Answers  One Dimensional Motion


1.

 The distance moved x divided by the time t equals the
constant velocity v. x/t = v or
x = vt = 15 m/s (2.0 s) = 30 m.
 Since x/t = v = constant, the distance moved is directly
proportional to the time elapsed. If you double the
time, you double the distance.
For t = 4.0 s, x = 2 x 30 m= 60 m.


2.

In general for constant acceleration,
the distance moved by the object x =
x_{o }+ v_{o}t + 1/2 at^{2}, where x_{o}
is the position of the object at t = 0 (the initial position),
v_{o }the velocity at t = 0 (the initial velocity),
a the acceleration, and t the time at which you wish to find
x. Let's arbitrarily take x_{o} = 0. Then
for an object initially at rest, v_{o} = 0, x
= 0 + 0 + 1/2 at^{2}.
 x(2.0 s) = 1/2 (10 m/s^{2})(2.0 s)^{2 }=
20 m
 x(4 s) = 1/2 (10 m/s^{2})(4 s)^{2} = 80
m. For constant acceleration and an initial position and
initial velocity of zero, x = 1/2 at^{2} or
x/t^{2} = 1/2 a = a/2 = a constant. For this case
x is directly proportional to the square of t. If you double
t, t^{2} goes up by a factor of four and x must
go up by a factor of 4 = 4 x 20 m = 80 m.
 For this part, we still take the initial position x_{o}
= 0, but now the initial velocity v_{o} = 4.0 m/s.
x(t) = 0 + v_{o}t + 1/2 at^{2}.
x(2.0 s) = 0 + (4.0 m/s)(2.0 s) + 1/2 (10 m/s^{2})(2.0
s)^{2} = 28 m
 x(4.0 s) = 0 + (4.0 m/s)(4.0 s) + 1/2 (10 m/s^{2})(4.0
s)^{2} = 96 m
With x = v_{o}t + 1/2 at^{2} and x/t^{2}
= v_{o}/t + a/2. Since t is a variable, the right
hand side of the above equation is not equal to a
constant and x is no longer proportional to the square of
the time, that is, 96 ≠ 4 x 28.


3.

By the definition of constant
acceleration, a = (v  v_{o})/(t  0), where v is the
velocity of the object at time t and v_{o }is the velocity
of the object at time t = 0.
 a = {(60  20) m/s}/(2.0 s  0) = 20 m/s^{2}
 Taking x_{o} = 0, x(t) = v_{o}t + 1/2
at^{2}.
x(2.0 s) = 20 m/s(2.0 s) + 1/2 (20 m/s^{2})(2.0
s)^{2} = 80 m.


4.

 The average velocity = x/t = (v + v_{i})/2, where
v = 15 m/s and v_{i} is the initial speed. The average
speed = x/t = 60 m/6.0 s = (15 m/s + v_{i})/2
or
10 m/s = (15 m/s + v_{i})/2 or
20 m/s = 15 m/s + v_{i }so v_{i }=
5 m/s.
 The acceleration a = (v  v_{i})/(t  t_{i})
= (15 m/s  5 m/s)/(6.0 s) = 5/3 m/s^{2}.
 v_{i}^{2} = v_{o}^{2}
+ 2a(x_{i}  x_{o}).
(5 m/s)^{2} = 0 + 2(5/3 m/s^{2})(x_{i}
 x_{o})
25 m^{2}/s^{2} = (10/3 m/s^{2})(x_{i}
 x_{o})
or
(x_{i}  x_{o}) = 7.5 m.
Let's check all of this. First we find the time for the
object to acquire a velocity v_{i} = 5.0 m/s with an
acceleration of 5/3 m/s^{2} starting from an initial
velocity of zero. v = v_{o} + at or
5.0 m/s = 0 + (5/3 m/s^{2})t or
t = 3.0 s.
At t = (6.0 + 3.0)s, x(9.0 s) = 1/2 (5/3 m/s^{2})(9.0
s)^{2 }= 67.5 m = 60 m + 7.5 m, where 60 m is the
distance given in the statement of the problem for the object
to go from 5 m/s to 15 m/s and 7.5 m is the distance moved from
the point where the object was at rest to where it was traveling
with 5 m/s.


