅m/s2.
*OQ1.9 Answer (d). Take the difference of the x coordinates of the ends of the
vector, head minus tail: –4 – 2 = –6 cm.
*OQ1.10 Answer (a). Take the difference of the y coordinates of the ends of the
vector, head minus tail: 1 − (−2) = 3 cm.
*OQ1.11 The answers are (a) yes (b) no (c) no (d) no (e) no (f) yes (g) no. Only
force and velocity are vectors. None of the other quantities requires a
direction to be described.
*OQ1.12 Answers (a), (b), and (c). The magnitude can range from the sum of the
individual magnitudes, 8 + 6 =14, to the difference of the individual
magnitudes, 8 − 6 = 2. Because magnitude is the “length” of a vector, it
is always positive.
*OQ1.13 Answer (a). The vector −2D1 will be twice as long as D1 and in the
opposite direction, namely northeast. Adding D2 , which is about
equally long and southwest, we get a sum that is still longer and due
east.
*OQ1.14 Answer (c). A vector in the second quadrant has a negative x
component and a positive y component.
*OQ1.15 Answer (e). The magnitude is 102 + 102 m/s.
*OQ1.16 Answer (c). The signs of the components of a vector are the same as the
signs of the points in the quadrant into which it points. If a vector
arrow is drawn to scale, the coordinates of the point of the arrow equal
the components of the vector. All x and y values in the third quadrant
are negative.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 1 3
ANSWERS TO CONCEPTUAL QUESTIONS
*CQ1.1 A unit of time should be based on a reproducible standard so it can be
used everywhere. The more accuracy required of the standard, the less
the standard should change with time. The current, very accurate
standard is the period of vibration of light emitted by a cesium atom.
Depending on the accuracy required, other standards could be: the
period of light emitted by a different atom, the period of the swing of a
pendulum at a certain place on Earth, the period of vibration of a
sound wave produced by a string of a specific length, density and
tension, and the time interval from full Moon to full Moon.
*CQ1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms.
*CQ1.3 Density varies with temperature and pressure. It would be necessary
to measure both mass and volume very accurately in order to use the
density of water as a standard.
*CQ1.4 Vectors A and B are perpendicular to each other.
*CQ1.5 (a) The book’s displacement is zero, as it ends up at the point from
which it started. (b) The distance traveled is 6.0 meters.
*CQ1.6 No, the magnitude of a vector is always positive. A minus sign in a
vector only indicates direction, not magnitude.
*CQ1.7 The inverse tangent function gives the correct angle, relative to the +x
axis, for vectors in the first or fourth quadrant, and it gives an incorrect
answer for vectors in the second or third quadrant. If the x and y
components are both positive, their ratio y/x is positive and the vector
lies in the first quadrant; if the x component is positive and the y
component negative, their ratio y/x is negative and the vector lies in
the fourth quadrant. If the x and y components are both negative, their
ratio y/x is positive but the vector lies in the third quadrant; if the x
component is negative and the y component positive, their ratio y/x is
negative but the vector lies in the second quadrant.
*CQ1.8 Addition of a vector to a scalar is not defined. Try adding the speed
and velocity, 8.0 m/s + (15.0 m/s î) : Should you consider the sum to
be a vector or a scaler? What meaning would it have?
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 4 Introduction and Vectors
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
4 3
P1.1 For either sphere the volume is V = π r and the mass is
3
4
m = ρV = ρ π r 3 . We divide this equation for the larger sphere by the
3
same equation for the smaller:
m ρ 4π r3 3 r3
= = =5
ms ρ 4π rs3 3 rs3
Then r = rs 3 5 = 4.50 cm ( 1.71) = 7.69 cm .
