practice 02 – – Due: Jan 29 2007, noon
1
Question 1, chap 2, sect 5.
part 1 of 1
10 points
A sailboat is initially moving at a speed
of 2 m
/
s.
A strong wind blows up and ac
celerates the boat forward with a constant
acceleration of 0
.
1 m
/
s
2
for 10 s.
What is the final speed of the sail boat?
Correct answer: 3 m
/
s (tolerance
±
1 %).
Explanation:
The speed of a boat with initial speed
v
0
with a constant acceleration
a
is given by
v
=
v
0
+
at
Therefore the speed of the boat after time 10 s
s is
v
= 2 m
/
s + (10 s) (0
.
1 m
/
s
2
)
= 3 m
/
s
Question 2, chap 2, sect 6.
part 1 of 1
10 points
An object is released from rest on a planet
that has no atmosphere.
The object falls
freely for 3 m in the first second.
What is the magnitude of the acceleration
due to gravity on the planet?
1.
6
.
0 m
/
s
2
correct
2.
1
.
5 m
/
s
2
3.
3
.
0 m
/
s
2
4.
10
.
0 m
/
s
2
5.
12
.
0 m
/
s
2
Explanation:
Let :
s
= 3 m
.
s
=
1
2
at
2
a
=
2
s
t
2
=
2 (3 m)
(1
.
0 s)
2
=
6 m
/
s
2
.
Question 3, chap 2, sect 5.
part 1 of 2
5 points
A motorist is traveling at 15 m
/
s when he
sees a deer in the road 39 m ahead.
If the maximum negative acceleration of
the vehicle is

7 m
/
s
2
, what is the maximum
reaction time Δ
t
of the motorist that will
allow him to avoid hitting the deer?
Correct answer: 1
.
52857 s (tolerance
±
1 %).
Explanation:
Let :
v
0
= 15 m
/
s
,
v
= 0
,
and
a
=

7 m
/
s
2
.
Basic Concept:
v
=
x
t
v
2
=
v
2
0
+ 2
ax.
Solution:
v
2
=
v
2
0
+ 2
ax
stop
x
stop
=

v
2
0
2
a
= 16
.
0714 m
.
x

x
stop
=
v
0
Δ
t.
This yields
Δ
t
=
x

x
stop
v
0
= 1
.
52857 s
.
Question 4, chap 2, sect 5.
part 2 of 2
5 points
If his reaction time is 1
.
71747 s, how fast
will he be traveling when he reaches the deer?
Correct answer: 6
.
29832 m
/
s (tolerance
±
1
%).
Explanation:
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practice 02 – – Due: Jan 29 2007, noon
2
Let:
Δ
t
= 1
.
71747 s
Solution:
x
slow
=
x

v
0
Δ
t.
We again use the relation
v
2
=
v
2
0
+ 2
ax
slow
,
which gives
v
=
radicalBig
v
2
0
+ 2
a
(
x

v
0
Δ
t
)
= 6
.
29832 m
/
s
.
Question 5, chap 2, sect 5.
part 1 of 1
10 points
Hint:
x
1
and
x
2
never have the same value.
The
x
coordinates of two objects moving
along the
x
axis are given below as a function
of time
t
.
x
1
= (4 m
/
s)
t
and
x
2
=

(82 m) + (16 m
/
s)
t

(1 m
/
s
2
)
t
2
.
Calculate the magnitude of the distance of
closest approach of the two objects.
1.
23 m
2.
46 m
correct
3.
33 m
4.
39 m
5.
47 m
6.
54 m
7.
25 m
8.
36 m
9.
29 m
10.
51 m
Explanation:
Given :
v
1
= 4 m
/
s
,
x
0
= 82 m
,
v
2
= 16 m
/
s
,
and
a
2
= 1 m
/
s
2
.
x
1
=
v
1
t
and
x
2
=

x
0
+
v
2
t

a
2
t
2
.
Δ
x
=
x
2

x
1
=

x
0
+ [
v
2

v
1
]
t

a
2
t
2
.
The distance of closest approach will be
when the derivative
d
Δ
x
dt
=
v
2

v
1

2
a
2
t
= 0
.
Therefore,
t
=
v
2

v
1
2
a
2
=
(16 m
/
s)

(4 m
/
s)
2 (1 m
/
s
2
)
= 6 s
.
At that time, we have
x
2
=

x
0
+
v
2
t

a
2
t
2
=

(82 m) + (16 m
/
s) (6 s)

(1 m
/
s
2
) (6 s)
2
=

22 m
x
1
=
at
= (4 m
/
s) (6 s)
= 24 m
Δ
x
=
x
2

x
1
= (

22 m)

(24 m)
=

46 m

Δ
x

= 46 m
.
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 Spring '08
 Turner
 Acceleration, Velocity, m/s, xstop

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