5.

Three equations are used frequently for
motion with constant acceleration: For motion in the ydirection,
 v(t) = v_{o} + at
 y(t) = y_{o} + v_{o}t + 1/2 at^{2,
} and
 v^{2}(y) = v_{o}^{2} + 2a(y 
y_{o})
The first is just the definition of constant acceleration, a
= (v  v_{o})/(t  0), rearranged algebraically. Eq.
1 gives the velocity of an object as a function of time. If
you know the initial velocity v_{o }and the acceleration
a of the object, you can find its velocity v at time t. The
second equation is derived from the area under the velocity
versus time curve and the definition of acceleration (See solution
to Problem #16). It gives the position y as a function of t.
If you know the initial position y_{o} of the object,
v_{o} and a, you can find y for any time t. The third
equation is found by solving for t in Eq. (1) and substituting
it into Eq. (2). It gives the velocity v as a function of height
y. If you know y_{o}, v_{o}, and a, you can
find y for any value of v.
 We know the height reached y, the time t for it to reach
this height, the acceleration a of the ball, and taking
y_{o} = 0, we can find the one unknown v_{o}
with:
40 m = 0 + v_{o}(2.0 s)  1/2 (10 m/s^{2})(4.0
s^{2})
40 m = v_{o}(2.0 s)  20 m or
v_{o} = 30 m/s
 v(2.0s) = v_{o} + at = 30 m/s  10 m/s^{2}(2
s) = 10 m/s
 We must interpret the meaning of "how much higher
will the ball go." The answer, of course, is "until
it stops rising." At the highest point the ball comes
momentarily to rest and v(t) = 0.
0 = v^{2 }= v_{o}^{2} + 2a(y  y_{o})
= (30 m/s)^{2}  20 m/s^{2}(y  0)
or
0 = 900 m^{2}  20 m(y) and
y = 45 m.
Added height = (45  40)m = 5 m.


6.

 We do not know the time, but taking y_{o} = 60
m and y = 0, we can use
v^{2}(y) = v_{o}^{2}
+ 2a_{y}(y  y_{o})
v^{2}(0) = (20 m/s)^{2 }  20 m/s^{2}(0
 60 m) = (400 + 1200)m^{2}/s^{2}
and v = 40 m/s, where the negative sign occurs because
we have taken up as positive.
 v(t) = v_{o} + at or
40 m/s = 20 m/s  10 m/s^{2} t and t = 2 s
 Now the initial and final positions are the same, but
the initial velocity v_{o} = +20 m/s. The algebra,
however, will be the same as in part (a) and the answer
is again v = 40 m/s when it hits the ground. The time does
change with v(t) = v_{o} + at and v_{o }=
+20 m/s:
40 m/s = 20 m/s 10 m/s^{2 }t and
now t = 6 s.


7.

 i. By definition, the slope of a plot of
x as a function of t at any time t is
dx/dt. Since v = dx/dt, the slope of a plot of x as a function
of t at any time t is the velocity at that instant.
 ii. If you plot y as a function of x, you
know that ∫y dx is the area under the curve of y versus
x. Similarly, ∫v dt is the area under the velocity
v versus time t curve up to time t. Since ∫dx = ∫v
dt, the area under the velocity v versus time t up to time
t is the distance moved.
 ii. By definition, the slope of v versus
t at any instant is dv/dt.
Since a = dv/dt, the slope of v versus t at any instant
is the acceleration at that instant.
 iii. ∫a dt is the area under the a
versus t curve. Since ∫dv = ∫a dt, the area
under the a versus t curve up to any time t is the change
in velocity in that time interval.


8.

The answers are written on the graph below for each interval.
Note that in #1 the slope of the x vs t curve is positive
and increasing in magnitude so the velocity is positive and
increasing and the acceleration is positive. In #3 the velocity
is positive, but decreasing in magnitude so the acceleration
is negative. In #5 the velocity is negative, but it is increasing
in magnitude so the acceleration is negative.


9.

