Recitation ClassWork 2 solutions
Physics 203
1.
Cutnell & Johnson 8
th
Edition, Chapter 2 Problem 68. Also, explain how you can tell from the graph (without doing
any calculations) when the bus has a zero acceleration and a nonzero acceleration.
REASONING
The average velocity for each segment is the slope of the line for that segment.
SOLUTION
Taking the direction of motion as positive, we have from the graph for segments
A
,
B
, and
C
,
1
1
10.0 km – 40.0 km
–2.0
10 km/h
1.5 h – 0.0 h
20.0 km – 10.0 km
1.0
10 km/h
2.5 h – 1.5 h
40.0 km – 20.0 km
40 km/h
3.0 h – 2.5 h
A
B
C
v
v
v
=
=
×
=
=
×
=
=
2.
Problem 65
REASONING AND SOLUTION
The average acceleration for each segment is the slope of that segment.
2
A
2
B
2
C
40 m/s
0 m/s
1.9 m/s
21 s
0 s
40 m/s
0 m/s
0 m/s
48 s
21 s
80 m/s
0 m/s
3.3 m/s
60 s
48 s
a
a
a
−
=
=
−
− 4
=
=
−
− 4
=
=
−
3.
Problem 6
REASONING
The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The
number of football fields is equal to this distance divided by the length
L
of one football field.
SOLUTION
The number of football fields is
(
)(
)
3
3
7.6
10
m /s
110
10
s
Number =
9.1
91.4 m
x
vt
L
L
−
×
×
=
=
=
4.
Problem 7
REASONING
In a race against elGuerrouj, Bannister would run a distance given by his average speed times
the time duration of the race (see Equation 2.1).
The time duration of the race would be elGuerrouj’s
winning time of 3:43.13 (223.13 s).
The difference between Bannister’s distance and the length of the race
is elGuerrouj’s winning margin.
SOLUTION
From the table of conversion factors on the page facing the front cover, we find that one mile
corresponds to 1609 m.
According to Equation 2.1, Bannister’s average speed is
Distance
1609 m
Average speed =
Elapsed time
239.4 s
=
Had he run against elGuerrouj at this average speed for the 223.13s duration of the race, he would have
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 Fall '08
 RICHARDSON, B
 Physics, Acceleration, Work, Velocity, m/s, Bannister

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