Consider a ball dropped from a height of 256 feet, modeled by the position equation, s(t) = −16t2 + 256

Question

Determine the average velocity of the ball from when the ball was released to when it hit

the ground.

iii) What does the mean value theorem tell us about the relationship between the average rate

of change over an interval (a, b), and the instantaneous rate of change at some point c in (a, b)?

iv) At what time t, is the instantaneous velocity of the ball equal to the average velocity calculated in (ii)

Answer 1

The way the equation is written, the ball hits the ground when s = 0 or 16t² = 256, t=4 sec

a) Average velocity for t = 0 to 4 is (s(4)-s(0))/(4-0) = -256 ft/4 s = -64 ft/s

b) The mean value theorem says that there is a point c in the interval where the instantaneous velocity is equal to the average velocity for the interval.

c) The instantaneous velocity is ds/dt = -32t

This is equal to the calculated average velocity at t = 2s (This should not be surprising as the velocity changes linearly with time - 1/2 the time, 1/2 the velocity change)

Please consider a tutor. Take care.