A particle moves in a circle in such a way that, its tangential deceleration is numerically equal to its radial acceleration. If the initial velocity of the particle is $V_{0}$ find the variation of its velocity with time

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$\frac{V_{0} t}{1 + \frac{V_{0} t}{R}}$

$\frac{V_{0}}{1 + \frac{V_{0} t}{R}}$

$\frac{V_{0}}{t + \frac{V_{0} t}{R}}$

answer is A.

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Detailed Solution

$- \frac{dV}{dt} = \frac{V^{2}}{R}; \int_{V_{0}}^{V} \frac{dV}{V^{2}} = - \int_{t = 0}^{t} \frac{dt}{R} \Rightarrow V = \frac{V_{0}}{1 + \frac{V_{0}}{R} t}$

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