1. (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.
2. In general for constant acceleration, the distance moved by the
object x = xo + vot + 1/2 at2, where xo is the position of the object at t =
0 (the initial position), vo 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 xo = 0. Then for an object initially at rest, vo = 0, x = 0
+ 0 + 1/2 at2.
(a) x(2.0 s) = 1/2 (10 m/s2)(2.0 s)2 = 20 m.
(b) x(4 s) = 1/2 (10 m/s2)(4 s)2 = 80 m. For constant acceleration and
an initial position and initial velocity of zero, x = 1/2 at2 or x/t2 = 1/2 a
= a/2 = a constant. For this case x is directly proportional to the
square of t. If you double t, t2 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 xo = 0, but now the
initial velocity vo = 4.0 m/s. x(t) = 0 + vot + 1/2 at2.
x(2.0 s) = 0 + (4.0 m/s)(2.0 s) + 1/2 (10 m/s2)(2.0 s)2 = 28 m
(d) x(4.0 s) = 0 + (4.0 m/s)(4.0 s) + 1/2 (10 m/s2)(4.0 s)2 = 96 m
With x = vot + 1/2 at2 and x/t2 = vo/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 - vo)/(t - 0), where
v is the velocity of the object at time t and vo is the velocity of the
object at time t = 0.
(a) a = {(60 - 20) m/s}/(2.0 s - 0) = 20 m/s2.
(b) Taking xo =0,
x(t) = vot + 1/2 at2.
x(2.0 s) = 20 m/s(2.0 s) + 1/2 (20 m/s2)(2.0 s)2 = 80 m.
4. a) The average speed = x/t = (v + vi)/t, where v = 15 m/s and vi is
the initial speed. The average speed = x/t = 60 m/6.0 s = (15 m/s +
vi)/2 or 10 m/s =
(15 m/s + vi)/2 and 20 m/s = (15 m/s + vi) so vi = 5 m/s.
(b) The acceleration a = (v - vi)/(t - ti) = (15 m/s - 5 m/s)/(6.0 s) =
5/3 m/s2.
(c) vi2 = vo2 + 2a(xi - xo). (5 m/s)2 = 0 + 2(5/3 m/s2)(xi - xo) 25 m2/s2 =
(10/3 m/s2)(xi - xo) or (xi - xo) = 7.5 m.
Let's check all of this. First we find the time for the object to acquire
a velocity vi = 5.0 m/s with an acceleration of 5/3 m/s2 starting from
an initial velocity of zero. v = vo + at or 5.0 m/s = 0 + (5/3 m/s2)t or t
= 3.0 s. At t = (6.0 + 3.0)s, x(9.0 s) = 1/2 (5/3 m/s2)(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 y-direction,
(1) v(t) = vo + at
(2) y(t) = yo + vot + 1/2 at2, and
(3) v2(y) = vo2 + 2a(y - yo).
The first is just the definition of constant acceleration, a = (v - vo)/(t 0), rearranged algebraically. Eq. 1 gives the velocity of an object as
a function of time. If you know the initial velocity vo 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 yo of the object, vo 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 yo, vo, 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 yo = 0, we can find
the one unknown vo with
40 m = 0 + vo(2.0 s) - 1/2 (10 m/s2)(4.0 s2)
40 m = vo(2.0 s) - 20 m or vo = 30 m/s
(b) v(2.0s) = vo + at = 30 m/s - 10 m/s2(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 = v2 = vo2 + 2a(y - yo) = (30 m/s)2 - 20 m/s2(y - 0), or
2
6. (a) We do not know the time, but taking yo = 60 m and y = 0, we
can use
v2(y) = vo2 + 2ay(y - yo)
v2(0) = (-20 m/s)2 - 20 m/s2(0 - 60 m)
= (400 + 1200)m2/s2 and
v = - 40 m/s,
where the negative sign occurs because we have taken up as
positive.
(b) v(t) = vo + at or -40 m/s = -20 m/s - 10 m/s2 t and t = 2 s.
(c) Now the initial and final positions are the same, but the initial
velocity
vo = +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) = vo + at and vo = +20 m/s, -40 m/s = +20
m/s -10 m/s2 t and now t = 6 s.
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.
10. (a) "Coming to rest in a distance of 20 m" means x(t) - xo = 20 m.
v2(x) = vo2 + 2a(x - xo).
0 = (10 m/s)2 + 2a(20 m), or(-40a) m = 100 m2/s2, or a = -2.5 m/s2.
(b) v(t) = vo + at 0 = 10 m/s - (2.5 m/s2)t. t = 4.0 s.
(c) and (d) are shown in the figure below.
11. The average velocity = 2.0m/0.10s = 20 m/s» v at top or bottom
of window.
v2 = vo2 + 2a(y - yo) or (20 m/s)2 = 0 + 2(-10 m/s2)(0 - yo) and yo = 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.
15. For the ball thrown downward, y(t) = yo + vot + 1/2 at2 or
taking up as +, y(t) = 40 m - (8.0 m/s)t - (5.0 m/s2)t2
(Equation 1)
For the ball thrown upward from the ground, y(t) = 0 + (12 m/s)t (5.0 m/s2)t2
(Equation 2)
When the balls collide the two y's are equal: 40 m - (8.0 m/s)t - (5.0
m/s2)t2 =
0 + (12 m/s)t - (5.0 m/s2)t2
a. 40 m - (8.0 m/s)t = (12 m/s)t
b. 40 m = (20 m/s)t and t = 2.0 s.
c.
d.
or
e. From Eq. 2, y(2.0s) = (12 m/s)(2.0 s) - (5.0 m/s2)(2.0s)2 = 4.0 m
From Eq. 1, y(t) = 40 m - (8.0 m/s)(2.0 s) - (5.0 m/s2)(2.0 s)2=
4.0 m.
f.
In general, v(t) = vo+ at. For the second ball, v(2.0 s) = 12 m/s
- (10 m/s2)(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.
9. See Figure 2 below:
a. 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/s2.
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/s2.
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/s2.
b. 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.
c.
d.
e.