A person walks 25.0^{circ} north of east for | vedclass

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(D) The person walks a displacement vector of magnitude $r = 3.18 \,km$ at an angle $	heta = 25.0^{\circ}$ north of east.To find the equivalent path

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The person walks a displacement vector of magnitude $r = 3.18 \,km$ at an angle $	heta = 25.0^{\circ}$ north of east.To find the equivalent path by walking due north and then due east,we resolve the displacement vector into its rectangular components.The component along the east direction (x-axis) is given by $x = r \cos 	heta = 3.18 	imes \cos 25.0^{\circ} \approx 3.18 	imes 0.9063 = 2.88 \,km$.The component along the north direction (y-axis) is given by $y = r \sin 	heta = 3.18 	imes \sin 25.0^{\circ} \approx 3.18 	imes 0.4226 = 1.34 \,km$.Therefore,the person must walk $1.34 \,km$ due north and $2.88 \,km$ due east to reach the same location.Thus,the correct option is $(d)$.

$A$ 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?

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