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A particle moves in a circle of radius R in such a way that its tangential deceleration is numerically equal to its radial acceleration. If the initial velocity of the particle is $ν_{0}$ , variation of its velocity with time t as $ν =$

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$\frac{ν_{0}}{1 + \frac{ν_{0} t}{R}}$

$\frac{ν_{0} t}{R + ν_{0}}$

$\frac{ν_{0}}{1 - \frac{ν_{0} t}{R}}$

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

answer is A.

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Detailed Solution

Given $\frac{- d ν}{d t} = \frac{ν^{2}}{R} \Rightarrow \int_{ν_{0}}^{ν} \frac{d ν}{ν^{2}} = - \int_{0}^{t} \frac{d t}{R} \Rightarrow ν = \frac{ν_{0}}{1 + \frac{ν_{0} t}{R}}$

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