Homework 3
−
Solution
CHANG
1 Take the starting pint of the patrolman as the origin. The final position for the
car and patrolman should be the same and it is
x
P
x
C
2000 m
Both the car and patrolman also should have the same total traveling time, and it can be
obtained from the car’s motion. Motion of the car is just single stage with starting position
x
0
200 m and constant velocity
C
36 m/s. The time of travel can be calculated to be
x
C
x
0
C
t
2000
200
36
t
t
50 s
The patrolman’s motion is multistaged. For the first stage, we can find the velocity and
position at the transition point
A
using
0
0and
x
0
0:
A
0
a
1
t
1
0
2.5
T
2.5
T
x
A
x
0
0
t
1
1
2
a
1
t
1
2
0
0
1
2
2.5
T
2
1.25
T
2
For the second stage, the patrolman’s motion is constant velocity and his final position is at
2000 m, so we have
x
P
x
A
A
t
2
1.25
T
2
2.5
T
t
2
The duration for the second stage is simply
t
2
t
−
t
1
50
−
T
So the first stage’s duration can now be found to be
2000
1.25
T
2
2.5
T
50
−
T
T
20 and 80
Since the total time is only 50 s, the only reasonable answer is
T
20 s
CHANG
2 Take the starting point of car
A
as the origin, we then have
x
0
0 for the car but
x
0
350 m for truck
B
at
t
0. Both vehicles have the same traveling time,
t
, and are at the
same position,
x
A
x
B
when they meet on the road. The position of the car can be written
using the constant acceleration motion with zero initial velocity:
x
A
x
0
0
t
1
2
at
2
1
2
at
2
1
2
10
a
2
5
t
2
m
The position of the truck can also be found from the constant velocity motion:
x
B
x
0
B
t
350
−
15
t
m
Note that the velocity of the truck is negative because it is moving in the opposite direction
of the car. We then set the positions equal to find the time of travel:
x
A
x
B
5
t
2
350
−
15
t
t
−
10 (discard),
t
7s
The distance can now be calculated to be
x
A
x
B
5
7
2
245 m
3.60 (a) The point where the snowball strikes the ground depends on the time of
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View Full Documentflight, which can be determined from the vertical motion. Take origin to be at the ground level
directly below the starting point, so
x
0
0and
y
0
14 m. The initial velocity of the snowball
is
0
7m/sand
0
−
40
∘
. Note that the angle is negative, indicating the initial velocity is
pointing below the
x
−
axis. We set up equations for the horizontal and vertical positions at the
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 Spring '08
 TSChang
 Physics

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