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Homework Assignment 3 Solutions
Problem 2.33: Airbag injuries.
During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if
they had hit the windshield or steering wheel. According to safety standards, the bags produce a
maximum acceleration of 60 g, but lasting for only 36 ms (or less).
How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration
of 60g?
Answer:
38.1
cm
Before we try to solve this problem, convert 36 ms into seconds:
36
ms
⋅
1
sec
1000
ms
=
0.036
sec
Then we need to use two kinematic equations. First we use
v
=
v
o
at
to find the initial velocity
v
o
(or velocity before the braking started). We assume v = 0 (because it
stops):
v
o
=−
at
v
o
=−−
60
⋅
9.8
m
/
s
2
⋅
0.036
s
v
o
=
21.2
m
/
s
Note that the acceleration is negative (i.e., a deceleration). Then, we use the second kinematic equation,
v
2
=
v
o
2
2a
x
,
to find the distance
x
traveled in 0.036 sec (remember v = 0 because it stops):
x
=
−
v
o
2
2a
x
=
−
21.2
m
/
s
2
2
−
60
9.8
m
/
s
2
x
=
0.381
m
x
=
38.1
cm
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View Full DocumentProblem 2.38
The "reaction time" of the average automobile driver is about 0.700 s. (The reaction time is the interval
between the perception of a signal to stop and the application of the brakes.) If an automobile can slow
down with an acceleration of
12.0
ft
/
s
2
, compute the total distance covered in coming to a stop after a
signal is observed
(a) from an initial velocity of 15.0 mi/h (in a school zone)
and
(b) from an initial velocity of 55.0 mi/h.
A) Answer: 10.8 m
First, convert
12.0
ft
/
s
2
into m/s:
12
ft
1
s
2
⋅
1
meter
3.28
ft
=
3.7
m
/
s
2
And convert 15.0 mi/h into m/s:
15.0
mi
1
hr
⋅
1609
meters
1
mi
⋅
1
hr
3600
sec
=
6.7
m
/
s
Then to solve, we need to break the problem into two parts. The first part is when the car is not braking
due to the fact that the driver has a reaction time. This distance
d
1
, is easy to calculate:
d
1
=
v
o
t
d
1
=
6.7
m
/
s
0.7
s
d
1
=
4.69
m
Then we need to use the kinematic equation
v
2
=
v
o
2
2a
x
to find the second distance
d
2
. In this equation,
x
is our
d
2
and
v
=
0
.
d
2
=
−
v
o
2
2a
d
2
=
−
6.7m
/
s
2
2
−
3.7
m
/
s
2
d
2
=
6.1
m
Finally, we get
d
total
by adding up
d
1
and
d
2
:
d
total
=
d
1
d
2
d
total
=
10.8
m
B) 99.9 m
First convert 55.0 m/h into m/s:
55.0
mi
1
hr
⋅
1609
meters
1
mi
⋅
1
hr
3600
sec
=
24.6
m
/
s
Then we solve it the same way that we solved A:
d
1
=
v
o
t
d
1
=
24.6
m
/
s
0.7
s
d
1
=
17.2
m
d
2
=
−
v
o
2
2a
d
2
=
−
24.6
m
/
s
2
2
−
3.7
m
/
s
2
d
2
=
81.8
m
d
total
=
d
1
d
2
d
total
=
99.9
m
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View Full DocumentProblem 2.73
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to
Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the
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 Spring '09
 Geller
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