Chapter 4.1, Problem 64E

To determine

To show: the acceleration of the given falling body is constant.

Answer to Problem 64E

The acceleration is constant for a falling body because variables do not change.

Explanation of Solution

Given information:

A body falling with the velocity $v=k\sqrt{s}\text{m}/\text{s}$ .

Calculation:

The derivative of the velocity $v^{\prime }=k\sqrt{s}$ gives the acceleration of the falling body,

  $\begin{array}{c}a=\frac{dv}{dt}\\ \frac{dv}{dt}=\frac{dv}{ds}\cdot \frac{ds}{dt}\end{array}$

We know that $\frac{ds}{dt}$ is the velocity given,

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

We know that $k$ is a constant. Therefore we can conclude that the acceleration of the given falling body is constant.