Question

II Ruth sets out to visit her friend Ward, who lives north and east of her. She starts by driving east, but after she comes to a detour that takes her south before going east again. She then drives east for and runs out of gas, so Ward flies there in his small plane to get her. What is Ward's displacement vector? Give your answer (a) in component form, using a coordinate system in which the -axis points north, and (b) as a magnitude and direction.

Short answer

a.The displacement vector of the ward is calculated as follows,

$\begin{array}{l}\Delta \overrightarrow{W}=\left(\right(38\text{mi})\widehat{i}-(15\text{mi}\left)\widehat{j}\right)-\left(\right(100\text{mi})\widehat{i}+(50\text{mi}\left)\widehat{j}\right)\\ =\left(\right(-62\text{mi})\widehat{i}-(65\text{mi}\left)\widehat{j}\right)\end{array}$

b. The magnitude of the displacement vector is 90 miles and the direction is ${46}^{°}$south of west.

Step-by-step solution

Step.1.

Consider ward as a particle. Ruth's initial location is taken as the origin of the coordinate system.

Part.(a).

Consider the following diagram,

Step.2.

The position vector of the Ruth is calculated as follows,

$\begin{array}{l}{\overrightarrow{k}}_2=\left(\right(30\text{mi})\widehat{i}+(0.0\text{mi}\left)\widehat{j}\right)+\left(\right(0.0\text{mi})\widehat{i}-(15\text{mi}\left)\widehat{j}\right)+\left(\right(8\text{mi})\widehat{i}+(0.0\text{mi}\left)\widehat{j}\right)\\ =\left(\right(38\text{mi})\widehat{i}-(15\text{mi}\left)\widehat{j}\right)\end{array}$

When Ward picks up Ruth they are at the same location. Therefore

$W_2=k_2=\left(\right(38\text{mi})\widehat{i}-(15\text{mi}\left)\widehat{j}\right)$Since,

$W_1=\left(\right(100\text{mi})\widehat{i}-(50\text{mi}\left)\widehat{j}\right)$

Therefore, the displacement vector of the ward is calculated as follows,

$\begin{array}{l}\Delta \overrightarrow{W}=\left(\right(38\text{mi})\widehat{i}-(15\text{mi}\left)\widehat{j}\right)-\left(\right(100\text{mi})\widehat{i}+(50\text{mi}\left)\widehat{j}\right)\\ =\left(\right(-62\text{mi})\widehat{i}-(65\text{mi}\left)\widehat{j}\right)\end{array}$

Part.(b)

Find the magnitude of the displacement vector,

$\begin{array}{l}\left|\Delta \overrightarrow{W}\right|=\sqrt{{(-62\text{mi})}^2+{(-65\text{mi})}^2}\\ =90\text{mi}\end{array}$

The direction is calculated as follows,

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{-65\text{mi}}{-62\text{mi}}\right)\\ ={46}^{°}\end{array}$

Therefore, the magnitude of the displacement vector is 90 miles and the direction is ${46}^{°}$south of west.