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Question

(B) The average velocity is defined as the total displacement divided by the total time taken.First,calculate the displacement in the east direction:$

Vedclass · Accepted Answer

$5 \, m/s$ due $N-37^\circ-E$

Vedclass · Answer

(B) The average velocity is defined as the total displacement divided by the total time taken.First,calculate the displacement in the east direction:$\vec{d}_1 = 15 \, m/s 	imes 2 \, s 	imes \hat{i} = 30 \hat{i} \, m$Next,calculate the displacement in the north direction:$\vec{d}_2 = 5 \, m/s 	imes 8 \, s 	imes \hat{j} = 40 \hat{j} \, m$Total displacement $\vec{d} = \vec{d}_1 + \vec{d}_2 = 30 \hat{i} + 40 \hat{j} \, m$Total time $t = 2 \, s + 8 \, s = 10 \, s$Average velocity $\vec{v}_{av} = \frac{\vec{d}}{t} = \frac{30 \hat{i} + 40 \hat{j}}{10} = 3 \hat{i} + 4 \hat{j} \, m/s$The magnitude of the average velocity is $|\vec{v}_{av}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \, m/s$.The direction is given by $	an 	heta = \frac{v_y}{v_x} = \frac{4}{3}$,which implies $	heta = 53^\circ$ north of east,or $37^\circ$ east of north.Thus,the average velocity is $5 \, m/s$ due $N-37^\circ-E$.

$A$ particle moves in the east direction with $15 \, m/s$ for $2 \, s$,then moves northward with $5 \, m/s$ for $8 \, s$. The average velocity of the particle is:

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