Physics 221
Fall 2008
Homework #2 Solutions
Ch. 2
Due Tues, Sept 9, 2008
2.1 A particle moving along the x-axis moves directly from position x = 0.0 m at time t = 0.0 s to
position x = 10.0 m by time t = 10.0 s, and then moves directly to position x = 0.0 m by time t =
20.0 s.
What is the average velocity of the particle
(a) between t = 0.0 s and t = 10.0 s?
vav-x =
x 10.0 m
=
= 1.00 m/s
t 10.0 s
(b) between t = 10.0 s and t = 20.0 s?
vav-x =
x 10.0 m
=
= 1.00 m/s
t
10.0 s
(c) between t = 0.0 s and t = 20.0 s?
vav-x =
x 0.00 m
=
= 0.00 m/s
t
20.0 s
The average speed vave of a particle over some time interval t is the total distance D traveled over
that time interval divided by the time interval. “Distance” is always positive or zero; it is what your
odometer reads on your car.
What is the average speed of the particle
(d) between t = 0.0 s and t = 10.0 s?
vave =
D 10.0 m
=
= 1.00 m/s
t 10.0 s
(e) between t = 10.0 s and t = 20.0 s?
vave =
D 10.0 m
=
= 1.00 m/s
t 10.0 s
(f) between t = 0.0 s and t = 20.0 s?
vave =
D 20.0 m
=
= 1.00 m/s
t 20.0 s
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(g) Compare the results in parts (a)-(c) with the respective results in (d)-(f).
differences.
Explain any
The answers in (a) and (d) are the same because the velocity and speed are both positive during
that time interval. The answers in (b) and (c) are different from those in (e) and (f),
respectively, because of the difference in the definitions of average speed and average velocity.
2.2 A particle moves along the x-axis. The initial position x0 and initial velocity v0x of the particle at
time t = 0 are both zero. The particle accelerates with an (instantaneous) acceleration given by
ax = 3.00 m/s2 – (1.00 m/s3)t where t is the time in seconds.
(a) Plot ax versus t for t = 0 to t = 10.0 s. Use a spreadsheet program if you can to make this and
the following plots.
The three plots for this problem were made using the program KaleidaGraph on a Macintosh.
All three requested plots are made on the same figure below. The numerical vertical scale
applies to all three plots.
Problem 2.2
20
x (m)
15
5
0
x
x
a,v,x
10
-5
a (m/s2)
x
-10
vx (m/s)
-15
-20
0
2
4
6
time (s)
8
10
(b) Derive an expression for the velocity vx versus t. Plot vx versus t for t = 0 to t = 10.0 s.
t
t
0
0
vx = v0 x + ax (t) dt = 0 + [(3.00 m/s 2 ) (1.00 m/s 3 )t] dt = (3.00 m/s 2 )t (0.500 m/s 3 )t 2
(c) Derive an expression for the position x versus t. Plot x versus t for t = 0 to t = 10.0 s.
t
t
0
0
x = x0 + vx (t) dt = 0 + [(3.00 m/s 2 )t (0.500 m/s 3 )t 2 ] dt = (1.50 m/s 2 )t 2 (0.167 m/s 3 )t 3
2
(d) What are the average velocity and the average acceleration of the particle over the time interval
from t = 0 to t = 10.0 s?
The initial velocity and position are both zero. From parts (c) and (b), the position at t = 10.0 s
is x = 17 m and the velocity at t = 10.0 s is vx = 20 m/s, so
vx-av =
x 17 m
=
= 1.7 m/s
t
10 s
and
ax-av =
vx 20 m/s
=
= 2.0 m/s 2 .
10 s
t
2.3 A particle is moving along the x-axis. The position x of the particle versus time is plotted in the
figure.
1.5
Problem 2.3
1.0
x (m)
0.5
0.0
-0.5
-1.0
-1.5
-1.0
-0.5
0.0
time (s)
0.5
1.0
(a) At what instants or intervals of time is the velocity vx positive? negative? zero? Explain.
