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The distance between two moving particles at any time is $a = 32 m .$ If v be their relative velocity and $V_{1} = 4 m / s a n d v_{2} = 8 m / s$ be the component of v along and perpendicular to a The time when they are closest to each other is (in seconds)(nearest integer)

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answer is 2.

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Detailed Solution

$t = \frac{(a) \sin α}{V_{r}} = \frac{(a)}{V_{r}} (\frac{V_{1}}{V_{r}}) = \frac{a . V_{1}}{{V_{r}}^{2}} = \frac{a . V_{1}}{{[\sqrt{V_{1}^{2} + V_{2}^{2}}]}^{2}} = \frac{(32)}{{[\sqrt{4^{2} + 8^{2}}]}^{2}} = \frac{(32) (4)}{80} = 1.6$

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