Chapter 4.2, Problem 53E

To determine

To show: The body’s acceleration is $32{\text{ ft/sec}}^2$ .

Answer to Problem 53E

The acceleration of the body is $32\text{ft}/s^2$ .

Explanation of Solution

Given Information: The velocity of a falling body is $v=8\sqrt{s-t+1}$ feet per second at the instant t .

Calculation:

The acceleration of the object is equal to the rate of change of velocity. The acceleration is,

  $\begin{array}{c}v=8\sqrt{s-t+1}\\ a=\frac{dv}{dt}\\ =\frac{d}{dt}\left(8\sqrt{s-t+1}\right)\\ =8\times \frac{1}{2}{\left(s-t+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1}\\ =\frac{4}{\sqrt{s-t+1}}\left(\frac{ds}{dt}-1\right)\\ =\frac{4}{\sqrt{s-t+1}}\left(v-1\right)\\ =\frac{4}{\frac{v-1}{8}}\left(v-1\right)\\ \text{=}\frac{\text{32}\left(v-1\right)}{\left(v-1\right)}\left(\sqrt{s-t+1}=\frac{v-1}{8}\right)\\ =32\end{array}$

• Here, a is the acceleration of the body.