Chapter 4.1, Problem 65E

To determine

To show: the acceleration of the meteorite is inversely proportional to $s^2$ .

Answer to Problem 65E

The acceleration of the meteorite entering the earth’s atmosphere is inversely proportion to $s^2$ .

Explanation of Solution

Given information:

A meteorite entering the earth’s atmosphere is inversely proportional to $\sqrt{s}$ .

Calculation:

The relation between distance $s$ and the velocity $v$ is given as $v=\frac{k}{\sqrt{s}}$ .

The derivative of the velocity gives the acceleration $a$ .

  $\begin{array}{l}a=\frac{d}{dt}\frac{k}{\sqrt{s}}\cdot \frac{ds}{dt}\\ a=k\frac{d}{dt}{\left(s\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cdot v\\ a=k\left(-\frac{1}{2}{\left(s\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\cdot \frac{k}{\sqrt{s}}\\ a=-k\frac{1}{2{\left(\sqrt{s}\right)}^3}\cdot \frac{k}{\sqrt{s}}\\ a=-\frac{k^2}{2s^2}\end{array}$

Thus we can show that the acceleration of the meteorite is inversely proportion to $s^2$ .