Chapter 7.1, Problem 9E

a.

To determine

To find: The velocity of the automobile after 9 seconds.

Answer to Problem 9E

  $63$ mph/sec.

Explanation of Solution

Given information:

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

Concept used:

Relation between velocity and acceleration is given by $v\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}a\left(t\right)dt}$ .

Calculation:

Given that $a\left(t\right)=1+3\sqrt{t}$ .

Find the velocity at any time $t$ .

  $\begin{array}{c}v\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}a\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}\left(1+3\sqrt{t}\right)dt}\\ =t+3\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\\ =t+2t^{\frac{3}{2}}\end{array}$

Now, substitute $t=9$ in the equation $v\left(t\right)=t+2t^{\frac{3}{2}}$ .

  $\begin{array}{c}v\left(9\right)=9+2{\left(9\right)}^{\frac{3}{2}}\\ =9+2\cdot 27\\ =63\end{array}$

Therefore, the velocity after 9 seconds is $63$ mph/sec.

Conclusion:

The velocity after 9 seconds is $0.5$ mph/sec.

b.

To determine

To find: The distance travelled in in 9 seconds.

Answer to Problem 9E

The distance travelled in 9 seconds is $234.9$ mph.

Explanation of Solution

Given information:

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

Concept used:

Relation between distance travelled and the velocity is given by $s\left(t\right)=\frac{d}{dt}\left(v\left(t\right)\right)$ .

Calculation:

The velocity formula from part (a) is $v\left(t\right)=t+2t^{\frac{3}{2}}$ .

Find the expression for the distance travelled at any time $t$ .

  $\begin{array}{c}s\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}v\left(t\right)dt}\\ ={\displaystyle \underset{0}{\overset{t}{\int }}\left(t+2t^{\frac{3}{2}}\right)dt}\\ =\frac{t^2}{2}+2\cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}}\\ =\frac{t^2}{2}+\frac{4}{5}{t}^{\frac{5}{2}}\end{array}$

Now, substitute $t=9$ in the equation $s\left(t\right)=\frac{t^2}{2}+\frac{4}{5}{t}^{\frac{5}{2}}$ .

  $\begin{array}{c}s\left(9\right)=\frac{9^2}{2}+\frac{4}{5}{9}^{\frac{5}{2}}\\ =\frac{81}{2}+\frac{4}{5}\cdot 243\\ =\frac{2349}{10}\\ =234.9\end{array}$

Therefore, the distance travelled in 9 seconds is $234.9$ mph.

Conclusion:

The distance travelled in 9 seconds is $234.9$ mph.