Thomas Calculus 13th [Solutions]

Section 3.4 The Derivative as a Rate of Change 139
3.
4.
3 2
s t 3t 3 t , 0 t 3
(a) displacement s s(3) s(0)
9 m, v
s 9
3 m/ sec
(b)
(c)
s
ds
dt
2
v 3t 6t 3 | v(0)| | 3| 3 m/ sec and | v (3)| | 12| 12 m/ sec; a
2
av
2
t
3
2
d s
dt
6t
6
2
a(0) 6 m/ sec and a(3) 12 m/ sec
2
2
2
v 0 3t 6t
3 0 t 2t
1 0 ( t 1) 0 t 1. For all other values of t in the interval
2
the velocity v is negative (the graph of v 3t 6t 3 is a parabola with vertex at t 1 which opens
downward the body never changes direction).
4
t 3
t
4
t
2 , 0 t 3
(a) s s(3) s(0) 9
m,
(b)
(c)
3 2
9
4
v
s 3
av m/ sec
t 3 4
v and | v(3)| 6 m/sec;
4
2
v t 3t 2t | (0)| 0 m/ sec
a 3t
6t
2 a(0) 2 m/ sec and
2
a(3) 11 m/ sec
3 2
v 0 t 3t 2t 0 t( t 2)( t 1) 0 t 0,1, 2 v t( t 2)( t 1) is positive in the interval for
0 t 1 and v is negative for 1 t 2 and v is positive for 2 t 3 the body changes direction at t 1
and at t 2.
2
5. s
25 5 , 1 t 5
2
t t
(a) s s(5) s(1)
20 m, v
20
av 5 m/ sec
4
(b) v 50 5
| v (1)| 45 m/sec and | v(5)|
1
150 10
2
m/ sec; a
a(1) 140 m/ sec and a(5)
4
m/sec
3 2
t t
5
4 3
t t
25
(c) v 0
50 5t
0 50 5t 0 t 10 the body does not change direction in the interval
3
6. s
25
t
t
5 , 4 t 0
(a) s s(0) s( 4) 20 m, v
20
av 5 m/sec
4
(b) v
25
| v ( 4)| 25 m/ sec and
50
2
| v(0)| 1 m/ sec; a a ( 4) 50 m/ sec and a(0)
2
m/ sec
2
3
( t 5)
( t 5)
5
(c) v 0
25
0 v is never 0 the body never changes direction
2
( t 5)
2
2
7.
8.
9.
3 2
s t 6t 9t and let the positive direction be to the right on the s-axis.
2
2
2
(a) v 3t 12t 9 so that v 0 t 4t 3 ( t 3)( t 1) 0 t 1 or 3; a 6t 12 a(1) 6 m/ sec
2
and a (3) 6 m/ sec . Thus the body is motionless but being accelerated left when t 1, and motionless
but being accelerated right when t 3.
(b) a 0 6t 12 0 t 2 with speed | v (2)| |12 24 9| 3 m/sec
(c) The body moves to the right or forward on 0 t 1, and to the left or backward on 1 t 2. The positions
are s(0) 0, s (1) 4 and s (2) 2 total distance | s(1) s(0)| | s(2) s (1)| |4| | 2| 6 m.
2
v t 4t 3 a 2t
4
2
2
2
(a) v 0 t 4t 3 0 t 1 or 3 a (1) 2 m/sec and a(3) 2 m/sec
(b) v 0 ( t 3) ( t 1) 0 0 t 1 or t 3 and the body is moving forward; v 0 ( t 3)( t 1) 0
1 t 3 and the body is moving backward
(c) velocity increasing a 0 2t 4 0 t 2; velocity decreasing a 0 2t 4 0 0 t 2
2
sm
1.86t vm
3.72t and solving 3.72t 27.8 t 7.5 sec on Mars; s j 11.44t v j 22.88t and
solving 22.88t 27.8 t 1.2 sec on Jupiter.
2
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