A particle moves in a circular path such that its speed v varies with distance as $v = α \sqrt{s}$ where $α$ is a positive constant. Find the acceleration of particle after tranversing a distance S?

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$α \sqrt{\frac{1}{2} + \frac{S^{2}}{R^{2}}}$

$α^{2} \sqrt{\frac{1}{2} + \frac{S^{2}}{R^{2}}}$

$α^{2} \sqrt{\frac{1}{4} - \frac{S^{2}}{R^{2}}}$

$α^{2} \sqrt{\frac{1}{4} + \frac{S^{2}}{R^{2}}}$

answer is B.

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Detailed Solution

$\begin{array}{l} v = α \sqrt{s} \Rightarrow a_{t} = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} \Rightarrow a_{t} = v \frac{dv}{ds} \\ = (α \sqrt{s}) [(α) \frac{1}{2} \times \frac{1}{\sqrt{s}}] = \frac{α^{2}}{2} \end{array}$

$a_{c} = \frac{v^{2}}{R} = \frac{α^{2} s}{R}$

$\Rightarrow a_{net} = \sqrt{a_{t}^{2} + a_{c}^{2}} = α^{2} \sqrt{\frac{1}{4} + \frac{s^{2}}{R^{2}}}$

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