The (instantaneous) velocity is dx/dt, which is the slope of the curve. Thus vx is positive for
1.0 s < t < 0.0 s, is negative for 0.0 s < t < 1.0 s, and is zero for t = 1.0 s, 0.0 s, and 1.0 s.
(b) At what instants or intervals of time is the acceleration ax positive? negative? zero? Explain.
The (instantaneous) acceleration is dvx/dt, which is the curvature d2x/dt2. The curvature is
positive if x(t) is bending upwards, is negative if x(t) is bending downwards, and is zero at an
inflection point. Thus ax is positive for 1.0 s t < 0.5 s and for 0.5 s < t 1.0 s, is negative
for 0.5 s < t < 0.5 s, and is zero for t = 0.5 s and 0.5 s.
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2.4 A person throws a ball vertically upward, in the positive y-direction, with an initial speed of 29.4
m/s. The release point is at y0 = 0. After release, the (constant) acceleration of the ball due to
gravity is ay = 9.80 m/s2. The vertical position of the ball versus time is designated as y(t). In this
problem ignore the possible influence of air resistance on the motion of the ball.
(a) Write down an expression for the position y of the ball versus time.
1
This is a constant acceleration problem, for which y = y0 + v0 yt + ayt 2 .
2
2
Here, y0 = 0, v0y = 29.4 m/s, and ay = 9.80 m/s as stated in the problem.
(b) Write down an expression for the velocity of the ball versus time. How long does it take for the
ball to reach its maximum height?
At the top of the trajectory, the ball is momentarily at rest. We have vy = v0y + ayt. Setting vy =
0 gives the time
v0 y
29.4 m/s
t=
=
= 3.00 s .
9.80 m/s 2
ay
(c) How high does the ball go before it starts coming back down?
From parts (a) and (b), we have
y = y0 + v0 yt +
1 2
ayt = 0 + (29.4 m/s)(3.00 s) + (4.90 m/s 2 )(3.00 s)2 = 44.1 m .
2
(d) How long does it take for the ball to reach y = 0 again after it is launched? Over what time
interval are the signs of vy and ay the same? Over what time interval are the signs of vy and ay
opposite to each other? What are the velocity and speed of the ball when the ball returns to y =
0?
Using y = y0 + v0 yt +
1 2
ayt from part (a), and setting y = y0 = 0, solving for the (nonzero) time
2
gives
2v0 y
2(29.4 m/s)
= 6.00 s .
9.80 m/s 2
ay
Note that this time is twice the time for the ball to reach its maximum height. Thus the time for
the ball to go up is the same as the time for the ball to come down.
t=
=
During the time interval from t = 0 to t = 3.00 s when the ball is going up, vy and ay have
opposite signs: vy is positive and ay is negative. From t = 3.00 s to t = 6.00 s when the ball is
coming down, vy and ay have the same negative sign.
To determine the velocity when the ball returns to y = 0, use the information that the ball has
zero velocity at the maximum height, v0y = 0, and it takes 3.00 s to return to y = 0 from there:
vy = v0 y + ayt = 0 + (9.80 m/s 2 )(3.00 s) = 29.4 m/s .
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Alternatively, one can use the time t = 6.00 s that the ball takes to go up and come down:
vy = v0y + ayt = 29.4 m/s + (9.80 m/s2)(6.00 s) = 29.4 m/s .
Thus the speed at which the ball returns to y = 0 is |vy| = 29.4 m/s. This speed is the same as the
launch speed, which we will see later is a consequence of the law of conservation of energy.
(e) Plot y, vy and ay versus time over the time interval for which the ball is at or above y = 0. Use a
spreadsheet program to make these plots if you can.
All three plots are placed on the same figure as follows. The requested time interval is 6.00 s.
The plots were made using the program KaleidaGraph on a Macintosh. The numerical scale on
the vertical axis applies to all three plots.
Problem 2.4(e)
50
40
y, vy, ay
30
y (m)
20
v (m/s)
y
10
0
-10
2
-20
-30
ay (m/s )
0
1
2
3
4
time (s)
5
6
Note that the velocity goes to zero at the top of the trajectory, the initial and final speeds are the
same, the velocity is zero and changes sign at the top of the trajectory, and the acceleration is
constant and negative.
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