Chapter 3, Problem 21P

(a)

To determine

The average speed the driver drives $30.0°$ south of the west for $48.0\text{min}$.

Answer to Problem 21P

The average speed the driver drives $30.0°$ south of the west for $48.0\text{min}$ $\underset{\_}{70\text{mi}/\text{h}}$

Explanation of Solution

Let $+x$ axis represent east and $+y$ axis represent north

Write the expression for velocity.

  $v=\frac{d}{t}$        (I)

Here, $v$ is the velocity, $d$ is the distance, and $t$ is the time.

Conclusion:

Substitute, $55\text{mi}/\text{h}$ for $v$, and $1.20\text{h}$ for $t$ in equation (I) to the distance travelled in $1.20\text{h}$.

  $\begin{array}{c}55\text{mi}/\text{h}=\frac{d_1}{1.20\text{h}}\\ d_1=66\text{mi}\end{array}$

The total distance is $122\text{mi}$, hence the remaining distance is.

  $\begin{array}{c}{d}_2=122\text{mi}-\text{66}\text{mi}\\ \text{=56}\text{mi}\end{array}$

Substitute, $\text{56}\text{mi}$ for $d$, and $48\text{min}$ for $t$ in equation (I) to find the speed for remaining $48\text{min}$.

  $\begin{array}{c}{v}_2=\frac{\text{56}\text{mi}}{48\text{min}\times \frac{1\text{h}}{60\text{min}}}\\ =70\text{mi}/\text{h}\end{array}$

Therefore, the average speed the driver drives $30.0°$ south of the west for $48.0\text{min}$ $\underset{\_}{70\text{mi}/\text{h}}$.

(b)

To determine

The average velocity of entire trip.

Answer to Problem 21P

The average velocity of entire trip is $\underset{\_}{59\text{mi}/\text{h}}$, in a direction $\underset{\_}{13.7°\text{south of west }}$.

Explanation of Solution

Write the expression for average velocity.

  $\left|{\overrightarrow{v}}_{av}\right|=\frac{\left|\Delta \overrightarrow{r}\right|}{\Delta t}$        (II)

Here, $\left|\Delta \overrightarrow{r}\right|$ is the displacement vector.

The total displacement can be divided into two, first for distance travelled in $1.20\text{h}$, and second is the distance travelled in $48\text{min}$.

Write the expression for displacement vector for first $1.20\text{h}$.

  $\begin{array}{c}\Delta {\overrightarrow{r}}_1=x_1\widehat{i}+y_1\widehat{j}\\ =-66\widehat{i}\end{array}$

Write the expression for displacement vector for second $48\text{min}$.

  $\begin{array}{c}\Delta {\overrightarrow{r}}_2=x_2\widehat{i}+y_2\widehat{j}\\ =-56\cos 30°\widehat{i}-56\sin 30°\widehat{j}\\ =-48.5\widehat{i}-28\widehat{j}\end{array}$

Hence the vector representation of total displacement vector is.

  $\begin{array}{c}\Delta \overrightarrow{r}=\Delta {\overrightarrow{r}}_1\widehat{i}+\Delta {\overrightarrow{r}}_2\widehat{j}\\ =\left(-66-48.5\right)\widehat{i}-28\widehat{j}\\ =114.5\widehat{i}-28\widehat{j}\end{array}$

Write the expression for the magnitude of displacement vector.

  $\left|\Delta \overrightarrow{r}\right|=\sqrt{\Delta {\overrightarrow{r}}_1{}^2+\Delta {\overrightarrow{r}}_2{}^2}$        (III)

Write the expression for the direction of displacement vector.

  $\theta ={\tan }^{-1}\frac{\Delta {\overrightarrow{r}}_2}{\Delta {\overrightarrow{r}}_1}$        (IV)

Conclusion:

Substitute, $114.5$ for $\Delta {\overrightarrow{r}}_1$, and $-28$ for $\Delta {\overrightarrow{r}}_2$ in equation (III) to find the magnitude of displacement vector.

  $\begin{array}{c}\left|\Delta \overrightarrow{r}\right|=\sqrt{{\left(-114.5\right)}^2+{\left(-28\right)}^2}\\ =117.9\text{mi}\end{array}$

Substitute, $114.5$ for $\Delta {\overrightarrow{r}}_1$, and $-28$ for $\Delta {\overrightarrow{r}}_2$ in equation (III) to find the direction of displacement vector.

  $\begin{array}{c}\theta ={\tan }^{-1}\frac{\left|-28\right|}{\left|-114.5\right|}\\ =13.7°\end{array}$

Substitute, $117.9\text{mi}$ for $\Delta \overrightarrow{r}$, and $2\text{h}$ for $\Delta t$ in equation (IV) to find the average velocity of entire trip.

  $\begin{array}{c}\left|{\overrightarrow{v}}_{av}\right|=\frac{117.9\text{mi}}{2\text{h}}\\ =59\text{mi}/\text{h}\end{array}$

Therefore, the average velocity of entire trip is $\underset{\_}{59\text{mi}/\text{h}}$, in a direction $\underset{\_}{13.7°\text{south of west }}$.