Chapter 2.4, Problem 32E

To determine

To find: The speed of the rock at $t=2\sec$.

Answer to Problem 32E

The speed of the rock at $t=2\sec$ is equal to $45.76\text{m}/\sec$.

Explanation of Solution

Given information:

The equation for the free fall at the surface of Jupiter is $s=11.44t^2\text{m}$ with $t$ in seconds. Assume a rock is dropped from the top of a $500\text{m}$ cliff.

Calculation:

The speed of the rock at $t=2\sec$ is the instantaneous rate of change of function for distance of rock at $t=2\sec$.

The formula for the instantaneous rate of change of distance with respect to time at $t\sec$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(t+h\right)-f\left(t\right)}{h}$.

The formula for the instantaneous rate of change of $f\left(t\right)=11.44t^2\text{m}$ at $t=2\sec$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$.

Substitute $2+h$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=11.44t^2\\ f\left(2+h\right)=11.44{\left(2+h\right)}^2\\ f\left(2+h\right)=11.44\left(4+h^2+4h\right)\\ f\left(2+h\right)=45.76+11.44h^2+45.76h\end{array}$

Substitute $2$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=11.44t^2\\ f\left(2\right)=11.44{\left(2\right)}^2\\ f\left(2\right)=11.44\left(4\right)\\ f\left(2\right)=45.76\end{array}$

Substitute $45.76+11.44h^2+45.76h$ for $f\left(2+h\right)$ and $45.76$ for $f\left(2\right)$ in the formula.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{45.76+11.44h^2+45.76h-45.76}{h}\\ =\underset{h\to 0}{\lim }\frac{11.44h^2+45.76h}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(11.44h+45.76\right)}{h}\\ =\underset{h\to 0}{\lim }\left(11.44h+45.76\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=11.44\left(0\right)+45.76\\ =45.76\end{array}$

Therefore, the speed of the rock at $t=2\sec$ is equal to $45.76\text{m}/\sec$.