Solutions Manual for Conceptual Physics 12th Edition Hewitt

SOLUTIONS MANUAL FOR
CONCEPTUAL PHYSICS 12TH
EDITION HEWITT

, Conceptual Physics 12th Edition Hewitt Solutions Manual




Solutions


3-1. (a) Distance hiked = b + c km. b km
(b) Displacement is a vector representing Paul’s change in
position. Drawing a diagram of Paul’s trip we can see that displacement –c km
his displacement is b + (–c) km east = (b –c) km east.
(c) Distance = 5 km + 2 km = 7 km; Displacement = (5 km – 2 km) east = 3 km east.

d x
3-2. (a) From v  v  .
t t
x
(b) v  . We want the answer in m/s so we’ll need to convert 30 km to meters and 8 min
t
to seconds:
x 30,000 m
30.0 km  1000
1 km
m
 30, 000 m; 8.0 min 160mins  480 s. Then v    63 m
s
.
t 480 s
Alternatively, we can do the conversions within the equation:
1000 m
x 30.0 km  1 km
v   63 m
s
.
t 8.0 min 160mins
In mi/h:
x 18.6 mi
1 mi
30.0 km  1.61 km
 18.6 mi; 8.0 min 601min
h
 0.133 h. Then v    140 mi
h
.
t 0.133 h
1 mi
x 30.0 km 60 min 1 mi x 30.0 km  1.61 km
Or, v   mi
 1 h  1.61 km  140 h . Or, v   1 h
 140 mi
h
.
t 8.0 min t 8.0 min 60 min
There is usually more than one way to approach a problem and arrive at the correct answer!

d L
3-3. (a) From v  v  .
t t
L 24.0 m
(b) v    40 ms .
t 0.60 s


d x
3-4. (a) From v  v  .
t t
x 0.30 m
(b) v    30 ms .
t 0.010 s

d 2 r
3-5. (a) v   .
t t
2 r 2 (400m)
(b) v    63 ms .
t 40s




© Paul G. Hewitt and Phillip R. Wolf
3-1


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