The position function of an object moving along a straight line is given bys = f (t). The average velocity of the object over the time interval [a, b] is the average rate of change of f over [a, b]; its (instantaneous) velocity at t = a is the rate of change of f at a. A ball is thrown straight up with an initial velocity of 112 ft/sec, so that its height (in feet) after t sec is given by s = f (t) = 112t - 16t2. (a) What is the average velocity of the ball over the following time intervals?[3,4] ft/sec[3,3.5] ft/sec[3,3.1] ft/sec (b) What is the instantaneous velocity at time t = 3?

a) Average velocity for the time interval [3, 4] : 0 ft/s

Average velocity for the time interval [3, 3.5]: 8 ft/s

Average velocity for the time interval [3, 3.1]: 14.4 ft/s

b) The instantaneous velocity at t = 3 s is 16 ft/s

Explanation:

The average velocity of the ball can be determined by the following expression:

Average velocity = Δh / Δt

Average velocity = (final height - initial height) / elapsed time

Then, for the time interval [3, 4] the initial height will be f (3) and the final height f (4):

f (t) = 112 · t - 16 · t²

f (3) = 112 · 3 - 16 · (3) ² = 192 ft

f (4) = 112 · 4 - 16 · (4) ² = 192 ft

The average velocity will be:

average velocity = (192 ft - 192 ft) / 4 s - 3 s = 0 ft/s

For the interval [3, 3.5]:

f (3) = 192 ft

f (3.5) = 112 · 3.5 - 16 · (3.5) ² = 196

Average velocity = (196 ft - 192 ft) / (3.5 s - 3 s) = 8 ft/s

For the interval [3, 3.1]:

f (3) = 192 ft

f (3.1) = 112 · 3.1 - 16 · (3.1) ² = 193.44 ft

Average velocity = (193.44 ft - 192 ft) / (3.1 s - 3 s) = 14.4 ft/s

b) When the elapsed time Δt is nearly 0, we obtain the instantaeous velocity:

d f (t) / dt = instanteneous velocity

d f (t) / dt = 112 - 32 · t

Evaluating the derivative of f (t) at t = 3:

d f (3) / dt = 112 - 32 · 3 = 16 ft/s

The instantaneous velocity at time t = 3 s is 16 ft/s