So (dds785) – HW04 – Radin – (54915)
1
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001
(part 1 of 4) 10.0 points
A Calculus student begins walking in a
straight line from the RLM building towards
the PCL Library. After
t
minutes his velocity
v
=
v
(
t
) is given (in multiples of 10 yards per
minute) by the function whose graph is
3
2
1
0
1
2
3
4
5
6
7
8
9
10
2
4
6
8
10
−
2
2
4
6
−
2
t
v
How far is the student from the RLM build
ing at time
t
= 5?
1.
dist = 55 yards
2.
dist = 65 yards
3.
dist = 35 yards
4.
dist = 45 yards
correct
5.
dist = 85 yards
Explanation:
The student is walking
towards the PCL
Library
whenever
v
(
t
)
>
0 and
towards RLM
whenever
v
(
t
)
<
0.
The distance he has
walked is given by 10 times the area between
the graph of
v
and the
t
axis since 1 unit is
equivalent to 10 yards.
Now at
t
= 5 this
area is 4
.
5 units, so the student has walked a
distance of 45 yards.
002
(part 2 of 4) 10.0 points
How far is the student from the RLM build
ing at time
t
= 7?
1.
dist = 95 yards
2.
dist = 145 yards
3.
dist = 125 yards
4.
dist = 105 yards
correct
5.
dist = 115 yards
Explanation:
The distance the student has walked is
given by the area between the graph of
v
and the
t
axis. Now at
t
= 7 this area is 10
.
5
units, so the student has walked a distance of
105 yards.
003
(part 3 of 4) 10.0 points
What is the total distance walked by the
student from time
t
= 0 to
t
= 10?
1.
dist = 155 yards
2.
dist = 165 yards
correct
3.
dist = 195 yards
4.
dist = 145 yards
5.
dist = 175 yards
Explanation:
The distance the student has walked is
given by the area between the graph of
v
and the
t
axis.
Now at
t
= 10 this area is
16
.
5 units (area is always positive remember),
so the student has walked a distance of 165
yards.
004
(part 4 of 4) 10.0 points
How far is the student from the RLM build
ing when he turns back?
1.
dist = 175 yards
2.
dist = 145 yards
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So (dds785) – HW04 – Radin – (54915)
2
3.
dist = 125 yards
4.
dist = 195 yards
5.
dist = 155 yards
correct
Explanation:
The student will turn back when his veloc
ity changes from postive to negative,
i.e.
, at
t
= 9. At this time he is 155 yards from the
RLM building.
keywords:
distance,
time, graph, velocity,
area
005
10.0 points
Determine the indefinite integral
I
=
integraldisplay
2 + 4
x
√
x
dx .
1.
I
= 4
x
1
/
2
+
8
3
x
3
/
2
+
C
correct
2.
I
= 2
x
1
/
2
+
8
3
x
3
/
2
+
C
3.
I
= 4
x
1
/
2
−
8
3
x
3
/
2
+
C
4.
I
= 2
x
1
/
2
−
8
3
x
3
/
2
+
C
5.
I
= 2
x
1
/
2
−
4
3
x
3
/
2
+
C
6.
I
= 4
x
1
/
2
+
4
3
x
3
/
2
+
C
Explanation:
After division we see that
2 + 4
x
√
x
=
2
√
x
+ 4
√
x
= 2
x
−
1
/
2
+ 4
x
1
/
2
.
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 Fall '08
 Cepparo
 Calculus, Derivative, Fundamental Theorem Of Calculus, dx

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