A particle (A) moves due north at 3,km/h and | vedclass

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(A) Given:Velocity of particle $A$,$\vec{v}_A = 3\hat{j}\,km/h$Velocity of particle $B$,$\vec{v}_B = -4\hat{i}\,km/h$The relative velocity of $A$ with

Vedclass · Accepted Answer

$5\,km/h, 37^{\circ}$ north of east

Vedclass · Answer

(A) Given:Velocity of particle $A$,$\vec{v}_A = 3\hat{j}\,km/h$Velocity of particle $B$,$\vec{v}_B = -4\hat{i}\,km/h$The relative velocity of $A$ with respect to $B$ is given by $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.$\vec{v}_{AB} = 3\hat{j} - (-4\hat{i}) = 4\hat{i} + 3\hat{j}\,km/h$.The magnitude of relative velocity is $|\vec{v}_{AB}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,km/h$.To find the direction,let $\alpha$ be the angle with the east direction (positive $x$-axis).$	an \alpha = \frac{v_{AB,y}}{v_{AB,x}} = \frac{3}{4}$.Since $	an 37^{\circ} = 3/4$,we have $\alpha = 37^{\circ}$.Thus,the direction is $37^{\circ}$ north of east.Therefore,the correct option is $A$.

$A$ particle $(A)$ moves due north at $3\,km/h$ and another particle $(B)$ moves due west at $4\,km/h$. The relative velocity of $A$ with respect to $B$ is $(\tan 37^{\circ} = 3/4)$.

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