See Figure 2 below:
 The acceleration at any instant equals the slope of v
versus t at that instant.
From t = 0 to t = 1.0 s, a = (15  5)m/s/(1  0 )s = 10
m/s^{2}.
From t = 1.0 s to 2.0 s, the velocity is constant and the
acceleration a = 0.
From t = 2.0 s to 3.0 s, a = (0  15.0)m/s/(3.0  2.0)s
= 15 m/s^{2}.
From t = 3.0 s to 4.0 s, the velocity equals zero and the
acceleration a = 0. From t = 4.0 s to 5.0 s, a = (15.0
 0)m/s/(5.0 4.0)s = 15 m/s^{2}.
 The distance moved by an object in time t equals the area
under the v vs t curve from t = 0 to t.
From t = 0 to t = 0.5 s, area equals the area of a triangle
of height
5.0 m/s and base 0.5 s and square with sides= 5.0 m/s and
0.5 s.
Area = (1/2)(5.0 m/s x 0.5 s) + (5 m/s x 0.5s) = 3.75 m.
From t = 0.5 s to t = 1.0 s,
area = (1/2 x 5.0 m/s x 0.5 s) + (10 m/s x 0.5 s)= 6.25
m.
From t = 0 to t = 1, x(1s) = (3.75 + 6.25)m =10 m.
From t = 1 to t = 2 s, area = 15 m/s x 1 s = 15 m.
x(2s) = (10 + 15) m = 25 m.
From t = 2 s to t = 3 s, area = 1/2(15 m/s)(1s) = 7.5 m.
x(3s) = (25 + 7.5)m = 32.5 m.
From t = 3 to 4 s, the area = 0.
x(4s) = 32.5 m.
From t = 4 to 5 s, area = 1/2(15 m/s)(1 s) = 7.5 m.
x(5s) = (32.5  7.5)m = 25 m.
 The slope of x vs t at any instant is the velocity at
that instant.
From t = 0 to t = 1 s, the slope and velocity are positive
and increasing. From t = 1 to t = 2 s, the slope and velocity
are constant and positive.
From t = 2 to t = 3 s, the slope and velocity are positive,
but decreasing. From t = 3 to t = 4 s, the slope and velocity
equal zero.
From t = 4 to t = 5 s, slope and velocity are negative,
but increasing.
All agree with v versus t in Fig. 2a.


10.

 "Coming to rest in a distance of 20 m" means x(t)  x_{o}
= 20 m.
v^{2}(x) = v_{o}^{2} + 2a(x  x_{o})
0 = (10 m/s)^{2} + 2a(20 m)
(40a) m = 100 m^{2}/s^{2
} a = 2.5 m/s^{2}.
 v(t) = v_{o} + at
0 = 10 m/s  (2.5 m/s^{2})t
t = 4.0 s.
 Shown in the figure below
 Shown in the figure below


11.

The average velocity of the flowerpot
equals:
2.0 m/0.10 s = 20 m/s ≈ v
at top or bottom of window.
v^{2}(y) = v_{o}^{2} + 2a(y  y_{o})
or
(20 m/s)^{2} = 0 + 2(10 m/s^{2})(0  y_{o}) or
y_{o }= 20 m. Since the distance between
floors is 4.0 m, it fell from an apartment 5 floors up or, since
she is on the fifth floor, from an apartment on the tenth floor.


12.



13.



14.

The change in velocity up to any time
equals the area under a versus t curve up to that time. The
distance moved in time t equals the area under the v versus
t curve up to that time t.
 Velocity as a function of time and position as a function
of time are shown in the figure below.
 Velocity as a function of time and position as a function
of time are shown in the figure above.
 The maximum velocity occurs for t = 1.0 s through 2.0
s and it is equal to
10 m/s, as illustrated in figure 3b above.
 The distance moved by the particle in 3.0 s is 20.0 m.
Notice that the slope of x vs t at any instant equals the
velocity v at that instant and the slope of v vs t at any
instant equals the acceleration a at that instant.


15.

