Two particles A and B move with constant velocity ${\bar{V}}_{1} a n d {\bar{V}}_{2}$ along two mutually perpendicular straight lines towards intersection point O as shown in figure. At moment $t = 0$ particles were located at distance $I_{1}$ and $I_{2}$ respectively from O. The time taken to have minimum distance between the particles is

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$\frac{(I_{1} V_{1} + I_{2} V_{2})}{(V_{1} + V_{2})}$

$\frac{I_{1} V_{2} + I_{2} V_{1}}{V_{1}^{2} + V_{2}^{2}}$

$\frac{(I_{1} V_{2} + I_{2} V_{1})}{(V_{1} + V_{2})}$

$\frac{I_{1} V_{1} + I_{2} V_{2}}{V_{1}^{2} + V_{2}^{2}}$

answer is A.

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Detailed Solution

$S^{2} = {(l_{1} - V_{1} t)}^{2} + {(l_{2} - V_{2} t)}^{2} S \to \min t = \frac{l_{1} v_{1} + l_{2} v_{2}}{V_{1}^{2} + V_{2}^{2}} \frac{d S}{d t} = 0 S_{\min} = \frac{l_{1} v_{2} - l_{2} v_{1}}{\sqrt{V_{1}^{2} + V_{2}^{2}}}$

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