Chapter 8.1, Problem 9E

a.

To determine

To determine the velocity of an automobile after $9$ seconds.

Answer to Problem 9E

The velocity after $9$ seconds is $63mph$

Explanation of Solution

Given:

An automobile accelerates from rest at $1+3\sqrt{t}$ for $9$ seconds.

Calculation:

  $\begin{array}{l}a\left(t\right)=\frac{dv}{dt}\\ v\left(t\right)={\displaystyle \int a\left(t\right)dt}\\ v\left(t\right)={\displaystyle \int \left(1+3\sqrt{t}\right)dt}\\ v\left(t\right)=t+2t^{3/2}+C\end{array}$

It is given that automobile accelerates from rest that is at $t=0$ , $v\left(t\right)=0$

Therefore

  $\begin{array}{l}v\left(t\right)=t+2t^{3/2}+C\\ 0=0+2{\left(0\right)}^{3/2}+C\\ C=0\end{array}$

So, $v\left(t\right)=t+2t^{3/2}$

  $v\left(9\right)=9+2{\left(9\right)}^{3/2}=63mph$

b.

To determine

To find the distance travel by an automobile in $9$ seconds.

Answer to Problem 9E

The distance travel by an automobile in $9$ seconds is $234.9m$

Explanation of Solution

Given:

An automobile accelerates from rest at $1+3\sqrt{t}$ for $9$ seconds.

Calculation:

From part (a)

  $v\left(t\right)=t+2t^{3/2}$

  $\begin{array}{l}v\left(t\right)=\frac{ds}{dt}\\ s={\displaystyle \int v\left(t\right)dt}\\ s={\displaystyle \underset{0}{\overset{9}{\int }}\left(t+2t^{3/2}\right)}dt\\ s={\left|\frac{t^2}{2}+\frac{4t^{5/2}}{5}\right|}_0^9\\ s=234.9m\end{array}$