Chapter 3, Problem 123P

(a)

To determine

The direction and distance the person should travel to reach the starting point.

Answer to Problem 123P

The person has to walk $6.3\text{km}$ at $29.6°$ angle with horizontal to reach the starting point.

Explanation of Solution

Draw the vector diagram showing the displacements of the person.

Represent the displacements of person as vectors. Let west direction is taken as -ve x-direction, east direction as +ve x-direction, north as +y-axis and south as -ve y-axis.

Express 2.00 km in west direction as a vector.

    $\overrightarrow{A}=-\left(2.00\text{km}\right)\widehat{i}$

Express 5.00 km in south-west direction as a vector.

    $\begin{array}{c}\overrightarrow{B}=-\left[\left(5.00\text{km}\right)\left(\cos 53°\right)\right]\widehat{i}-\left[\left(5.00\text{km}\right)\left(\sin 53°\right)\right]\widehat{j}\\ =-\left(3.0\text{km}\right)\widehat{i}-\left(4.0\text{km}\right)\widehat{j}\end{array}$

Express 1.00 km in north-west direction as a vector.

    $\begin{array}{c}\overrightarrow{C}=-\left[\left(1.00\text{km}\right)\left(\cos 60°\right)\right]\widehat{i}+\left[\left(1.00\text{km}\right)\left(\sin 60°\right)\right]\widehat{j}\\ =-\left(0.5\text{km}\right)\widehat{i}+\left(0.87\text{km}\right)\widehat{j}\end{array}$

The net displacement is given by the vector addition.

    $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$

Here, $\overrightarrow{R}$ is the net displacement vector.

Rewrite the above equation by substituting the equations for $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$.

  $\begin{array}{c}\overrightarrow{R}=-\left(2.00\text{km}\right)\widehat{i}+-\left(3.0\text{km}\right)\widehat{i}-\left(4.0\text{km}\right)\widehat{j}+-\left(0.5\text{km}\right)\widehat{i}+\left(0.87\text{km}\right)\widehat{j}\\ =\left(-5.5\text{km}\right)\widehat{i}-\left(3.13\text{km}\right)\widehat{j}\end{array}$

Write the equation to find the magnitude of $\overrightarrow{R}$.

    $\left|\overrightarrow{R}\right|=\sqrt{x^2+y^2}$

Here, $\left|\overrightarrow{R}\right|$ is the magnitude, $x$ is the x-component and $y$ is the y-component.

Write the equation to find the direction of $\overrightarrow{R}$.

    $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Here, $\theta$ is the direction.

Conclusion:

Substitute $-5.5\text{km}$ for $x$ and $-3.13\text{km}$ for $y$ in the equation for $\left|\overrightarrow{R}\right|$.

    $\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{{\left(-5.5\text{km}\right)}^2+{\left(3.13\text{km}\right)}^2}\\ =\sqrt{40.05{\text{km}}^{\text{2}}}\\ =6.3\text{km}\end{array}$

Substitute $-5.5\text{km}$ for $x$ and $-3.13\text{km}$ for $y$ in the equation for $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-3.13\text{km}}{-5.5\text{km}}\right)\\ ={\tan }^{-1}\left(0.569\right)\\ =29.6°\end{array}$

Thus, the person has to walk $6.3\text{km}$ at $29.6°$ angle with horizontal to reach the starting point.

(b)

To determine

The time taken by the person to reach starting from the final point at the speed of $5.00\text{m}/\text{s}$.

Answer to Problem 123P

Time taken will be $21\min$.

Explanation of Solution

Write the equation to find the time taken.

    $t=\frac{d}{v}$

Here, $t$ is the time taken and $d$ is the net displacement.

Conclusion:

Substitute $6.3\text{km}$ for $d$ and $5.00\text{m}/\text{s}$ for $v$ in the above expression.

    $\begin{array}{c}t=\frac{6.3\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)}{5.00\text{m}/\text{s}}\\ =1.26\times {\text{10}}^{\text{3}}\text{s}\left(\frac{\text{1}\text{min}}{60\text{s}}\right)\\ =21\min \end{array}$

Here, the time taken will be $21\min$.