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Al Ali, Ahmed – Homework 3 – Due: Sep 10 2007, 11:00 pm – Inst: Dr Mike Stokes
1
This printout should have 26 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
This is ONLINE HOMEWORK No 3
11.00 pm in Texas is 8.00 am the next morn
ing in Abu Dhabi
001
(part 1 oF 1) 10 points
A rocket sled For testing equipment under
large accelerations starts at rest and acceler
ates according to the expression
a
= (1
.
9 m
/
s
3
)
t
+ (2
.
7 m
/
s
2
)
.
How Far does the sled move in the time
interval
t
= 0 to
t
= 2
.
7 s?
Correct answer: 16
.
0745 m.
Explanation:
Let :
α
= 1
.
9 m
/
s
3
,
β
= 2
.
7 m
/
s
2
,
and
t
= 2
.
7 s
.
Basic Concepts:
a
=
d v
dt
=
d
dt
µ
d x
dt
¶
Solution:
We are given
a
=
α t
+
β ,
thus
v
=
Z
t
0
(
α t
0
+
β
)
dt
0
=
α
2
t
2
+
β t
=
d x
dt
x
=
Z
t
0
‡
α
2
t
0
2
+
β t
0
·
dt
0
=
α
6
t
3
+
β
2
t
2
=
(1
.
9 m
/
s
3
)
6
(2
.
7 s)
3
+
(2
.
7 m
/
s
2
)
2
(2
.
7 s)
2
= 16
.
0745 m
.
keywords:
002
(part 1 oF 2) 5 points
A truck travels beneath an airplane that is
moving 130 km
/
h at an angle oF 29
◦
to the
ground.
a) How Fast must the truck travel to stay
beneath the airplane?
Correct answer: 113
.
701 km
/
h.
Explanation:
v
y
v
x
130
km
/
h
29
◦
Note:
±igure is not drawn to scale.
Basic Concept:
v
x
=
v
(cos
θ
)
Given:
v
= 130 km
/
h
θ
= 29
◦
Solution:
We need the plane’s horizontal component:
v
x
= (130 km
/
h)(cos29
◦
)
= 113
.
701 km
/
h
003
(part 2 oF 2) 5 points
b) What is the magnitude oF the vertical com
ponent oF the velocity oF the plane?
Correct answer: 63
.
0253 km
/
h.
Explanation:
Basic Concept:
v
y
=
v
(sin
θ
)
Solution:
v
y
= (130 km
/
h)(sin29
◦
)
= 63
.
0253 km
/
h
keywords:
004
(part 1 oF 2) 5 points
Tammy leaves the o²ce, drives 31 km due
north, then turns onto a second highway and
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View Full DocumentAl Ali, Ahmed – Homework 3 – Due: Sep 10 2007, 11:00 pm – Inst: Dr Mike Stokes
2
continues in a direction of 31
◦
north of east
for 94 km.
What is her total displacement from the
oFce?
Correct answer: 113
.
131 km.
Explanation:
Basic Concepts
Choose a coordinate system with the positive
x
axis representing 0
◦
and the positive
y
axis
representing 90
◦
.
Solution
S
1
S
S
S
2
S
1
S
2
S
2n
S
n
S
2e
S
e
θ
α
±or the vector
s
2
angled at
θ
north of east,
the north component
s
2
n
is the side opposite
the angle
θ
and the east component
s
2
e
is the
side adjacent.
The north component of the ²nal displace
ment is
s
n
=
s
1
+
s
2
n
=
s
1
+
s
2
sin
θ
The east component of the ²nal displacement
is
s
e
=
s
2
e
=
s
2
cos
θ
The total displacement is the hypotenuse of
the right triangle formed by
s
n
and
s
e
, so
s
=
q
s
2
n
+
s
2
e
005
(part 2 of 2) 5 points
At what angle is her displacement? (Consider
east to be 0
◦
and north 90
◦
.)
Correct answer: 44
.
5846
◦
.
Explanation:
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 Spring '09
 NAJAFZADEH
 mechanics, Work

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