P1.2 (a) Modeling the Earth as a sphere, we find its volume as
π r = π ( 6.37 × 106 m ) = 1.08 × 1021 m 3
4 3 4 3
3 3
Its density is then
m 5.98 × 1024 kg
ρ= = = 5.52 × 103 kg/m 3
V 1.08 × 1021 m 3
(b) This value is intermediate between the tabulated densities of
aluminum and iron. Typical rocks have densities around 2000 to
3000 kg/m3. The average density of the Earth is significantly
higher, so higher-density material must be down below the
surface.
m
P1.3 Let V represent the volume of the model, the same in ρ = , for both.
V
mgold
Then ρiron = 9.35 kg/V and ρgold = .
V
ρgold mgold
Next, = and
ρiron 9.35 kg
⎛ 19.3 × 103 kg/m 3 ⎞
mgold = ( 9.35 kg ) ⎜ = 22.9 kg
⎝ 7.87 × 10 kg/m ⎟⎠
3 3
P1.4 The volume of a spherical shell can be calculated from
4
V = Vo − Vi = π ( r23 − r13 )
3
m
From the definition of density, ρ = , so
V
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Introduction and Vectors
CHAPTER OUTLINE
1.1 Standards of Length, Mass, and Time
1.2 Dimensional Analysis
1.3 Conversion of Units
1.4 Order-of-Magnitude Calculations
1.5 Significant Figures
1.6 Coordinate Systems
1.7 Vectors and Scalars
1.8 Some Properties of Vectors
1.9 Components of a Vector and Unit Vectors
1.10 Modeling, Alternative Representations, and Problem-Solving Strategy
* An asterisk indicates an item new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
*OQ1.1 The answer is yes for (a), (c), and (e). You cannot add or subtract a
number of apples and a number of jokes. The answer is no for (b) and
(d). Consider the gauge of a sausage, 4 kg/2 m, or the volume of a
cube, (2 m)3. Thus we have (a) yes (b) no (c) yes (d) no (e) yes.
*OQ1.2 41 € ≈ 41 € (1 L/1.3 €)(1 qt/1 L)(1 gal/4 qt) ≈ (10/1.3) gal ≈ 8 gallons,
answer (c).
*OQ1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041
kg (e) 0.27 kg. Then the ranking is c = e > d > a > b.
*OQ1.4 Answer (c). The vector has no y component given. It is therefore 0.
*OQ1.5 The population is about 6 billion = 6 × 109. Assuming about 100 lb per
person = about 50 kg per person (1 kg has the weight of about 2.2 lb),
1
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 2 Introduction and Vectors
the total mass is about (6 × 109)(50 kg) = 3 × 1011 kg, answer (d).
*OQ1.6 The number of decimal places in a sum of numbers should be the same
as the smallest number of decimal places in the numbers summed.
21.4 s
15 s
17.17 s
4.003 s
57.573 s = 58 s, answer (d).
*OQ1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The
meterstick measurement and (c) can both be 4.24 cm. Only (d) does not
overlap. Thus (a), (b), and (c) all agree with the meterstick
measurement.
*OQ1.8 Mass is measured in kg; acceleration is measured in m/s2. Force =
mass × acceleration, so the units of force are answer (a) kg⋅m/s2.
*OQ1.9 Answer (d). Take the difference of the x coordinates of the ends of the
vector, head minus tail: –4 – 2 = –6 cm.
*OQ1.10 Answer (a). Take the difference of the y coordinates of the ends of the
vector, head minus tail: 1 − (−2) = 3 cm.
*OQ1.11 The answers are (a) yes (b) no (c) no (d) no (e) no (f) yes (g) no. Only
force and velocity are vectors. None of the other quantities requires a
direction to be described.
*OQ1.12 Answers (a), (b), and (c). The magnitude can range from the sum of the
individual magnitudes, 8 + 6 =14, to the difference of the individual
magnitudes, 8 − 6 = 2. Because magnitude is the “length” of a vector, it
is always positive.
*OQ1.13 Answer (a). The vector −2D1 will be twice as long as D1 and in the
opposite direction, namely northeast. Adding D2 , which is about
equally long and southwest, we get a sum that is still longer and due
east.
