Chapter 1, Problem 51P

A person going for a walk follows the path shown in Figure P1.51. The total trip consists of four straight-line paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?

Figure P1.51

To determine

The person’s resultant displacement and the direction in the starting point.

Answer to Problem 51P

The person’s resultant displacement in the starting point is $240\text{m}$ and the direction is $237°$.

Explanation of Solution

The following figure shows the person going a walk follows the path.

The person going for a walk in the first straight path is ${\overrightarrow{d}}_1=100\text{m}\widehat{i}$.

The person going for a second straight line path is ${\overrightarrow{d}}_2=-300\text{m}\widehat{i}$.

Write the expression for third path of the person moves in x and y component.

  ${\overrightarrow{d}}_3=-d_{3x}\cos \theta \widehat{i}-d_{3y}\sin \theta \widehat{j}$        (I)

Here, ${\overrightarrow{d}}_3$ is the distance of third path of the person moves, $d_{3x}$ is the distance of the x component, $d_{3y}$ is the distance of the y component, and $\theta$ is the angle makes with the distance.

Write the expression for fourth path of the person moves in x and y component.

  ${\overrightarrow{d}}_4=-d_{4x}\cos \theta \widehat{i}+d_{4y}\sin \theta \widehat{j}$        (II)

Here, ${\overrightarrow{d}}_4$ is the distance of fourth path of the person moves, $d_{4x}$ is the distance of the x component, $d_{4y}$ is the distance of the y component, and $\theta$ is the angle makes with the distance.

Write the expression for net displacement vectors.

  $\overrightarrow{R}={\overrightarrow{d}}_1+{\overrightarrow{d}}_2+{\overrightarrow{d}}_3+{\overrightarrow{d}}_4$        (III)

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

Write the expression for the magnitude of the resultant displacement.

  $\left|\overrightarrow{R}\right|=\sqrt{R_x^2+R_y^2}$        (IV)

Here, $\left|\overrightarrow{R}\right|$ is the magnitude of the resultant displacement, $R_x$ is the x component of the distance, and $R_y$ is the y component of the distance.

Write the expression for the direction of the displacement.

  $\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$        (V)

Here, $\theta$ is the direction for the resultant displacement.

Conclusion:

Substitute $150\text{m}$ for $d_{3x}$, and $d_{3y}$, $30°$ for $\theta$ in the equation (I) to find ${\overrightarrow{d}}_3$.

  $\begin{array}{c}{\overrightarrow{d}}_3=-\left(150\text{m}\right)\cos \left(30°\right)\widehat{i}-\left(150\text{m}\right)\sin \left(30°\right)\widehat{j}\\ =\left(130\widehat{i}-75.0\widehat{j}\right)\text{m}\end{array}$

Substitute $200\text{m}$ for $d_{4x}$, and $d_{4y}$, $60°$ for $\theta$ in the equation (II) to find ${\overrightarrow{d}}_4$.

  $\begin{array}{c}{\overrightarrow{d}}_4=-\left(200\text{m}\right)\cos \left(60°\right)\widehat{i}+\left(200\text{m}\right)\sin \left(60°\right)\widehat{j}\\ =\left(-100\widehat{i}+173\widehat{j}\right)\text{m}\end{array}$

Substitute $100\text{m}\widehat{i}$ for ${\overrightarrow{d}}_1$, $-300\text{m}\widehat{i}$ for ${\overrightarrow{d}}_2$, $\left(130\widehat{i}-75.0\widehat{j}\right)\text{m}$ for ${\overrightarrow{d}}_3$, and $\left(-100\widehat{i}+173\widehat{j}\right)\text{m}$ for ${\overrightarrow{d}}_4$ in equation (III) to find $\overrightarrow{R}$.

  $\begin{array}{c}\overrightarrow{R}=100\text{m}\widehat{i}-300\text{m}\widehat{i}+\left(130\widehat{i}-75.0\widehat{j}\right)\text{m}+\left(-100\widehat{i}+173\widehat{j}\right)\text{m}\\ =\left(-130\widehat{i}-202\widehat{j}\right)\text{m}\end{array}$

Substitute $-130\text{m}$ for $R_x$ and $-202\text{m}$ for $R_y$ in equation (IV) to find $\left|\overrightarrow{R}\right|$.

  $\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{{\left(-130\text{m}\right)}^2+{\left(-202\text{m}\right)}^2}\\ =240\text{m}\end{array}$

Substitute $-130\text{m}$ for $R_x$ and $-202\text{m}$ for $R_y$ in equation (V) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-202\text{m}}{-130\text{m}}\right)\\ =57.2°\\ \theta =180°+57.2°\\ =237°\end{array}$

Therefore, the person’s resultant displacement in the starting point is $240\text{m}$ and the direction is $237°$.