Chapter 3, Problem 44P

(a)

To determine

The distance from the home to the second town.

Answer to Problem 44P

The distance from the home to the second town is $76.1\text{km}$.

Explanation of Solution

The speed of the person due east is $90.0\text{km}/\text{h}$, the speed of the person due south of west is $76.0\text{km}/\text{h}$, the time taken towards east is $80.0\text{m}$, time taken towards west of south direction is $45.0\min$ and direction in west of south is $30.0°$.

Write the expression to calculate the displacement vector.

$\overrightarrow{r}=\left(v_1{t}_1-v_2{t}_2\cos \theta \right)\text{east}-\left(v_2{t}_2\sin \theta \right)\text{south}$

Here, $\overrightarrow{r}$ is the displacement vector, $v_1$ is the speed of the person due east, $t_1$ is the time taken to move due east, $v_2$ is the speed of the person due west of south, $t_2$ is the time taken to move due west of south and $\theta$ is the direction in west of south.

Substitute $90.0\text{km}/\text{h}$ for $v_1$, $76.0\text{km}/\text{h}$ for $v_2$, $80.0\text{m}$ for $t_1$, $45.0\min$ for $t_2$ and $30.0°$ for $\theta$ in the above equation to calculate $\overrightarrow{r}$.

$\begin{array}{c}\overrightarrow{r}=\left(90.0\text{km}/\text{h}\left(80.0\text{m}\right)\left(\frac{1\text{h}}{60\min }\right)-76.0\text{km}/\text{h}\left(45.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\cos 30.0°\right)\text{east}-\\ \left(76.0\text{km}/\text{h}\left(45.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\sin 30.0°\right)\text{south}\\ =\left(\text{120}\text{km}-49.4\text{km}\right)\text{east}-\left(28.5\text{km}\right)\text{south}\\ =\left(\text{70}\text{.6}\text{km}\right)\text{east}-\left(28.5\text{km}\right)\text{south}\end{array}$

Write the expression to calculate the magnitude of the displacement.

$\begin{array}{c}r=\sqrt{{\left(\text{70}\text{.6}\text{km}\right)}^2+{\left(28.5\text{km}\right)}^2}\\ =76.1\text{km}\end{array}$

Conclusion:

Therefore, the distance from the home to the second town is $76.1\text{km}$.

(b)

To determine

The average velocity of the person.

Answer to Problem 44P

The average velocity of the person is $101.5\text{km}/\text{h}$ and direction is $22.0°$ north of west.

Explanation of Solution

The time taken towards west of south direction is $45.0\min$ and the displacement is $76.1\text{km}$.

Write the expression to calculate the average velocity.

$v_{av}=\frac{r}{t_2}$

Here, $v_{av}$ is the average velocity.

Substitute $45.0\min$ for $t_2$ and $76.1\text{km}$ for r in the above equation to calculate $v_{av}$.

$\begin{array}{c}{v}_{av}=\frac{76.1\text{km}}{45.0\min \left(\frac{1\text{h}}{60\min }\right)}\\ =101.5\text{km}/\text{h}\end{array}$

Write the expression to calculate the direction.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{76.0\text{km}/\text{h}\left(45.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\sin 30.0°}{-90.0\text{km}/\text{h}\left(80.0\text{m}\right)\left(\frac{1\text{h}}{60\min }\right)+76.0\text{km}/\text{h}\left(45.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\cos 30.0°}\right)\\ ={\tan }^{-1}\left(\frac{28.5}{-70.6}\right)\\ =-22.0°\end{array}$

The direction is $22.0°$ north of west.

Conclusion:

Therefore, the average velocity of the person is $101.5\text{km}/\text{h}$ and direction is $22.0°$ north of west.

(c)

To determine

The average velocity during the first two legs of trip.

Answer to Problem 44P

The average velocity during the first two legs of trip is $32.6\text{km}/\text{h}$ and the direction is $22.0°$ north of west.

Explanation of Solution

The time taken towards west of south direction is $140.0\min$ and the displacement is $76.1\text{km}$.

Write the expression to calculate the average velocity.

$v_{av}=\frac{r}{t}$

Here, $v_{av}$ is the average velocity and t is the time taken.

Substitute $140.0\min$ for t and $76.1\text{km}$ for r in the above equation to calculate $v_{av}$.

$\begin{array}{c}{v}_{av}=\frac{76.1\text{km}}{140.0\min \left(\frac{1\text{h}}{60\min }\right)}\\ =32.6\text{km}/\text{h}\end{array}$

The direction in this case is just opposite the above part (b). Thus, the direction is $22.0°$ south of east.

Conclusion:

Thus, the average velocity during the first two legs of trip is $32.6\text{km}/\text{h}$ and the direction is $22.0°$ north of west.

(d)

To determine

The average velocity over the entire trip.

Answer to Problem 44P

The average velocity over the entire trip is $0\text{km}/\text{h}$.

Explanation of Solution

The time taken for the overall is $185.0\min$ and the displacement is $0\text{km}$.

Write the expression to calculate the average velocity.

$v_{av}=\frac{r}{t}$

Here, $v_{av}$ is the average velocity and t is the time taken.

Substitute $185.0\min$ for t and $0\text{km}$ for r in the above equation to calculate $v_{av}$.

$\begin{array}{c}{v}_{av}=\frac{0\text{km}}{185.0\min \left(\frac{1\text{h}}{60\min }\right)}\\ =0\text{km}/\text{h}\end{array}$

Conclusion:

Thus, the average velocity over the entire trip is $0\text{km}/\text{h}$.

(e)

To determine

The average velocity for the trip

Answer to Problem 44P

The average velocity for the trip is $63.3\text{km}/\text{h}$.

Explanation of Solution

The time taken for the overall is $240.0\min$ and the displacement is $253.1\text{km}$.

Write the expression to calculate the average velocity.

$v_{av}=\frac{r}{t}$

Here, $v_{av}$ is the average velocity and t is the time taken.

Substitute $240.0\min$ for t and $253.1\text{km}$ for r in the above equation to calculate $v_{av}$.

$\begin{array}{c}{v}_{av}=\frac{253.1\text{km}}{240.0\min \min \left(\frac{1\text{h}}{60\min }\right)}\\ =63.3\text{km}/\text{h}\end{array}$

Conclusion:

Thus, the average velocity for the trip is $63.3\text{km}/\text{h}$.