A person walks 25.0^{circ} north of east for 3.18 ,km . How far would she have to walk due north and

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3-2.Motion in Plane

normal

A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

Towards north $2.88 \,km$ and towards east $1.34 \,km$

Towards north $2.11 \,km$ and towards east $2.11 \,km$

Towards north $1.25 \,km$ and towards east $1.93 \,km$

Towards north $1.34 \,km$ and towards east $2.88 \,km$

(KVPY-2016)

Solution

(d)

Displacement of person is

From above figure, distance travelled along north direction is $A B=O B \sin 25^{\circ}=3.18 \times \sin 25^{\circ}=1.34 \,km$ Distance travelled along east direction is $O A=3.18 \times \cos 25^{\circ}=2.88 \,km$

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