*OQ1.14 Answer (c). A vector in the second quadrant has a negative x
component and a positive y component.
*OQ1.15 Answer (e). The magnitude is 102 + 102 m/s.
*OQ1.16 Answer (c). The signs of the components of a vector are the same as the
signs of the points in the quadrant into which it points. If a vector
arrow is drawn to scale, the coordinates of the point of the arrow equal
the components of the vector. All x and y values in the third quadrant
are negative.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 1 3
ANSWERS TO CONCEPTUAL QUESTIONS
*CQ1.1 A unit of time should be based on a reproducible standard so it can be
used everywhere. The more accuracy required of the standard, the less
the standard should change with time. The current, very accurate
standard is the period of vibration of light emitted by a cesium atom.
Depending on the accuracy required, other standards could be: the
period of light emitted by a different atom, the period of the swing of a
pendulum at a certain place on Earth, the period of vibration of a
sound wave produced by a string of a specific length, density and
tension, and the time interval from full Moon to full Moon.
*CQ1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms.
*CQ1.3 Density varies with temperature and pressure. It would be necessary
to measure both mass and volume very accurately in order to use the
density of water as a standard.
*CQ1.4 Vectors A and B are perpendicular to each other.
*CQ1.5 (a) The book’s displacement is zero, as it ends up at the point from
which it started. (b) The distance traveled is 6.0 meters.
*CQ1.6 No, the magnitude of a vector is always positive. A minus sign in a
vector only indicates direction, not magnitude.
*CQ1.7 The inverse tangent function gives the correct angle, relative to the +x
axis, for vectors in the first or fourth quadrant, and it gives an incorrect
answer for vectors in the second or third quadrant. If the x and y
components are both positive, their ratio y/x is positive and the vector
lies in the first quadrant; if the x component is positive and the y
component negative, their ratio y/x is negative and the vector lies in
the fourth quadrant. If the x and y components are both negative, their
ratio y/x is positive but the vector lies in the third quadrant; if the x
component is negative and the y component positive, their ratio y/x is
negative but the vector lies in the second quadrant.
*CQ1.8 Addition of a vector to a scalar is not defined. Try adding the speed
and velocity, 8.0 m/s + (15.0 m/s î) : Should you consider the sum to
be a vector or a scaler? What meaning would it have?
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 4 Introduction and Vectors
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
4 3
P1.1 For either sphere the volume is V = π r and the mass is
3
4
m = ρV = ρ π r 3 . We divide this equation for the larger sphere by the
3
same equation for the smaller:
m ρ 4π r3 3 r3
= = =5
ms ρ 4π rs3 3 rs3
Then r = rs 3 5 = 4.50 cm ( 1.71) = 7.69 cm .
P1.2 (a) Modeling the Earth as a sphere, we find its volume as
π r = π ( 6.37 × 106 m ) = 1.08 × 1021 m 3
4 3 4 3
3 3
Its density is then
m 5.98 × 1024 kg
ρ= = = 5.52 × 103 kg/m 3
V 1.08 × 1021 m 3
(b) This value is intermediate between the tabulated densities of
aluminum and iron. Typical rocks have densities around 2000 to
3000 kg/m3. The average density of the Earth is significantly
higher, so higher-density material must be down below the
surface.
m
P1.3 Let V represent the volume of the model, the same in ρ = , for both.
V
mgold
Then ρiron = 9.35 kg/V and ρgold = .
V
ρgold mgold
Next, = and
ρiron 9.35 kg
⎛ 19.3 × 103 kg/m 3 ⎞
mgold = ( 9.35 kg ) ⎜ = 22.9 kg
⎝ 7.87 × 10 kg/m ⎟⎠
3 3
P1.4 The volume of a spherical shell can be calculated from
4
V = Vo − Vi = π ( r23 − r13 )
3
m
From the definition of density, ρ = , so
V
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
