Question:

A girl walks 4km4km towards west,then she walk 3km3km in a direction 30°30°east of north and stops.Determine the girls displacement from her initial point to departure.

Updated On: Sep 21, 2023
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Solution and Explanation

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
OA=4i^\overrightarrow{OA}=-4\hat{i}
AB=i^ABcos60°+j^ABsin60°\overrightarrow{AB}=\hat{i}|\overrightarrow{AB}|cos60°+\hat{j}|\overrightarrow{AB}|sin 60°
=i^3×12+j3×32=\hat{i}3\times\frac{1}{2}+j^{3}\times\frac{\sqrt{3}}{2}
=32i^+332j^=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
By the triangle law of vector addition,we have:
OB=OA+AB\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}
=(4i^)+(32i^+333j^)=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{3}\hat{j})
=(4+32)i^+332j^=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
=(8+32)i^+332j^=(-8+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
=52i^+332j^=\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
Hence,the girl's displacement from her initial point of departure is
52i^+332j^\frac{-5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}

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