Q. 2 - Calculus

Question

A watermelon dropped from the top of a $50$-foot building has height given by $s\left(t\right)=50-16t^2$ feet and t seconds. Calculate each of the following:

Part (a): The average rate of change of the watermelon over its entire fall, over the first half of its fall, and over the second half of its fall.

Part (b): The average rate of change over the last second, the last half-second, and the last quarter-second of its fall.

Step-by-Step Solution

Answer

Part (a): The average rate of change of the watermelon over its entire fall is $32t$, over the first half is $8t$ and over the second half is $8t$.

Part (b): The average rate of change over the last second is $32t$, the last half-second is $8t$, and the last quarter-second of its fall is $2t$.

Part (a) Step 1. Given information.

Consider the given question,

Height of the building is $50$ foot.

Height given by $s\left(t\right)=50-16t^2$.

Part (a) Step 2. Calculate the average rate of change of the watermelon over its entire fall.

The average rate of change of watermelon is calculated by differentiating the given equation with respect to the time as follows,

$$\begin{gathered} \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times 16t \\ =32t \end{gathered}$$

For the first half of its fall, the equation of the height is given below,

$$\begin{gathered} h\left(t\right)=50-16{\left(\frac{t}{2}\right)}^2 \\ h\left(t\right)=50-4t^2 \end{gathered}$$

So, the average rate of change of watermelon over the first half is given below,

$$\begin{gathered} \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times 4t \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=8t \end{gathered}$$

Similarly, the average rate of change of watermelon over the second half is given below,

$$\begin{gathered} \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times 4t \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=8t \end{gathered}$$

Part (b) Step 1. Calculate the average rate of change of the watermelon over its last seconds.

The average rate of change of watermelon is calculated by differentiating the given equation with respect to the time for the last  seconds as follows,

$$\begin{gathered} \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times 16t \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=32t \end{gathered}$$

For the last half of the second, the average rate of change is given below,

$$\begin{gathered} h\left(t\right)=50-16{\left(\frac{t}{2}\right)}^2 \\ h\left(t\right)=50-4t^2 \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times 4t \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=8t \end{gathered}$$

Similarly, for the last quarter second of the average rate of change of watermelon is given below,

$$\begin{gathered} h\left(t\right)=50-16{\left(\frac{t}{4}\right)}^2 \\ h\left(t\right)=50-t^2 \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=0+2\times t \\ \frac{d\left\{h\left(t\right)\right\}}{dt}=2t \end{gathered}$$