A particle is moving on a circular path with speed given by v= ` alphat`, where `alpha` is a constant. Find the acceleration of the particle at th instant when it has covered the ` n^("th")` fraction of the . Circle.

Explanation

The speed of particle is time dependent , the motion is non-uniform cicular motoin. Let the speed of particle be ` v_(0)` at the instant it covers the ` n^(th)` fraction of circle.
Tangential Acceleration
The tangential acceleration is the rate at which the speed changes with time.
` a_(r) = (dv)/(dt) = d/(dt) ( alphat) = alpha m//s^(2)`

The particle has a constant tangential acceleration ` alpha m//s^(2)` . The initial speed of particle is zero and becomes ` v_(0)` after travelling a distance s along the circle . Then
` v_(0)^(2) =0^(2) + 2alphas` (all scalars)
` Rightarrow v_(0) = sqrt(2alpha(2pir)n) ( therefore ` s is the ` n^("th")` fraction of circle)
Centripetal Acceleration
When the speed is ` V_(0) `
` a_(R) = (v_(0)^(2))/r = ( 4 pi ralpha n)/r = 4 pi alpha n m//s^(2)`
The acceleration of particle is
` a= sqrt(a_(T)^(2) +a_(R)^(2)) = sqrt(alpha^(2) + ( 4pi alphan)^(2)) = alphasqrt(1 + 19pi^(2)n^(2)) m//^(2)`