Physics 1111
Fall 2009
PS #2 Solutions
Page 1 of 11
Physics 1111
Problem Set #2 Solutions
Problem 1:
A ball is dropped from rest.
After
T
seconds, the ball has fallen a distance
d
.
Relative to this location, how much farther does it fall after another
4
T
seconds?
t
4
d
t
D
The ball is dropped from rest, so
v
i
= 0, and it is in free fall. We can also
assume it’s dropped from an initial position of
x
i
= 0. With this information,
we can get a relation between the ball’s position and the time using one of
the kinematic equations:
x
f
=
x
i
+
v
i
t
+
1
2
at
2
=
⇒
d
=
1
2
gT
2
.
In words, this says “the ball’s total distance covered is proportional to the
square of the total time.”
Now, if the ball falls for a
total
time of
T
+ 4
T
= 5
T
seconds, the
total
distance
D
it falls is
D
=
1
2
g
(5
T
)
2
= 25
parenleftbigg
1
2
gT
2
parenrightbigg
= 25
d.
So in this case the ball has fallen
24
d
farther
.
(This problem is just one example of what is called a “scaling relation”,
showing that the kinematic equations are useful for more than just plugging
in numbers. You’ll be seeing a lot of these!)
Problem 2:
A particle moves along the
x
axis according to the equation
x
=
46
t
+
8
t
2
(where
t
is in seconds and
x
is in meters). (a) Calculate the average velocity of the particle during the
first
3
.
0
s of its motion. (b) Calculate the instantaneous velocity of the particle at
t
= 3
.
0
s. (c)
Calculate the instantaneous acceleration of the particle at
t
= 3
.
0
s.
(a) The average velocity after a time
t
is simply the total displacement divided by the
time. The initial position is zero (since
x
(
t
= 0) = 0), so we can write
v
=
Δ
x
Δ
t
=
x

0
t
=
46
t
+ 8
t
2
t
= 46 + 8
t.
So in particular, after
t
= 3
.
0 s, the average velocity is
v
= 46 + 8(3
.
0) =
70 m/s
.
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Physics 1111
Fall 2009
PS #2 Solutions
Page 2 of 11
(b) There are several ways to answer this question.
One way is to use calculus, taking
the time derivative of the
x
(
t
) equation to get
v
(
t
). But we don’t ever need calculus
in this class! Another approach is to use the definition of “instantaneous velocity” as
the small
t
limit of the average velocity; by calculating average velocity for very small
time intervals around
t
= 3
.
0 s, we obtain an (approximate) answer. This approach is
laborious, though, and not very satisfying.
Here’s another way. Notice from part (a) that we can compute the average velocity
between
t
= 0 and any other time. By letting
t
go to zero in our above expression, we
can compute the
instantaneous
velocity at
t
= 0, which is simply
v
i
= 46 m/s. Now,
assuming the acceleration is constant (more on this below), we can write
v
=
1
2
(
v
i
+
v
f
)
and solve for the final instantaneous velocity at
t
= 3
.
0 s:
v
i
+
v
f
= 2
v
=
⇒
v
f
= 2
v

v
i
= 2(70 m/s)

(46 m/s) =
94 m/s
.
(c) Well, now that we know the initial and final velocities, we can compute the
average
acceleration over the time interval:
a
=
Δ
v
Δ
t
=
94

46
3
.
0
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 Physics, Pythagorean Theorem, Acceleration, Velocity, m/s, Solutions Page

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