oldmidterm 01 – FRENNEA, KYLE – Due: Feb 12 2007, 4:00 am
1
Question 1
part 1 of 2
10 points
The velocity
v
(
t
) of some particle is plotted
as a function of time on the graph below.
The scale on the horizontal axis is 3 s per
grid square and on the vertical axis 6 m
/
s per
grid square.
Initially, at
t
= 0 the particle is at
x
0
=
20 m.
0
1
2
3
4
5
6
0
1
2
3
4
5
6
7
8
9
v
(
t
)
time
×
(3 s)
velocity
×
(6 m
/
s)
What is the position
x
of the particle at
time
t
= 12 s?
Correct answer: 200 m (tolerance
±
1 %).
Explanation:
Looking at the
v
(
t
) plot we see that over
time
t
= 4
×
3 s = 12 s, the particle’s velocity
decreases from the initial
v
0
= 4
×
6 m
/
s =
24 m
/
s to final
v
f
= 1
×
6 m
/
s = 6 m
/
s. The
v
(
t
) line is straight, which indicates constant
deceleration rate, hence the average velocity
is given by
¯
v
=
v
0
+
v
f
2
=
(24 m
/
s) + (6 m
/
s)
2
= 15 m
/
s
.
Consequently,
the
particle’s
displacement
during this time is simply
Δ
x
=
t
¯
v
= (12 s) (15 m
/
s)
= 180 m
,
and its final position
x
=
x
0
+ Δ
x
= (20 m) + (180 m)
=
200 m
.
Question 2
part 2 of 2
10 points
What is the particle’s acceleration?
Correct answer:
−
1
.
5
m
/
s
2
(tolerance
±
1
%).
Explanation:
The average acceleration of the particle is
¯
a
=
Δ
v
Δ
t
=
v
f
−
v
0
t
=
(6 m
/
s)
−
(24 m
/
s)
(12 s)
=
−
1
.
5 m
/
s
2
.
The acceleration is the slope of the velocity
vs
time graph. Since the
v
(
t
) line is straight,
the acceleration is constant, hence
a
= ¯
a
=
−
1
.
5 m
/
s
2
.
Question 3
part 1 of 1
10 points
A ranger in a national park is driving at
28 mi
/
h when a deer jumps into the road
198 ft ahead of the vehicle. After a reaction
time of
t
the ranger applies the brakes to
produce and acceleration of
−
9
.
14 ft
/
s
2
.
What is the maximum reaction time al
lowed if she is to avoid hitting the deer?
Correct answer: 2
.
57489 s (tolerance
±
1 %).
Explanation:
Given :
v
i
= 28 mi
/
h = 41
.
0667 ft
/
s
,
v
f
= 0
,
and
a
=
−
9
.
14 ft
/
s
2
.
The total distance covered is defined by
v
i
t
+ Δ
x
stop
= Δ
x ,
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oldmidterm 01 – FRENNEA, KYLE – Due: Feb 12 2007, 4:00 am
2
where
t
is the reaction time.
v
2
f
=
v
2
i
+ 2
a
Δ
x
= 0
since
v
f
= 0 mi/h, so
Δ
x
stop
=
−
v
2
i
2
a
.
Thus
v
i
t
= Δ
x
−
Δ
x
stop
t
=
Δ
x
v
i
−
Δ
x
stop
v
i
=
Δ
x
v
i
+
v
2
i
2
a
v
i
=
Δ
x
v
i
+
v
i
2
a
=
198 ft
41
.
0667 ft
/
s
+
41
.
0667 ft
/
s
2 (
−
9
.
14 ft
/
s
2
)
=
2
.
57489 s
.
Question 4
part 1 of 1
10 points
The height of a helicopter above the ground
is given by
h
=
c t
3
, where
c
= 1
.
6 m
/
s
3
,
h
is
in meters, and
t
is in seconds. The helicopter
takes off at
t
= 0 s.
After 2 s it releases a
small mailbag.
The acceleration of gravity is 9
.
8 m
/
s
2
.
How long after its release does the mailbag
reach the ground?
Correct answer: 4
.
499 s (tolerance
±
1 %).
Explanation:
Given :
t
= 2 s
.
Under free fall,
h
(
t
) =
y
(
t
) =
y
0
+
v
0
t
+
1
2
a t
2
.
The initial height of the mailbag is the
height of the helicopter 2 s after takeoff
h
0
=
h
(
t
) = (1
.
6 m
/
s
3
) (2 s)
3
= 12
.
8 m
,
and it starts its free fall motion from this
point. Its initial velocity is equal to the veloc
ity of the helicopter at that time
v
0
=
dh
dt
= 3
c t
2
= 3 (1
.
6 m
/
s
3
) (2 s)
2
= 19
.
2 m
/
s
.
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 Spring '08
 Turner
 Acceleration, Velocity, Correct Answer, m/s, KYLE

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