A particle moves with deceleration along the circle of radius R so that at any moment of time t is tangential and normal acceleration are equal in magnitude .At the initial moment t=0 the speed of the particle equals $V_{0}$ then

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The speed of the particle as a function of the distance covered will be ${V=V}_{0} e^{-s/R}$

The total acceleration of the particle as function of velocity and distance covered $a= \sqrt{2} \frac{v^{2}}{R}$

The speed of the particle as a function of the distance covered will be ${V=V}_{0} e^{s/R}$

The total acceleration of the particle as function of velocity and distance covered $a= \sqrt{3} \frac{v^{2}}{R}$

answer is A.

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Detailed Solution

$a_{t} = a_{r} - V \frac{d V}{d S} = \frac{V^{2}}{R} \Rightarrow - \frac{d V}{V} = \frac{d S}{R}$

$\int_{v_{0}}^{V} \frac{d v}{v} = \frac{- 1}{R} \int_{0}^{s} d s$

$I n (\frac{V}{V_{0}}) = \frac{- S}{R} \Rightarrow V = V_{0} e^{- s / R}$

$a_{r} = \frac{V^{2}}{R}, a_{t} = V \frac{d V}{d S}$

$a_{t} = V_{0} e^{- S / R} (V_{0} \times e^{- S / R} \times \frac{- 1}{R})$

$\frac{- V_{0}^{2}}{R} e^{- 2 S / R} = \frac{- V^{2}}{R}$

$a_{n e t} = \sqrt{a_{r}^{2} + a_{t}^{2}} = \sqrt{2} \frac{V^{2}}{R}$

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