Chapter 11.3, Problem 72E

a.

To determine

To find: a formula for the instantaneous rate of change of the coin.

Answer to Problem 72E

  $s\text{'}(t)=-32t$ .

Explanation of Solution

Given information: A coin is dropped from the top of a 120-foot building . The height

or displacement s (in feet) of the coin can be modeled by the position function $s(t)=-16t^2+120$, where t is the time in second since it was dropped.

Calculation:

  $\begin{array}{l}s(t)=-16t^2+120\\ s\text{'}(t)=-32t\end{array}$

b.

To determine

To find: the average rate of change of the coin after the first two second of free fall.

Answer to Problem 72E

  $\text{Average}\text{rate}\text{of}\text{change}=32\text{ft/sec}$

Explanation of Solution

Given information: $s(t)=-16t^2+120$

Calculation:

  $\begin{array}{l}s(0)=120,s(3)=56\\ \text{Average}\text{rate}\text{of}\text{change}=\frac{120-56}{2}=\frac{64}{2}=32\text{ft/sec}\text{.}\end{array}$

c.

To determine

To find: the velocity of the coin as it impacts the ground.

Answer to Problem 72E

The velocity of the coin as it impacts the ground is $-32\sqrt{\frac{15}{2}}$ .

Explanation of Solution

Given information: $s(t)=-16t^2+120$

Calculation:

  $\begin{array}{l}\text{At}\text{ground}s(t)=0\\ s(t)=-16t^2+120=0\Rightarrow t=\sqrt{\frac{15}{2}}\\ s\text{'}(t)=-32t\Rightarrow s\text{'}(\sqrt{\frac{15}{2}})=-32\sqrt{\frac{15}{2}}\text{ft/sec}\end{array}$

d.

To determine

To find: the time when the velocity of the coin is -70 feet per second.

Answer to Problem 72E

The time when the velocity of the coin is -70 feet per second is approximately 2.19 seconds.

Explanation of Solution

Given information: $s(t)=-16t^2+120$

Calculation:

  $\begin{array}{l}s(t)=-16t^2+120\\ s\text{'}(t)=-32t=-70\Rightarrow t\approx 2.19\text{seconds}\end{array}$

e.

To determine

To graph: the model using a graphing utility.

Explanation of Solution

Given information: $s(t)=-16t^2+120$

Calculation:

The graph of the model using the graphing utility is shown below.

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