Business Calc Homework w answers_Part_18 - 86 Section 3.4...

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

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
}
?