Answer:
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
