Chapter 3.2, Problem 52E

To determine

To find : the velocity and acceleration of a body when $t=2$ sec.

Answer to Problem 52E

The velocity of body is $36$ m/sec.

The acceleration of body is $6.75$ m/sec square.

Explanation of Solution

Given information :

The position function of a body is given by $s={\left(4+6t\right)}^{3/2}$ where s in meters and t in seconds.

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

For velocity, differentiate $s={\left(4+6t\right)}^{3/2}$ with respect to $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}{\left(4+6t\right)}^{3/2}\\ v=\frac{3}{2}{\left(4+6t\right)}^{3/2-1}\cdot \frac{d}{dt}\left(4+6t\right)\\ =\frac{3}{2}{\left(4+6t\right)}^{1/2}\cdot \left(6\right)\\ =9{\left(4+6t\right)}^{1/2}\end{array}$

Substitute $t=2$ in $v=9{\left(4+6t\right)}^{1/2}$ .

  $\begin{array}{c}v\left(2\right)=9{\left(4+6\cdot 2\right)}^{1/2}\\ =9{\left(16\right)}^{1/2}\\ =9{\left(4^2\right)}^{1/2}\\ =36\end{array}$

The velocity of body is $36$ m/sec.

For acceleration, differentiate $v=9{\left(4+6t\right)}^{1/2}$ with respect to $t$ .

  $\begin{array}{c}\frac{dv}{dt}=\frac{d}{dt}9{\left(4+6t\right)}^{1/2}\\ a=9\cdot \frac{1}{2}{\left(4+6t\right)}^{1/2-1}\cdot \frac{d}{dt}\left(4+6t\right)\\ =\frac{9}{2}{\left(4+6t\right)}^{-1/2}\cdot \left(6\right)\\ =27{\left(4+6t\right)}^{-1/2}\end{array}$

Substitute $t=2$ in $a=27{\left(4+6t\right)}^{-1/2}$ .

  $\begin{array}{c}a\left(2\right)=27{\left(4+6\cdot 2\right)}^{-1/2}\\ =27{\left(16\right)}^{-1/2}\\ =27{\left(4^2\right)}^{-1/2}\\ =6.75\end{array}$

The acceleration of body is $6.75$ m/sec square.