Question

A girl walks \(4km\) towards west,then she walk \(3km\) in a direction \(30°\)east of north and stops.Determine the girls displacement from her initial point to departure.

Answer 1

Let O and B be the initial and final positions of the girl respectively.
Then,the girl's position can be shown as:

Now,we have
\(\overrightarrow{OA}=-4\hat{i}\)
\(\overrightarrow{AB}=\hat{i}|\overrightarrow{AB}|cos60°+\hat{j}|\overrightarrow{AB}|sin 60°\)
\(=\hat{i}3\times\frac{1}{2}+j^{3}\times\frac{\sqrt{3}}{2}\)
\(=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
By the triangle law of vector addition,we have:
\(\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\)
\(=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{3}\hat{j})\)
\(=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
\(=(-8+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
\(=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)
Hence,the girl's displacement from her initial point of departure is
\(\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}\)