Answers  One Dimensional Motion


1.

(a) The average speed = distance traveled/time
= 100 m/10 s = 10 m/s. (b) time = distance/average speed =
1610 m/10 m/s = 161 s = 161 s x (1 min/60 s) = 2.68 min.


2.

If y = kx, y/x = k = a constant. y is
directly proportional to x. (a) If you double x, you double
y. (b) If you halve x, you halve y. (c) A plot of y as a function
of x yields a straight line through the origin.


3.

If y = kx^{2}, y/x^{2}
= k = a constant. y is directly proportional to x^{2}.
(a) If you double x, y goes up by a factor of 4. (b) If you
halve x, y is reduced by a factor of four. (c) A plot of y
as a function of (i) x is a parabola; (ii) the square
of x is a straight line through the origin.


4.

(a) 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. (b) 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.


5.

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}.
(a) x(2.0 s) = 1/2 (10 m/s^{2})(2.0
s)^{2 }= 20 m.
(b) 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.
(c) 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
(d) 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.


6.

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) a = {(60  20) m/s}/(2.0 s  0) = 20 m/s^{2}.
(b) 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.


7.

Distance traveled = speed x time = 3.0
x 10^{8} m/s x 20/3 s = 20 x 10^{8 }m. Room
lengths = distance traveled/ "room length"
= 20 x 10^{8} m/20 m = 10^{8}.


8.

Time to travel distance = distance/speed
= 60 ft/132 ft/s = 0.454 s.


9.

(a) The average speed = x/t = (v + v_{i})/t,
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 and 20 m/s = (15 m/s + v_{i})
so v_{i} = 5 m/s.
(b) The acceleration a = (v  v_{i})/(t
 t_{i}) = (15 m/s  5 m/s)/(6.0 s) = 5/3 m/s^{2}.
(c) 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.


10.

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.



11. 
(a) For constant acceleration v = v_{i}
+ at. For the car to come to a stop, the final velocity v
= 0. 0 = 80 ft/s  (20 ft/s^{2})t or t = 4.0 s.
(b) In general for constant acceleration x = x_{o}
+ v_{i}t + 1/2 at^{2}. The distance moved
= x  x_{o} = v_{i}t + 1/2 at^{2}
= 80 ft/s(4.0 s) + 1/2 (20 ft/s^{2})(16s ^{2})
= 16 ft.
(c) During the (1 1/2 + 1/2) s = 2.0 s before the brake is
applied, the car traveling at a speed of 80 ft/s goes 80 ft/s(2.0
s) = 160 m. After the brake is applied the car travels 160
ft. The total distance traveled is (160 + 160)ft = 320 ft.
(d) Number of "car lengths" traveled = distance traveled/ "length
of car"
= 320 ft/15 ft = 21.3.


12.

The magnitude 10 m/s of the particle
is the same at t = 1.0 s and t = 5.0 s. If we take to the
right to be positive, the velocity is to the right at t =
1.0 s , but to the left at t = 5.0 s. For motion along a straight
line, we are free to use a positive sign to indicate motion
in one direction and a negative sign to indicate motion in
the opposite direction. We usually take to the right as positive
and to the left as negative.


13. 
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.


14.

Three equations are used frequently for motion with constant
acceleration: For motion in the ydirection,
(1) v(t) = v_{o} + at
(2) y(t) = y_{o} + v_{o}t + 1/2 at^{2},
and
(3) 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. 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.
(a) 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
(b) v(2.0s) = v_{o} + at = 30 m/s  10 m/s^{2}(2
s) = 10 m/s
(c) 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.


15.

(a) 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.
(b) v(t) = v_{o} + at or 40 m/s = 20 m/s  10 m/s^{2}
t and t = 2 s.
(c) 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. 

16.

(a) "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), or(40a) m = 100 m^{2}/s^{2,
}or a = 2.5 m/s^{2}.
(b) v(t) = v_{o} + at 0 = 10 m/s  (2.5 m/s^{2})t.
t = 4.0 s.
(c) and (d) are shown in the figure below. 


17.

The average velocity = 2.0m/0.10s = 20
m/s» v at top or bottom of window.
v^{2} = v_{o}^{2} + 2a(y  y_{o})
or (20 m/s)^{2} = 0 + 2(10 m/s^{2})(0
 y_{o}) and 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.


18.

 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.


19.

(a) In Fig. 19a 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}.
(b) In Fig. 19b, at t = 1 s, the velocity is zero (the slope
of x vs t at t = 1 s 1s zero). For t > 1s, the acceleration
is negative because the slope 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}.
(c) In Fig. 19c, 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}.
(d) In Fig. #19d, 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 = 2s,
v
= 4 m/s, a = {(4)  (3)}cm/s/(2  1)s= 1/2 cm/s^{2}.
(e) The speed, the magnitude of the velocity, at t = 1 s is
increasing for all cases except (c).



