TAM212
Introductory Dynamics
Spring 2005
S. Balachandar
Homework 4 (1.45, 1.53)
1.45
(a)
We are given acceleration as,
a
=
¨
x
ˆ
i
=
12
ˆ
it m
/
s
2
Integrating to get velocity
v
=
˙
x
ˆ
i
=
(6
t
2
+
C
1
)
ˆ
i m
/
s
Using, ˙
x
ˆ
i
=
2
ˆ
i m
/
s
at
t
=
1
s
,
2
=
6
+
C
1
C
1
=

4
∴
˙
x
=
6
t
2

4
Integrating, once again, we get the displacement
x
=
2
t
3

4
t
+
C
2
3
=
16

8
+
C
2
, using
x
=
3 at
t
=
2
s
C
2
=

5
∴
x
=
2
t
3

4
t

5
Now, to find the displacement vector from
t
=
0
s
to
t
=
5
s
x

t
=
5

x

t
=
0
=
(250

20

5)

(0

0

5)
=
230
m
∴
r
5

r
0
=
230
ˆ
i m
(b)
The displacement vector from
t
=
0
s
to
t
=
5
s
by itself will not give us sufficient information as
to the total distance traveled (Hint:
the particle could have reversed its direction during this time
)
We employ the fact that particle is in rectilinear motion and try and find the time at which the
velocity becomes 0.
˙
x
=
0
6
t
2

4
=
0
t
=
r
2
3
s
1
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So the total distance traveled is
x
Total
=
x

t
=
√
2
3

x

t
=
0
+
x

t
=
5

x

t
=
√
2
3
=

7
.
18

(

5)

+

225

(

7
.
18)

=
2
.
18
+
232
=
234
m
1.53
The speeder is going at a speed
˙
x
s
=
70
mph
=
70(88
/
60) ft/s
=
103 ft/s
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 Spring '10
 richard
 Velocity, displacement vector, Introductory Dynamics

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