1. If a ball is thrown vertically upward with a velocity of 80 feet per second, its height after t seconds is
s(t) = 80t − 16t2 . Answer the following questions with correct units.
(a) What is the velocity of the ball after 1 sec?
(b) What is the maximum height of the reached by the ball?
(c) What is the velocity of the ball when it is 96 feet above the ground on its way up?
(d) With what velocity does the ball hit the ground?
Solution:
(a) Velocity, v(t) = s0 (t) = 80 − 32t, so the velocity after 1 sec, v(1) = 80 − 32(1) = 48 ft per second.
(b) At maximum height, v(t) = 0. That is, 80 − 32t = 0 =⇒ t =
s(2.5) = 80(2.5) − 16(2.5)2 = 100 feet.
80
32
= 2.5. Now the maximum height
(c) We set height, s(t) = 96. That is, 80t − 16t2 = 80 =⇒ −16t2 + 80t − 96 = 0 =⇒ −16(t2 − 5t + 6) = 0.
Now we factor the polynomial to solve the equation for time, t taken to get the height of 96 feet, −16(t −
2)(t − 3) = 0 =⇒ t = 2, 3 seconds. So it takes 2 seconds on its way up and 3 seconds on its way down to get
the height of 96 feet respectively. Now the required velocity is v(2) = 80 − 32(2) = 16 feet per second.
(d) As in above we need to determine time, t for the ball to hit the ground. For we set height s(t) = 0 that
is 80t − 16t2 = 0 or 16t(5 − t) = 0 =⇒ t = 0, 5. Note that t = 0 refers the time when the ball was on the
ground before it is thrown. So it takes 5 seconds to get to the ground. (We could have just doubled the time
to reach the maximum height found in part (b) to find the time of flight.) In any case the velocity at impact
is v(t) = 80 − 32(5) = −80 feet per second.
2. A particle moves according to a law of motion s(t) = t3 − 12t2 + 36t, where t is measured in seconds and s in
feet. Answer the following questions with correct units.
(a) What is the velocity of the particle after 3 sec?
(b) When is the particle at rest?
(c) When is the particle moving in the positive direction?
(d) Find the total distance traveled during the first 8 sec.
(e) What is the acceleration after 2 sec?
Solution:
(a) Velocity, v(t) = s0 (t) = 3t2 − 24t + 36. So the velocity after 3 sec, v(3) = 3(3)2 − 24(3) + 36 = −9 f t/sec.
(b) Set v(t) = 0. That is, 3t2 − 24t + 36 = 0 =⇒ 3(t2 − 8t + 12) = 0 =⇒ 3(t − 2)(t − 6) = 0 =⇒ t = 2, 6 sec.
(c) We need to check the velocity in the time intervals (0, 2), (2, 6) (6, ∞). Pick a representative from
each interval and plug that into the velocity: v(1) = 3(1)2 − 24(1) + 36 = positive number, so the particle is moving in the positive direction for 0 < t < 2, that is, in the first 2 seconds. Similarly, v(3) =
3(3)2 − 24(3) + 36 = negative number, so the particle is moving in the negative direction for 2 < t < 6. Also
v(7) = 3(7)2 − 24(7) + 36 = positive number, so the particle is moving in the positive direction for t > 6.
Hence the particle is moving in the positive direction in the first 2 seconds and after 6 seconds.
(d) The total distance traveled in during the first 8 seconds = |s(2) − s(0)| + |s(6) − s(2)| + |s(8) − s(6)| =
|32 − 0| + |0 − 32| + |32 − 0| = 32 + 32 + 32 = 96 f t.
(e) Acceleration, a(t) = v 0 (t) = s”(t) = 6t−24. So the acceleration after 2 sec, a(2) = 6(2)−24 = −12 f t/sec2 .