Chapter 3, Problem 91P

(a)

To determine

The distance and the direction in which H should travel to return to his starting point.

Answer to Problem 91P

H should travel $6.33\text{ km}$ at $29.6°\text{ north of east}$ to return to his starting point.

Explanation of Solution

The travel of H is sketched in figure 1.

Compute the displacement using the component method.

Write the equation for displacement in $x$ direction.

$\Delta x=x_1+x_2+x_3$ (I)

Here, $\Delta x$ is the net displacement in $x$ direction and $x_1,x_2,x_3$ are the $x$ components of his successive displacements

Refer to figure 1 and write the values of $x$ components of different displacements.

$\begin{array}{c}{x}_1=-2.00\text{ km}\\ x_2=\left(5.00\text{ km}\right)\cos 233°\\ x_3=\left(1.00\text{ km}\right)\cos 120°\end{array}$

Put the value of $x_1,x_2,x_3$ in equation (I) to find $\Delta x$ .

$\Delta x=-2.00\text{ km}+\left(5.00\text{ km}\right)\cos 233°+\left(1.00\text{ km}\right)\cos 120°$

Write the equation for displacement in $y$ direction.

$\Delta y=y_1+y_2+y_3$ (II)

Here, $\Delta y$ is the net displacement in $y$ direction and $y_1,y_2,y_3$ are the $y$ components of his successive displacements

Refer to figure 1 and write the values of $y$ components of different displacements.

$\begin{array}{c}{y}_1=0\\ y_2=\left(5.00\text{ km}\right)\sin 233°\\ y_3=\left(1.00\text{ km}\right)\sin 120°\end{array}$

Put the value of $y_1,y_2,y_3$ in equation (II) to find $\Delta y$ .

$\begin{array}{c}\Delta y=0+\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°\\ =\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°\end{array}$

Write the equation for the magnitude of the net displacement.

$\Delta r=\sqrt{{\left(\Delta x\right)}^2+{\left(\Delta y\right)}^2}$ (III)

Here, $\Delta r$ is the magnitude of the net displacement

Write the equation for the direction of net displacement.

$\theta ={\tan }^{-1}\frac{\Delta y}{\Delta x}$ (IV)

Conclusion:

Substitute $-2.00\text{ km}+\left(5.00\text{ km}\right)\cos 233°+\left(1.00\text{ km}\right)\cos 120°$ for $\Delta x$ and $\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°$ for $\Delta y$ in equation (III) to find $\Delta r$ .

$\begin{array}{c}\Delta r=\sqrt{\begin{array}{l}{\left(-2.00\text{ km}+\left(5.00\text{ km}\right)\cos 233°+\left(1.00\text{ km}\right)\cos 120°\right)}^2\\ +{\left(\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°\right)}^2\end{array}}\\ =6.33\text{ km}\end{array}$

Substitute $\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°$ for $\Delta y$ and $-2.00\text{ km}+\left(5.00\text{ km}\right)\cos 233°+\left(1.00\text{ km}\right)\cos 120°$ for $\Delta x$ in equation (IV) to find $\theta$ .

$\begin{array}{c}\theta ={\tan }^{-1}\frac{\left(5.00\text{ km}\right)\sin 233°+\left(1.00\text{ km}\right)\sin 120°}{-2.00\text{ km}+\left(5.00\text{ km}\right)\cos 233°+\left(1.00\text{ km}\right)\cos 120°}\\ =29.6°\end{array}$

Both the components are negative so that $180°$ must be added to the value of $\theta$ .

$\begin{array}{c}\theta =29.6°+180°\\ =209.6°\\ =29.6°\text{ below the }-x\text{ axis}\\ =29.6°\text{ south of west}\end{array}$

H must travel in the opposite direction to return to his starting point.

Therefore, H should travel $6.33\text{ km}$ at $29.6°\text{ north of east}$ to return to his starting point.

(b)

To determine

The time taken for the trip of H if he returns directly to his return point with a speed of $5.00\text{ m/s}$ .

Answer to Problem 91P

The time taken for the trip of H if he returns directly to his return point with a speed of $5.00\text{ m/s}$ is $21.1\text{ min}$ .

Explanation of Solution

Write the equation for the time taken.

$\Delta t=\frac{\Delta r}{v}$ (V)

Here, $\Delta t$ is the time taken for the motion and $v$ is the speed of H

Conclusion:

Substitute $6.33\text{ km}$ for $\Delta r$ and $5.00\text{ m/s}$ for $v$ in equation (V) to find $\Delta t$ .

$\begin{array}{c}\Delta t=\frac{6.33\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)}{5.00\text{ m/s}}\\ =\frac{6.33\times {10}^3\text{ m}}{5.00\text{ m/s}}\\ =1266\text{ s}\left(\frac{1\text{ min}}{60\text{ s}}\right)\\ =21.1\text{ min}\end{array}$

Therefore, the time taken for the trip of H if he returns directly to his return point with a speed of $5.00\text{ m/s}$ is $21.1\text{ min}$ .