homework 03 – GADHIA, TEJAS – Due: Feb 7 2007, 4:00 am
1
Question 1
part 1 of 2
10 points
A particle travels to the right at a constant
rate of 5
.
1 m
/
s. It suddenly is given a vertical
acceleration of 2
.
8 m
/
s
2
for 4
.
7 s.
What is its direction of travel after the
acceleration with respect to the horizontal?
Answer between
−
180
◦
and +180
◦
.
Correct answer: 68
.
8168
◦
(tolerance
±
1 %).
Explanation:
Basic Concepts
The direction of the motion depends only on
the horizontal and vertical components of the
velocity at any moment.
For the horizontal motion,
a
x
= 0, so the ve
locity remains the same throughout the mo
tion, and
v
0
x
=
v
x
=
v
Solution
For the vertical motion,
v
0
y
= 0, so
v
y
=
v
0
y
+
a
y
t
=
a
y
t
v
x
v
y
v
opposite the angle
θ
and the horizontal com
ponent
v
x
is the side adjacent to the angle,
so
tan
θ
=
v
y
v
x
θ
= arctan
v
y
v
x
Question 2
part 2 of 2
10 points
What is the speed at this time?
Correct answer: 14
.
1137 m
/
s (tolerance
±
1
%).
Explanation:
so
v
=
radicalBig
v
2
x
+
v
2
y
Question 3
part 1 of 2
10 points
A particle at rest undergoes an acceleration
of 1
.
5 m
/
s
2
to the right and 4 m
/
s
2
up.
a) What is its speed after 7
.
7 s?
Correct answer: 32
.
8944 m
/
s (tolerance
±
1
%).
Explanation:
Basic Concepts
The
direction
of
the
motion
depends
only
on
the
horizontal
and
vertical
com
ponents
of
the
velocity
at
any
moment.
v
x
v
y
v
Solution:
For the horizontal motion,
v
0
x
= 0, so
v
x
=
v
0
x
+
a
x
t
=
a
x
t
For the vertical motion,
v
0
y
= 0, so
v
y
=
v
0
y
+
a
y
t
=
a
y
t
The resultant velocity is the hypotenuse of
the triangle formed by the components, so
v
=
radicalBig
v
2
x
+
v
2
y
Question 4
part 2 of 2
10 points
b) What is its direction with respect to
the horizontal at this time? Answer between
−
180
◦
and +180
◦
.
Correct answer: 69
.
444
◦
(tolerance
±
1 %).
Explanation:
The vertical component
v
y
is the side oppo
site the angle
θ
and the horizontal component
v
x
is the side adjacent to the angle, so
tan
θ
=
v
y
v
x
θ
= arctan
v
y
v
x
The angle
θ
must be expressed in degrees.
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homework 03 – GADHIA, TEJAS – Due: Feb 7 2007, 4:00 am
2
Question 5
part 1 of 2
10 points
A particle has constant acceleration vector
vector
a
with components
a
x
= 2
.
2 m
/
s
2
and
a
y
=
−
3
.
1 m
/
s
2
.
Its initial velocity vector
vector
v
0
(at
time
t
0
= 0) has components
v
0
x
= 3 m
/
s and
v
0
y
= 2
.
5 m
/
s.
What is the particle’s speed at time
t
=
3
.
3 s?
Correct answer: 12
.
846
m
/
s (tolerance
±
1
%).
Explanation:
Basic Concepts:
For a motion with a constant acceleration
vector, the particle’s velocity vector changes
with time according to
vector
v
(
t
) =
t
×
vector
a
+
vector
v
0
.
(1)
And speed is the magnitude of the velocity
vector,
v
=
bardbl
vector
v
bardbl
=
radicalBig
(
v
x
)
2
+ (
v
y
)
2
.
(2)
Solution:
In vector notations, the velocity of the par
ticle at time
t
is given by eq. (1). To evaluate
this equation, it is easiest to switch to compo
nent notations, thus:
v
x
=
t
×
a
x
+
v
0
x
= (3
.
3 s)
×
(2
.
2 m
/
s
2
) + 3 m
/
s
= 10
.
26 m
/
s
,
(3)
and
v
y
=
t
×
a
y
+
v
0
y
= (3
.
3 s)
×
(
−
3
.
1 m
/
s
2
) + 2
.
5 m
/
s
=
−
7
.
73 m
/
s
.
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 Spring '08
 Turner
 Acceleration, Work, Correct Answer

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