 For the ball thrown downward, y(t) = y_{o }+ v_{o}t
+ 1/2 at^{2} or
taking up as +,
y(t) = 40 m  (8.0 m/s)t  (5.0 m/s^{2})t^{2
} (Equation
1)
For the ball thrown upward from the ground,
y(t) = 0 + (12 m/s)t  (5.0 m/s^{2})t^{2}
(Equation
2)
When the balls collide the two y's are equal:
40 m  (8.0 m/s)t  (5.0 m/s^{2})t^{2
}=
0 + (12 m/s)t  (5.0 m/s^{2})t^{2} or
40 m  (8.0 m/s)t = (12 m/s)t
40 m = (20 m/s)t and
t = 2.0 s.
 From Eq. 2, y(2.0s) = (12 m/s)(2.0 s)  (5.0 m/s^{2})(2.0s)^{2}
= 4.0 m
From Eq. 1, y(t) = 40 m  (8.0 m/s)(2.0 s)  (5.0 m/s^{2})(2.0
s)^{2}= 4.0 m.
 In general, v(t) = v_{o}+ at.
For the second ball, v(2.0 s) = 12 m/s  (10 m/s^{2})(2.0
s) = 8.0 m/s.
Since the velocity at 2.0 s of the second ball is negative,
it is on its way down when the first ball collided with
it.


16.

In Fig. 4 below, I have used v’
and t’ as a variable velocity and time, respectively,
to distinguish between the velocity v at time t.
 The distance moved by the object in time t equals the
area under the velocity as a function of time curve up to
time t. In Fig. 4a, the area consists of the area of a triangle
of base t and height (v  v_{o}) and a rectangle
of length t and height v_{o}.
In time t, the distance traveled x = 1/2 (v  v_{o})t
+ v_{o}t = (v_{o }+ v)/2t.
The average velocity v_{av}= x/t = (v + v_{o})/2.
If the object were to move in a straight line with the average
velocity for the same time t as the object that moved with
a variable velocity, it would have the same displacement.
In Fig. 4b, we see that the triangles labeled Δ_{1
}and_{ } Δ_{2}
are congruent because their angles are equal and the heights,
(v_{av} – v_{o}) and (v – v_{av}),
are equal. Thus the rectangle with height v_{av}
and length t has area v_{av}t which equals 1/2 (v
 v_{o})t + v_{o}t, the area used
in Fig. 4a.
 Since Δ_{1 }and_{
} Δ_{2}
are congruent, their bases are equal. For this to be true,
the time at which the velocity equals the average velocity
occurs at t/2 or halfway through the interval.
In Problem #5, I mentioned that you could find x(t) from
Fig. 4.
Since a = (v  v_{o})/t or
(v  v_{o}) = at, we write:
1/2 (v  v_{o})t + v_{o}t =
1/2 at^{2} + v_{o}t = v_{o}t +
1/2 at^{2}
= distance traveled
If there is an initial displacement of x_{o},
x(t) = x_{o} + v_{o}t + 1/2 at^{2}.


17.

 In Fig. 17a below, at t = 1 s, the velocity is zero
(the slope of x vs t at t = 1 s is zero).
The acceleration is positive because the slope of x versus
t after t = 1 s is positive and increasing. For example,
imagine the slope of x vs t at t = 2 s is 1 cm/s, then a
= [(1  0)cm/s]/(2  1) s = +1/2 cm/s^{2}.
 In Fig. 17b above, at t = 1 s, the velocity is zero
(the slope of x vs t at t = 1 s is zero).
The acceleration is negative because the slope after t =
1s is less than zero and increasing. For example imagine
the slope of x vs t at t = 2 s
is 1 m/s, then a = [(1  0)cm/s]/(2  1)s = 1/2 cm/s^{2}.
 In Fig. 17c, at t = 1 s, the velocity is negative
(the slope of x vs t at t = 1 s is negative).
The acceleration is positive because the slope of x vs t
after t = 1 s is negative and decreasing. For example imagine
the slope of x vs t at
t = 1 s is 1 cm/s and at t = 2 s it is 0. Then a = {0 
(1)cm/s}/(2  1)s
= 1/2 cm/s^{2}.
 In Fig. #17d, at t = 1 s, the velocity is negative
(the slope of x vs t at t = 1 s is negative).
The acceleration is negative because the slope of x vs t
is negative and increasing. For example, imagine at t =
1 s, v = 3 m/s and at t = 2 s,
v = 4 m/s, then a = {(4)  (3)}cm/s/(2  1)s = 1/2 cm/s^{2}.
 The speed, the magnitude of the velocity, is increasing
at t = 1 s for all cases except (c).


