Chapter 11.3, Problem 72E

(a).

To determine

To calculate: The formula for the instantaneous rate of change of the balloon.

Answer to Problem 72E

The formula of instantaneous rate of change of the balloon is $s`\left(t\right)=-32t$

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+120$

Formula used: $\frac{d}{dx}\left(x^n\right)=n{x}^{n+1}$

Differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

Calculation:

  $\begin{array}{l}\frac{d}{dt}s=-16\left(2t^{2-1}\right)+0\\ \Rightarrow s`\left(t\right)=-16\times 2t+0\\ \Rightarrow s`\left(t\right)=-32t\end{array}$

Hence, the equation of rate change of the balloon is $s`\left(t\right)=-32t$

(b).

To determine

To calculate: Average rate of change of the balloon after the first three seconds.

Answer to Problem 72E

The average speed is 16 feet per second.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+64t+80$

Formula used: $\text{Then average speed = }\frac{\text{Total time taken}}{\text{Total distance travelled}}$

Calculation:

Total distance after two second:

  $\begin{array}{l}s\left(4\right)=-16{\left(2\right)}^2+120\\ =-16\times 4+120\\ =-64+120\\ =56\text{ feet}\end{array}$

Total time, when $\text{s}\left(t\right)=0$

Therefore, $s\left(t\right)=-16t^2+120=0$

  $\begin{array}{l}s\left(t\right)=-16t^2+120=0\\ \Rightarrow 16t^2=120\\ \Rightarrow t^2=\frac{120}{16}\\ \Rightarrow t=\sqrt{\frac{120}{16}}=\sqrt{7.5}\\ \Rightarrow t=2.74\text{sec}\left(\text{approximately}\right)\end{array}$

Hence, total time, $t=2.74\text{ second}$ (approximately)

Therefore,

  $\begin{array}{l}\text{Then average speed = }\frac{\text{Total time taken}}{\text{Total distance travelled}}\\ =\frac{\text{56 feet}}{\text{2}\text{.74 sec}}\\ =20.43\raisebox{1ex}{$\text{feet}$}\!\left/ \!\raisebox{-1ex}{$\text{sec}$}\right.\end{array}$

Hence, the average speed after two second is 20.43 feet per second.

(c).

To determine

To calculate: The velocity of the coin as it hit the ground.

Answer to Problem 72E

The velocity of the coin as it hit the ground is 32.32 feet per second.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+120$

Coin will impact the ground, when $s\left(t\right)=0$

Therefore, $s\left(t\right)=-16t^2+120=0$

Calculation:

  $\begin{array}{l}s\left(t\right)=-16t^2+120=0\\ \Rightarrow 16t^2=120\\ \Rightarrow t^2=\frac{120}{16}\\ \Rightarrow t=\sqrt{\frac{120}{16}}=\sqrt{7.5}\\ \Rightarrow t=2.74\text{sec}\left(\text{approximately}\right)\end{array}$

Hence, $t=2.74\text{ second}$

Formula used: $\frac{d}{dx}\left(x^n\right)=n{x}^{n+1}$

Differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

  $\begin{array}{l}\frac{d}{dt}s=-16\left(2t^{2-1}\right)+0\\ \Rightarrow s`\left(t\right)=-16\times 2t+0\\ \Rightarrow s`\left(t\right)=-32t\end{array}$

Put, $t=2.74\text{ second}$ in above equation, we get

  $\begin{array}{l}\Rightarrow s`\left(t\right)=-32\times 2.74+120\\ =-87.68+120\\ =32.32\\ \therefore s`\left(t\right)=32.32\raisebox{1ex}{$\text{feet}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.\end{array}$

Hence, velocity at time of impact is 32.32 feet per second.

(d).

To determine

To calculate: The time when the coin`s velocity − 70 feet per second.

Answer to Problem 72E

At 2.19 seconds the velocity of the coin is -70 feet per second.

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+120$

To get the velocity, differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

Formula used: $\frac{d}{dx}\left(x^n\right)=n{x}^{n+1}$

Differentiate the equation of displacement ( s ) with respect to $\text{'}t\text{'}$

  $\begin{array}{l}\frac{d}{dt}s=-16\left(2t^{2-1}\right)+0\\ \Rightarrow s`\left(t\right)=-16\times 2t+0\\ \Rightarrow s`\left(t\right)=-32t\end{array}$

Put, $s`\left(t\right)=-70\text{ feet per second}$ in above equation, we get

  $\begin{array}{l}\Rightarrow s`\left(t\right)=-70=-32t\\ \Rightarrow t=\frac{-70}{32}\\ \Rightarrow t=2.19\text{ sec}\end{array}$

Therefore, in 2.19 seconds, the velocity of the coin is -70 feet per second.

(e).

To determine

To draw: a graph to verify the result of parts (a) and (b)

Answer to Problem 72E

The required graph:

  

Explanation of Solution

Given information: The displacement s, $s\left(t\right)=-16t^2+120$

The above quadratic equation has two solutions one is negative and the second one is positive and equal to $\sqrt{7.5}$ seconds.

Therefore, it takes $\sqrt{7.5}$ seconds of the object to hit the ground after it has been thrown upward. The graphical meaning to the answers to parts a and d are shown below.

If $t=0$ then, $s\left(0\right)=0$

Calculation:

Hence,

  $\begin{array}{l}s\left(0\right)=-16\times 0+120\\ =120\\ \therefore s\left(0\right)=120\end{array}$

It means the height of the coin from the ground just before throwing, since the initial is 120 feet.