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11.
Picture the Problem
: The car is accelerated horizontally in
the direction opposite its motion in order to slow it down
from 16.0 m/s to 9.50 m/s.
Strategy:
Use equation 51 and the definition of
acceleration to determine the net force on the car as it slows
down.
Then use equation 210 to find the distance traveled
by the car as it slows down.
Solution:
1. (a)
Use equation 51 and the
definition
of acceleration to find the net force on the
car:
(
29
(
29
(
29
(
9.50 16.0 m/s
ˆ
950 kg
51
1.20 s
5.1 kN opposite to the direction of motion
m
m
t

∆
=
=
=
= 
∆
=
v
F
a
x
r
r
r
2.
(b)
Use equation 210 to find the distance
traveled by the car as it slows down:
(
29
(
29
(
29
1
1
0
2
2
16.0
9.50 m/s 1.20 s
15.3
x
v
v
t
∆ =
+
∆ =
+
=
Insight:
We must consider “950 kg” as having only two significant figures because the zero is
ambiguous.
That limits the net force to two significant figures, even though the acceleration
(
29
2
ˆ
5.42 m/s

x
has three significant figures.
18.
Picture the Problem
: The free body diagrams for the car and
the trailer is shown at right.
The diagram assumes there is no
friction.
Strategy:
In order to determine the forces acting on an object,
you must consider only the forces acting on that object and
the motion of that object alone.
For the trailer there is only
one force
1
F
r
exerted on it by the car, and it has the same
acceleration (1.85 m/s
2
) as the car.
For the car there are two
forces acting on it, the engine
2
F
r
and the trailer
1

F
r
.
Apply
Newton’s Second and Third Laws as appropriate to find the
requested forces.
Solution:
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This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.
 Spring '11
 BEAN

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