Cal Poly State University, SLO
D. Niebuhr
Physics Department
Winter 2008
Phys 141
Homework Solutions
Chapter 2
Assigned Problems:
2, 3, 4, 6, 8, 16, 18, 21, 24, 30, 32, 52, 70, 78, 82
2.2.
Solve:
Diagram
Position
Velocity
Acceleration
(a)
(b)
(c)
Negative
Negative
Positive
Positive
Negative
Negative
Positive
Negative
Negative
2.3.
Solve:
A forgetful physics professor goes for a walk on a straight country road. Walking at a constant
speed, he covers a distance of 300 m in 300 s. He then stops and watches the sunset for 100 s. Finding that it was
getting dark, he walks faster back to his house covering the same distance in 200 s.
2.4.
Solve:
Forty miles into a car trip north from his home in El Dorado, an absentminded English professor
stopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he was
supposed to go on this trip on Sunday. Absentmindedly he missed his exit and stopped after one hour of driving
at another rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly,
confused and tired as he was, and reached El Dorado in two hours.
2.6.
Model:
We will consider Larry to be a particle.
Visualize:
Solve:
Since Larry’s speed is constant, we can use the following equation to calculate the velocities:
f
i
s
f
i
s
s
v
t
t

=

(a)
For the interval from the house to the lamppost:
1
200 yd
600 yd
200 yd/min
9:07 9:05
v

=
= 

For the interval from the lamppost to the tree:
2
1200 yd
200 yd
333 yd/min
9:10
9:07
v

=
= +

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View Full DocumentCal Poly State University, SLO
D. Niebuhr
Physics Department
Winter 2008
Phys 141
(b)
For the average velocity for the entire run:
avg
1200 yd
600 yd
120 yd/min
9:10
9:05
v

=
= +

2.8.
Model:
The bicyclist is a particle.
Visualize:
Please refer to Figure Ex2.8.
Solve:
The slope of the positionversustime graph at every point gives the velocity at that point. The slope
at
t
=
10 s is
100 m
50 m
2.5 m/s
20 s
s
v
t
∆

=
=
=
∆
The slope at
t
=
25 s is
100 m 100 m
0 m/s
10 s
v

=
=
The slope at
t
=
35 s is
0 m 100 m
10 m/s
10 s
v

=
= 
2.16.
Model:
Represent the jet plane as a particle
Visualize:
Solve:
(a)
Since we don’t know the time of acceleration, we will use
2
2
1
0
1
0
2
2
2
2
2
1
0
1
2 (
)
(400 m/s)
(300 m/s)
8.75 m/s
2
2(4000 m)
v
v
a x
x
v
v
a
x
=
+



⇒
=
=
=
(b)
The acceleration of the jet is approximately equal to
g
,
the acceleration due to gravity.
2.18.
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 Winter '06
 staff
 Acceleration, Work, Physics Department PHYS, SLO Physics Department, SLO Physics Department Phys, D. Niebuhr Winter, Cal Poly State

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