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A particle moving on a straight line so that its distance S from a fixed point at any time ‘t’ is proportion to t n . If V be the velocity and ‘a’ the acceleration at any time, then $\frac{n a s}{n - 1} =$ =

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$v^{2}$

$V^{3}$

answer is B.

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Detailed Solution

$s α t^{n} \Rightarrow s = {kt}^{n} v = \frac{ds}{dt} = {nkt}^{n - 1} a = \frac{dv}{dt} = n (n - 1) {kt}^{n - 2} \frac{nas}{n - 1} = \frac{n . n (n - 1) {kt}^{n - 2} . {kt}^{n}}{n - 1} = {(n k t^{n - 1})}^{2} = v^{2}$

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