A woman walksin the direction$\mathbf{30}\mathbf{°}$east of north, then$\mathbf{175}\mathbf{}\mathbf{m}$directly east. Find (a) the magnitude and (b) the angle of her final displacement from the starting point. (c) Find the distance she walks. (d) Which is greater, that distance or the magnitude of her displacement?

A woman walks 250 m in the direction $30°$east of north, and then 175 m east.

Using a vector diagram, we can find the resultant of vectors and the angle between them. It is possible to find the total displacement and distance traveled by the person or object using simple vector addition and simple mathematical operations. We need to resolve the given vectors along the x and y-axis and add them vectorially to find the total displacement.

Formulae:

The components of the vector $\overrightarrow{A}$along x and Y axis are,

$$\begin{gathered} A_x=\overrightarrow{A}\cos \theta \\ {A}_y=\overrightarrow{A}\sin \theta \end{gathered}$$

(where, $\theta$is the angle made by vector $\overrightarrow{A}$with x-axis.)

Let’s consider that vector$\overrightarrow{A}$ is of magnitude 250 m and vector$\overrightarrow{B}$is of magnitude 175 m and vector$\overrightarrow{R}$is the resultant vector of them.

The components of vector$\overrightarrow{A}$ along x and y-axis are respectively,

$$\begin{gathered} \overrightarrow{\mathrm{A}}\sin 30°=250(0.5) \\ =125\mathrm{m} \\ \overrightarrow{\mathrm{A}}\cos 30°=250(0.866) \\ =216.5\mathrm{m} \end{gathered}$$

So, the x component of the vector $\overrightarrow{R}$is,

$$\begin{gathered} {\overrightarrow{R}}_x=\overrightarrow{A}\sin (30°)+\overrightarrow{B} \\ =125+175 \\ =300 \end{gathered}$$

Similarly, the y component of vector $\overrightarrow{R}$is,

$$\begin{gathered} \overrightarrow{R}=\overrightarrow{A}\cos (30°) \\ =216.5\mathrm{m} \end{gathered}$$

Therefore,the resultant displacement of women is,

$$\begin{gathered} \overrightarrow{R}=\sqrt{{R_x}^2+{R_y}^2} \\ =\sqrt{{300}^2+216.5^2} \\ =370\mathrm{m} \end{gathered}$$

Therefore, the Magnitude of the displacement is 370 m.

From the vector diagram,

$$\begin{gathered} \tan \alpha =\frac{{\overrightarrow{R}}_y}{{\overrightarrow{R}}_x} \\ \alpha ={\tan }^{-1}\left(\frac{{\overrightarrow{R}}_y}{{\overrightarrow{R}}_x}\right) \\ ={\tan }^{-1}\left(\frac{216.5}{300}\right) \\ =36° \end{gathered}$$

Therefore, the angle made by the displacement of woman with positive x-axis is $36°$.

It means, the angle of her displacement from the starting point is $36°$north of due east.

The distance walked by the woman is addition of the magnitude of $\overrightarrow{A}\mathrm{and}\overrightarrow{B}$. Therefore, the total distance is,

$$\begin{gathered} \left|\overrightarrow{A}\right|+\left|\overrightarrow{B}\right|=250+175 \\ =425\mathrm{m} \end{gathered}$$

Therefore, the total distance walked by the woman is 425 m.

As, 425 > 370 , the distance walked by the woman is greater than the net displacement.