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madding (cmm3632) – HW02 – markert – (56475)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
IF the acceleration oF an object is zero at some
instant in time, what can be said about its
velocity at that time?
1.
It is not changing at that time.
correct
2.
It is negative.
3.
It is zero.
4.
Unable to determine.
5.
It is positive.
Explanation:
The acceleration
a
=
Δ
v
Δ
t
= 0
Δ
v
= 0
.
002
10.0 points
An object travels 6 m in the frst second oF
travel, 6 m again during the second second oF
travel, and 6 m again during the third second.
What is the approximate average accelera
tion oF the object during this time interval?
1.
5 m/s
2
2.
9 m/s
2
3.
4 m/s
2
4.
3 m/s
2
5.
6 m/s
2
6.
7 m/s
2
7.
0 m/s
2
correct
8.
Unable to determine
9.
8 m/s
2
Explanation:
Because the distance traveled each second
is constant, its velocity is constant, so the
acceleration oF the object is zero.
003
10.0 points
A record oF travel along a straight path is as
Follows:
(a) Start From rest with constant accelera
tion oF 2
.
53 m
/
s
2
For 19
.
9 s;
(b) Constant velocity oF 50
.
347 m
/
s For the
next 0
.
864 min;
(c) Constant negative acceleration oF
−
10
.
5 m
/
s
2
For 3
.
91 s.
What was the total displacement
x
For the
complete trip?
Correct answer: 3227
.
54 m.
Explanation:
This trip is divided into three sections:
acceleration From rest:
x
a
=
1
2
a t
2
=
1
2
(2
.
53 m
/
s
2
) (19
.
9 s)
2
= 500
.
953 m ;
constant velocity motion:
x
b
=
v t
= (50
.
347 m
/
s)(0
.
864 min)
60 s
1 min
= 2609
.
99 m ;
and deceleration:
x
c
=
v t
+
1
2
a t
2
= (50
.
347 m
/
s)(3
.
91 s)
+
1
2
(
−
10
.
5 m
/
s
2
)(3
.
91 s)
2
= 116
.
594 m
.
ThereFore,
x
tot
=
x
a
+
x
b
+
x
c
= 500
.
953 m + 2609
.
99 m
+ 116
.
594 m
=
3227
.
54 m
.
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View Full Documentmadding (cmm3632) – HW02 – markert – (56475)
2
004
(part 1 of 2) 10.0 points
The initial speed of a body is 7 m
/
s.
What is its speed after 2
.
64 s if it accelerates
uniformly at 1
.
55 m
/
s
2
?
Correct answer: 11
.
092 m
/
s.
Explanation:
Let :
v
0
= 7 m
/
s
a
1
= 1
.
55 m
/
s
2
,
and
t
= 2
.
64 s
.
v
=
v
0
+
a
1
t ,
= 7 m
/
s + (1
.
55 m
/
s
2
) (2
.
64 s)
=
11
.
092 m
/
s
.
005
(part 2 of 2) 10.0 points
What is its speed after 2
.
64 s if it accelerates
uniformly at
−
1
.
55 m
/
s
2
?
Correct answer: 2
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 Spring '08
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