If a particle moves such that the displacement is proportional to the square of the velocity acquired , then its acceleration is

Question

Dr. Abubacker · Accepted Answer

(d)
Since , $\ \ \ \ \ \ \ s \propto v^{2}\  \Rightarrow v = k\sqrt{s}$
Now, acceleration $\frac{dv}{dt} = k\frac{1}{2\sqrt{s}}\ \frac{ds}{dt}$
=$k\frac{1}{2\sqrt{s}} \bullet k\sqrt{s}$ $\left\lbrack \because v = \frac{ds}{dt} = k\sqrt{s} \right\rbrack$
$= \frac{k^{2}}{2} =$constant

Q. If a particle moves such that the displacement is proportional to the square of the velocity acquired , then its acceleration is

Explanation

Reviewed by Dr. Abubacker

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