Pickering, Tracey – Homework 6 – Due: Feb 3 2006, noon – Inst: Drummond
1
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printout
should
have
9
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
A jet airliner moving initially at 636 mph to
the east where there is no wind moves into a
region where the wind is blowing at 853 mph
in a direction 15
◦
north of east.
What is the new speed of the aircraft with
respect to the ground?
Correct answer: 1476
.
53 mph.
Explanation:
V
jet
V
wind
V
φ
θ
The new velocity of the aircraft is the sum
of the initial velocity and the velocity of the
wind:
~v
f
=
~v
i
+
~v
wind
=
v
x
ˆ
ı
+
v
y
ˆ
where
v
x
=
v
i
+
v
wind
cos
φ
v
y
=
v
wind
sin
φ
with
φ
being the angle between
v
wind
and
v
jet
.
So the speed is

~v

=
q
v
2
x
+
v
2
y
.
002
(part 2 of 2) 10 points
What is the direction of the aircraft (use coun
terclockwise from due East as the positive an
gular direction, between the limits of

180
◦
and +180
◦
)?
Correct answer: 8
.
59917
◦
.
Explanation:
The direction is
θ
= arctan
v
y
v
x
¶
.
003
(part 1 of 1) 10 points
A dart gun is fired while being held hori
zontally at a height of 0
.
865 m above ground
level, and at rest relative to the ground. The
dart from the gun travels a horizontal dis
tance of 6
.
03 m. A child holds the same gun
in a horizontal position while sliding down a
30
.
4
◦
incline at a constant speed of 2
.
89 m
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
30
.
4
◦
2
.
89 m
/
s
x
x
1
.
26 m
What horizontal distance
x
will the dart
travel if the child fires the gun forward when
it is 1
.
26 m above the ground?
Correct answer: 6
.
39023 m.
Explanation:
Let :
x
0
= 6
.
03 m
,
y
0
= 0
.
865 m
,
v
0
= 2
.
89 m
/
s
θ
= 30
.
4
◦
,
and
y
1
= 1
.
26 m
First, find the initial velocity of the dart
when shot horizontally, at rest, 0
.
865 m above
the ground. From
y
=
v
y
0
t
+
1
2
g t
2
,
with
v
y
0
=0, and down defined as positive, we
have
t
=
r
2
y
0
g
=
s
2 (0
.
865 m)
(9
.
8 m
/
s
2
)
= 0
.
420155 s
,
and
x
=
v
x
0
t
, thus
v
x
0
=
x
0
t
=
(6
.
03 m)
(0
.
420155 s)
= 14
.
3518 m
/
s
.
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Pickering, Tracey – Homework 6 – Due: Feb 3 2006, noon – Inst: Drummond
2
Since we now know the initial horizontal
velocity, we can find how far the dart will go
if it is shot horizontally,
y
1
= 1
.
26 m above
the ground while sliding down the board at
v
0
= 2
.
89 m
/
s
.
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 Fall '08
 Turner
 Acceleration, Work, Velocity, m/s, Pickering

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