26. (a)
To graph the velocity, we estimate the slopes at several
points as follows, then connect the points to create a
smooth curve.
To graph the acceleration, we estimate the slope of the
velocity graph at several points as follows, and then
connect the points to create a smooth curve.
(b)
}
d
d
s
t
}
5
30
t
2
3
t
2
[0, 15] by [
2
300, 100]
}
d
dt
2
2
s
}
5
30
2
6
t
[0, 15] by [
2
100, 50]
The graphs are very similar.
27. (a)
Solving 160
5
490
t
2
gives
t
5 6
}
4
7
}
.
It took
}
4
7
}
of a second. The average velocity was
sec = 280 cm/sec.
(b)
v
5
s
9
(
t
)
5
980
t
a
5
s
0
(
t
)
5
980
The velocity was
s
9
1
}
4
7
}
2
5
560 cm/sec.
The acceleration was
s
0
1
}
4
7
}
2
5
980 cm/sec
2
.
(c)
Since there were about 16 flashes during
}
4
7
}
of a second,
the light was flashing at a rate of about 28 flashes per
second.
28.
Graph C is position, graph A is velocity, and graph B is
acceleration.
A is the derivative of C because it is positive, negative, and
zero where C is increasing, decreasing, and has horizontal
tangents, respectively. The relationship between B and A is
similar.
29.
Graph C is position, graph B is velocity, and graph A is
acceleration.
B is the derivative of C because it is negative and zero
where C is decreasing and has horizontal tangents,
respectively.
A is the derivative of B because it is positive, negative, and
zero where B is increasing, decreasing, and has horizontal
tangents, respectively.
30. (a)
}
d
d
y
t
}
5
}
d
d
t
}
3
6
1
1
2
}
1
t
2
}
2
2
4
5
}
d
d
t
}
3
6
1
1
2
}
6
t
}
1
}
1
t
4
2
4
}
24
5
}
d
d
t
}
1
6
2
t
1
}
2
1
4
}
t
2
2
5
0
2
1
1
}
1
t
2
}
5
}
1
t
2
}
2
1
(b)
The fluid level is falling fastest when
}
d
d
y
t
}
is the most
negative, at
t
5
0, when
}
d
d
y
t
}
5 2
1. The fluid level is
falling slowest at
t
5
12, when
}
d
d
y
t
}
5
0.
(c)
[0, 12] by [
2
2, 6]
y
is decreasing and
}
d
d
y
t
}
is negative over the entire
interval.
y
decreases more rapidly early in the interval,
and the magnitude of
}
d
d
y
t
}
is larger then.
}
d
d
y
t
}
is 0 at
t
5
12, where the graph of
y
seems to have a horizontal
tangent.
160 cm
}
1
}
4
7
}
2
a
(km/hr
2
)
50
25
–25
–50
t
(hours)
15
10
v
(km/hr)
100
–100
–200
–300
t
(hours)
15
5
86
Section 3.4
t
(hours)
0
2.5
5
7.5
10
12.5
15
v
(km/hour
2
)
30
15
0
2
15
2
30
2
45
2
60
t
(hours)
0
2.5
5
7.5
10
12.5
15
v
(km/hour)
0
56
75
56
0
2
94
2
225
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31. (a)
}
d
d
V
r
}
5
}
d
d
r
}
1
}
4
3
}
p
r
3
2
5
4
p
r
2
When
r
5
2,
}
d
d
V
r
}
5
4
p
(2)
2
5
16
p
cubic feet of
volume per foot of radius.
(b)
The increase in the volume is
}
4
3
}
p
(2.2)
3
2
}
4
3
}
p
(2)
3
<
11.092 cubic feet.
32.
For
t
.
0, the speed of the aircraft in meters per second
after
t
seconds is
}
2
9
0
}
t
. Multiplying by
}
360
1
0
h
sec
} ?
}
1
1
00
k
0
m
m
}
,
we find that this is equivalent to 8
t
kilometers per hour.
Solving 8
t
5
200 gives
t
5
25 seconds. The aircraft takes
25 seconds to become airborne, and the distance it travels
during this time is
D
(25)
<
694.444 meters.
33.
Let
v
0
be the exit velocity of a particle of lava. Then
s
(
t
)
5
v
0
t
2
16
t
2
feet, so the velocity is
}
d
d
s
t
}
5
v
0
2
32
t
.
Solving
}
d
d
s
t
}
5
0 gives
t
5
}
3
v
2
0
}
. Then the maximum height, in
feet, is
s
1
}
3
v
2
0
}
2
5
v
0
1
}
3
v
2
0
}
2
2
16
1
}
3
v
2
0
}
2
2
5
}
v
6
0
4
2
}
. Solving
}
v
6
0
4
2
}
5
1900 gives
v
0
<
6
348.712. The exit velocity was
about 348.712 ft/sec. Multiplying by
}
360
1
0
h
sec
}
?
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 Spring '08
 ALL
 Derivative, Velocity, Cos, 1 sec

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