A particle moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in moduli. At the initial moment `t=0` the speed of the particle equals `v_(0)`, then th speed of the particle as a function of the distance covered S will be

Options

A. `v=v_(0) e^(-S//R)`
B. `v=v_(0)e^(S//R)`
C. `v=v_(0) e^(-R//S)`
D. `v=v_(0) e^(R//S)`

Answer

A

Explanation

Given `(dv)/(dt)=v^(2)/rrArr(dv)/(ds)=v^(2)/r, -underset(v_(0))overset(v)(int)1/vdv=underset(o)overset(s)(int)(ds)/r`
`rArr ln[v_(0)/v]=S/rrArr v_(0)/v=e^(S//r)`
`rArr v_(0)=ve^(S//r)rArrv=v_(0)e^(-S//r)`