A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
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:
$\overrightarrow{\mathrm{OA}}=-4 \hat{i}$
$\overrightarrow{\mathrm{AB}}=\hat{i}|\overrightarrow{\mathrm{AB}}| \cos 60^{\circ}+\hat{j}|\overrightarrow{\mathrm{AB}}| \sin 60^{\circ}$
$=\hat{i} 3 \times \frac{1}{2}+\hat{j} 3 \times \frac{\sqrt{3}}{2}$
$=\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$
By the triangle law of vector addition, we have:
$\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}$
$=(-4 \hat{i})+\left(\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}\right)$
$=\left(-4+\frac{3}{2}\right) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$
$=\left(\frac{-8+3}{2}\right) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$
$=\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$
Hence, the girl’s displacement from her initial point of departure is
$\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$
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