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Two particles A and B are moving in xy-plane. Particle A is moving with a velocity of $2 \sqrt{2} \hat{i}$ m/s and particle B is moving with a velocity of magnitude 3 m/s along a straight line which makes an angle of 45⁰ with positive x-axis. Then velocity of B relative to A is

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$\sqrt{5}$ m/s

4 $\sqrt{3}$ m/s

3 m/s

$\sqrt{3}$ $2\sqrt{3}$ m/s

answer is D.

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Detailed Solution

${\vec{v}}_{B / A} = {\vec{v}}_{B} - {\vec{v}}_{A} = 3 (\cos 45^{0} \hat{i} + \sin 45^{0} \hat{j}) - 2 \sqrt{2} \hat{i} m / s \Rightarrow |{\vec{v B}}_{/ A}| = \sqrt{5} m / s$

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