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tiL12214_appd_633-642.indd Page 633 09/10/10 8:35 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles

APPENDIX D

Solutions for Follow-Up Example Exercises

Note: Solutions that involve calculations of measurements are Example 2.4, p. 30
rounded up or down to conform to the rules for significant fig- m _ _
v f − v i
v i = 0 a = ∴ v f = at + v i
s
ures as described in appendix A. t
v f = ? m _
( )

= 5 (6 s)
m _
a = 5 s 2

2
CHAPTER 1 s m _ s _

t = 6 s = (5)(6) ×
2
Example 1.2, p. 9 s 1
m _
m = 15.0 g ρ = = 30
m _

V
V = 4.50 cm 3 s
15.0 g
ρ = ? _

=
4.50 cm 3 Example 2.6, p. 32
m _
v f − v i
g
_

= 3.33 v i = 25.0 a = _
s
t

cm 3 m _
v f = 0 m _ m _

s __

0 − 25.0

= s s
CHAPTER 2 t = 10.0 s 10.0 s
a = ?
−25.0 m _
1 _

Example 2.2, p. 28 = _ ×

s
− 10.0 s
v = 8.00 km/h
m _
t = 10.0 s = −2.50
d = ? s 2
The bicycle has a speed of 8.00 km/h and the time factor is 10.0 s, so
Example 2.9, p. 43
km/h must be converted to m/s: F _

m = 20 kg F = ma ∴ a =
m
m _ F = 40 N
0.2778

kg . m
− _ _ a = ? _
km
s
×

v = 8.00 40
_ h _
2
km
s

h = 20 kg
km
m _ h _ _ kg . m 1 _

= (0.2778)(8.00) × × 40 _ _

s km h = s ×
20
2
kg
m _

= 2.22 m _

s = 2
2
s
− _
d
v =

t
Example 2.11, p. 44
− dt _ m = 60.0 kg

v t =
t
w = 100.0 N
−
d = t g = ?
v
m _ w _

= (2.22 )(10.0 s) w = mg ∴ g =
s m
kg . m
m _ s _ _

= (2.22)(10.0) × 100.0

s 1 2
s

= __
= 22.2 m 60.0 kg
100.0 _
_ kg . m 1 _

= ×
60.0 s 2 kg
m _

= 1.67
s 2
D-1 633