A girl walks 4 , km towards west,then she wal | vedclass

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Question

(A) Let $O$ be the origin $(0,0)$ representing the initial position of the girl.The girl walks $4 \, km$ towards the west. So,the position of point $A

Vedclass · Accepted Answer

$\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$

Vedclass · Answer

(A) Let $O$ be the origin $(0,0)$ representing the initial position of the girl.The girl walks $4 \, km$ towards the west. So,the position of point $A$ is $(-4, 0)$,which can be represented as $\overrightarrow{OA} = -4 \hat{i}$.From point $A$,she walks $3 \, km$ in a direction $30^{\circ}$ east of north. This means the angle with the positive $y$-axis (North) is $30^{\circ}$ towards the east. The angle with the positive $x$-axis (East) is $90^{\circ} - 30^{\circ} = 60^{\circ}$.The displacement vector $\overrightarrow{AB}$ can be written as:$\overrightarrow{AB} = 3(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j})$$= 3(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = \frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.The total displacement $\overrightarrow{OB}$ is given by the vector sum:$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}$$= -4 \hat{i} + (\frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j})$$= (-4 + \frac{3}{2}) \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$$= \frac{-8+3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$$= \frac{-5}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.Thus,the girl's displacement from her initial point is $\frac{-5}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}$.

$A$ girl walks $4 \, km$ towards west,then she walks $3 \, km$ in a direction $30^{\circ}$ east of north and stops. Determine the girl's displacement from her initial point of departure.

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