Question

A silver dollar is dropped 960 feet from the top floor of a hotel at time T=0. The position as a function of time for this object is S (t)= -16t^2+960. A) find the average rate of change for 1≤t≤4.B) find the instantaneous rate of change at t=3C) how long will it take for the dollar to hit the ground ?D) find the instantaneous rate of change of the dollar just before it hits the ground?

Answer 1

Equation: S (t)= -16t^2+960

A) The average rate of change of a function over an interval (a,b) is given by:

$avarege\text{ rate of change = }\frac{change\text{ in y}}{change\text{ in x}}=\frac{s(b)-s(a)}{b-a}$

Then:

$average\text{ rate of change = }\frac{s(4)-s(1)}{4-1}$

For s(4)

$s(4)=-16(4)^2+960=-16(16)+960=-256+960=704$

For s(1)

$s(1)=-16(1)^2+960=-16+960=944$

Therefore average rate of change is:

$\frac{704-944}{4-1}=\frac{-240}{3}=-80$

Answer: average rate of change = - 80 ft/s

B) This is just the value of the derivative when t = 3, so

$\begin{gathered} s(t)=-16t^2+960 \\ s^{\prime}(t)=-2(16)t^{2-1}+0=-32t \\ s(3)=-32(3)=-96 \end{gathered}$

Answer: the instantaneous rate of change = - 96 ft/s

C) This is when s = 0, so:

$\begin{gathered} -16t^2+960=0 \\ -16t^2+960-960=0-960 \\ -16t^2=-960 \\ \frac{-16t^2}{-16}=\frac{-960}{-16} \\ t^2=60 \\ t=\sqrt[]{60} \\ t=7.75 \end{gathered}$

Answer: time = 7.75 s

D) This happens when t = 7.75, so:

$s^{\prime}(7.75)=-32(7.75)=-248$

Answer: instantaneous rate of change of the dollar just before it hits the ground = - 248 ft/s