reyes (mr29667) – Homework 3 – chelikowsky – (59005)
1
This printout should have 17 questions.
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001
(part 1 oF 2) 10.0 points
A particle at rest undergoes an acceleration
oF 3
.
3 m
/
s
2
to the right and 4
.
2 m
/
s
2
up.
a) What is its speed aFter 6
.
4 s?
Correct answer: 34
.
1846 m
/
s.
Explanation:
Basic Concepts
The direction oF the motion depends only
on the horizontal and vertical components oF
the velocity at any moment.
v
x
v
y
v
Solution:
±or the horizontal motion,
v
0
x
= 0, so
v
x
=
v
0
x
+
a
x
t
=
a
x
t
±or 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
=
r
v
2
x
+
v
2
y
002
(part 2 oF 2) 10.0 points
b) What is its direction with respect to the
horizontal at this time?
Answer between
−
180
◦
and +180
◦
.
Correct answer: 51
.
8428
◦
.
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.
003
(part 1 oF 2) 10.0 points
A particle travels to the right at a constant
rate oF 6
.
2 m
/
s. It suddenly is given a vertical
acceleration oF 2 m
/
s
2
For 3
.
9 s.
What is its direction oF travel aFter the
acceleration with respect to the horizontal?
Answer between
−
180
◦
and +180
◦
.
Correct answer: 51
.
5198
◦
.
Explanation:
Basic Concepts
The direction oF the motion depends only on
the horizontal and vertical components oF the
velocity at any moment.
±or 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
±or 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
p
v
y
v
x
P
= arctan
p
7
.
8 m
/
s
6
.
2 m
/
s
P
= 51
.
5198
◦
004
(part 2 oF 2) 10.0 points
What is the speed at this time?
Correct answer: 9
.
96393 m
/
s.
Explanation:
so
v
=
r
v
2
x
+
v
2
y
=
r
(6
.
2 m
/
s)
2
+ (7
.
8 m
/
s)
2
= 9
.
96393 m
/
s
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2
005
(part 1 of 2) 10.0 points
A cannon Fres a 0
.
24 kg shell with initial
velocity
v
i
= 8
.
9 m
/
s in the direction
θ
= 55
◦
above the horizontal.
Δ
x
Δ
h
8
.
9 m
/
s
55
◦
Δ
y
y
The shell’s trajectory curves downward be
cause of gravity, so at the time
t
= 0
.
494 s the
shell is below the straight line by some verti
cal distance Δ
h
. Your task is to calculate the
distance Δ
h
in the absence of air resistance.
On what does Δ
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 Fall '09
 Chelikowsky
 Acceleration, Velocity, Correct Answer, initial velocity, m/s, Reyes

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