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(D) Given that the magnitude of radial acceleration $(a_r = v^2/R)$ and tangential acceleration $(a_t = dv/dt)$ are equal:$dv/dt = v^2/R$Rearranging t

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$\frac{R}{V_0} (1 - e^{-2 \pi})$

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(D) Given that the magnitude of radial acceleration $(a_r = v^2/R)$ and tangential acceleration $(a_t = dv/dt)$ are equal:$dv/dt = v^2/R$Rearranging the terms for integration:$\int_{V_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R}$$[-1/v]_{V_0}^{v} = t/R \Rightarrow 1/V_0 - 1/v = t/R \Rightarrow t = R(1/V_0 - 1/v) \dots(1)$Also,$a_t = v(dv/ds) = v^2/R$,which implies $dv/v = ds/R$.Integrating for one full revolution ($s = 0$ to $s = 2\pi R$):$\int_{V_0}^{v} \frac{dv}{v} = \int_{0}^{2\pi R} \frac{ds}{R}$$\ln(v/V_0) = 2\pi \Rightarrow v = V_0 e^{2\pi} \dots(2)$Substituting $v$ from $(2)$ into $(1)$:$t = R/V_0 (1 - 1/e^{2\pi}) = R/V_0 (1 - e^{-2\pi})$.

$A$ particle is moving along a circular path of radius $R$ in such a way that at any instant the magnitude of radial acceleration and tangential acceleration are equal. If at $t = 0$ the velocity of the particle is $V_0$,the time period of the first revolution of the particle is:

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