Physics 5th Edition Solution Manual by James S. Walker | 2026/2027 Complete Problem Solutions Guide

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Chapter 1: Introduction to Physics

Answers to Even-Numbered Conceptual Questions


2. The quantity T + d does not make sense physically, because it adds together variables that have different physical
dimensions. The quantity d/T does make sense, however; it could represent the distance d traveled by an object in the
time T.
4. The frequency is a scalar quantity. It has a numerical value, but no associated direction.
7 17 8 9
6. (a) 10 s; (b) 10,000 s; (c) 1 s; (d) 10 s; (e) 10 s to 10 s.




Solutions to Problems and Conceptual Exercises

1. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
1 gigadollars
Solution: (a) Convert the units: $152,000,000   0.152 gigadollars
1109 dollars
1 teradollars
(b) Convert the units again: $152,000,000   1.52 10 4 teradollars
11012 dollars
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.


2. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.

1.0 10 6 m
Solution: (a) Convert the units: 85  m   8.5 105 m
m

1.0 10 6 m 1000 mm
(b) Convert the units again: 85  m    0.085 mm
m 1m
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.

3. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.

Gm 1109 m
Solution: Convert the units: 0.3   3 108 m/s
s Gm
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.




Copyright © 2017 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.
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Q&A solutions guide- Solution Manual for Physics
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Edition)for
byPhysics
James Solution
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(5th
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Manual
by James
for Physics
S. Walker.pdf
(5th Edition) by James S. WalkerQ&A solutions guide

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(5th Edition) by James S. WalkerQ&A solutions guide
Chapter 1: Introduction to Physics James S. Walker, Physics, 5th Edition


4. Picture the Problem: This problem is about the conversion of units.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
teracalculation 11012 calculations 110 9 s
Solution: Convert the units: 136.8  
s teracalculation ns
 136,800 calculations/ns  1.368  10 calculations/ns
5



Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.


5. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.

Solution: 1. (a) Substitute 1 2
dimensions for the variables: x at
2
1 [L] 2
[L] 2
[T] [L] The equation is dimensionally consistent.
2 [T]
2. (b) Substitute dimensions v
for the variables:
t
x
 L T  1
T    Not dimensionally consistent
 L T
3. (c) Substitute dimensions 2x
t
for the variables: a
 L
T  T    T   Dimensionally consistent
2

 L T 
2


Insight: The number 2 does not contribute any dimensions to the problem.


6. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.

Solution: 1. (a) Substitute dimensions x  L  1  T Yes
  
for the variables: v  L  T  1 T 

a  L   T  1  T    T 
2
1
2. (b) Substitute dimensions for the variables:    No
v  L  T  1 T  T 
2x  L 1
T   T 
2
3. (c) Substitute dimensions for the variables:    Yes
 L T  1 T 
2 2
a

v 2  L   T   L  T    L  L
2 2 2 2

4. (d) Substitute dimensions for the variables:     No
 L T   L T   L
2 2
a

Insight: When squaring the velocity you must remember to square the dimensions of both the numerator (meters) and
the denominator (seconds).




Copyright © 2017 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.
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Q&A solutions guide- Solution Manual for Physics
Solution
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byPhysics
James Solution
S.
(5th
Walker
Edition)
Manual
by James
for Physics
S. Walker.pdf
(5th Edition) by James S. WalkerQ&A solutions guide

, 2026/2027: exam testbank solutions manual Q&A
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Physics
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(5th Edition) by James S. WalkerQ&A solutions guide
Chapter 1: Introduction to Physics James S. Walker, Physics, 5th Edition


7. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.

Solution: 1. (a) Substitute dimensions   L 
  T      
vt   T  L Yes
for the variables:
 
  L  2
2. (b) Substitute dimensions for the variables: 1
a t 2  12  2   T    L Yes
2
 T  
 
  L   L No
3. (c) Substitute dimensions for the variables: 2a t  2  2   T  
 T  
  T
v 2  L   T   L T   L  L Yes
2 2 2 2

4. (d) Substitute dimensions for the variables:    
 L T   L T   L
2 2
a
Insight: When squaring the velocity you must remember to square the dimensions of both the numerator (meters) and
the denominator (seconds).


8. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.

Solution: 1. (a) Substitute dimensions   L  2
for the variables:
1
a t 2  12  2   T    L No
2
 T  
 
  L   L
2. (b) Substitute dimensions for the variables: at   2  T   Yes
 T 


 T 
2x 2  L
3. (c) Substitute dimensions for the variables:   T  No
 L T 
2
a

  L   L   L Yes
2

4. (d) Substitute dimensions for the variables: 2a x  2  2   L  
 T  T T
2
 
Insight: When taking the square root of dimensions you need not worry about the positive and negative roots; only the
positive root is physical.


9. Picture the Problem: This is a dimensional analysis question.
Strategy: Manipulate the dimensions in the same manner as algebraic expressions.

Solution: Substitute dimensions for the variables: v 2  2a x p
  L    L
2

     L
p

  T     T  
2

 L   L
2 p 1
therefore p  1
Insight: The number 2 does not contribute any dimensions to the problem.




Copyright © 2017 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.
1–3
Q&A solutions guide- Solution Manual for Physics
Solution
(5thManual
Edition)for
byPhysics
James Solution
S.
(5th
Walker
Edition)
Manual
by James
for Physics
S. Walker.pdf
(5th Edition) by James S. WalkerQ&A solutions guide