nunez (djn358) – HW04 – Radin – (54915)
1
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beFore answering.
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
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
= 4?
1.
dist = 90 yards
correct
2.
dist = 100 yards
3.
dist = 110 yards
4.
dist = 80 yards
5.
dist = 130 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
= 4 this
area is 9 units, so the student has walked a
distance oF 90 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 = 200 yards
correct
2.
dist = 190 yards
3.
dist = 210 yards
4.
dist = 240 yards
5.
dist = 220 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 20
units, so the student has walked a distance oF
200 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 = 245 yards
correct
2.
dist = 225 yards
3.
dist = 255 yards
4.
dist = 275 yards
5.
dist = 235 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
24
.
5 units (area is always positive remember),
so the student has walked a distance oF 245
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 = 265 yards
2.
dist = 215 yards
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2
3.
dist = 195 yards
4.
dist = 245 yards
5.
dist = 225 yards
correct
Explanation:
The student will turn back when his veloc
ity changes from postive to negative,
i.e.
, at
t
= 8. At this time he is 225 yards from the
RLM building.
keywords:
distance, time, graph, velocity,
area
005
10.0 points
Determine the indeFnite integral
I
=
i
2
−
6
x
√
x
dx .
1.
I
= 4
x
1
/
2
+ 4
x
3
/
2
+
C
2.
I
= 4
x
1
/
2
−
4
x
3
/
2
+
C
correct
3.
I
= 2
x
1
/
2
+ 2
x
3
/
2
+
C
4.
I
= 2
x
1
/
2
+ 4
x
3
/
2
+
C
5.
I
= 2
x
1
/
2
−
4
x
3
/
2
+
C
6.
I
= 4
x
1
/
2
−
2
x
3
/
2
+
C
Explanation:
After division we see that
2
−
6
x
√
x
=
2
√
x
−
6
√
x
= 2
x
−
1
/
2
−
6
x
1
/
2
.
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 Spring '09
 GOGOLEV
 Derivative, Fundamental Theorem Of Calculus, dx, Núñez

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