badruddin (ssb776) – HW04 – Radin – (57410)
1
This
printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
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
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
= 4?
1.
dist = 60 yards
2.
dist = 110 yards
3.
dist = 90 yards
4.
dist = 70 yards
correct
5.
dist = 80 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 7 units, so the student has walked a
distance of 70 yards.
002
(part 2 of 4) 10.0 points
How far is the student from the RLM build
ing at time
t
= 8?
1.
dist = 240 yards
2.
dist = 230 yards
correct
3.
dist = 270 yards
4.
dist = 220 yards
5.
dist = 250 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
= 8 this area is 23
units, so the student has walked a distance of
230 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 = 270 yards
correct
2.
dist = 250 yards
3.
dist = 280 yards
4.
dist = 260 yards
5.
dist = 300 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
27 units (area is always positive remember),
so the student has walked a distance of 270
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 = 250 yards
2.
dist = 230 yards
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
badruddin (ssb776) – HW04 – Radin – (57410)
2
3.
dist = 280 yards
4.
dist = 260 yards
correct
5.
dist = 300 yards
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 260 yards from the
RLM building.
keywords:
distance,
time, graph, velocity,
area
005
10.0 points
The graph of
f
is shown in the figure
2
4
6
8
2
4
6
If
F
is an antiderivative of
f
and
integraldisplay
8
1
f
(
x
)
dx
=
217
8
,
find the value of
F
(8)

F
(0).
1.
F
(8)

F
(0) =
63
2
2.
F
(8)

F
(0) =
119
4
3.
F
(8)

F
(0) =
245
8
4.
F
(8)

F
(0) =
231
8
5.
F
(8)

F
(0) = 28
correct
Explanation:
We already know that the area under the
graph on the interval 1
≤
x
≤
8 is equal
to
217
8
, alternatively, by the Fundamental
Theorem of Calculus we can say that
F
(8)

F
(1) =
217
8
.
On the other hand,
integraldisplay
8
0
f
(
x
)
dx
=
integraldisplay
1
0
f
(
x
)
dx
+
integraldisplay
8
1
f
(
x
)
dx.
Thus we need to find
integraldisplay
1
0
f
(
x
)
dx
=
F
(1)

F
(0)
.
Now
integraldisplay
1
0
f
(
x
)
dx
=
integraldisplay
1
0
7
4
x dx
=
7
8
bracketleftBig
x
2
bracketrightBig
1
0
=
7
8
.
This is the end of the preview.
Sign up
to
access the rest of the document.