A particle is moving on a circle of radius `R` such that at every instant the tangential and radial accelerations are equal in magnitude. If the velocity of the particle be `v_(0)` at `t=0`, the time for the completion of the half of the first revolution will be

Options

A. `R//v_(0)`
B. `(R//v_(0))(1-e^(-x))`
C. `(R//v_(0))e^(-x)`
D. `(R//v_(0))(1-e^(-2x))`

Explanation

`omega^(2)R=R(domega)/(dt)`, `omega=(d theta)/(dt)=(omega_(0))/(1-omega_(0)t)`
`rArrT=(R )/(v_(0))(1-e^(-x))`