Two particles A and B move with velocities $v_{1}$ and $v_{2}$ respectively along the x & y axis. The initial separation between them is ‘ $d$ ’ as shown in the fig. Find the least distance between them during their motion.

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$\frac{d . v_{1}}{\sqrt{v_{1}^{2} + v_{2}^{2}}}$

$\frac{d . v_{2}}{\sqrt{v_{1}^{2} + v_{2}^{2}}}$

$\frac{d . v_{2}^{2}}{v_{1}^{2} + v_{2}^{2}}$

$\frac{d . v_{1}^{2}}{v_{1}^{2} + v_{2}^{2}}$

answer is C.

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Detailed Solution

$⊥$ distance = $d \sin θ = \frac{d V_{1}}{\sqrt{{V_{1}}^{2} + {V_{2}}^{2}}}$

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