An object moves along the circle with normal acceleration proportional to `t^(alpha)`, where t is the time and `alpha` is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to 

Options

A. `t^(alpha -1)`
B. `t^(alpha//2)`
C. `t^((1+alpha)/(2))`
D. `t^(2 alpha)`

Answer

A

Explanation

Given, Normal acceleration `= t^(alpha)`
We know,
Normal acceleration `= a_(n) = (v^2 )/(R ) = t^(alpha)`
`therefore v^2 = (t^(alpha))R`
`K.E. = (1)/(2) mv^2 = (m)/(2) (t^(alpha) R)`
Power `= (d)/(dt) (K.E.) = (d)/(dt) ( (m)/(2) t^(alpha) R)`
`P = (mR)/(2) ( alpha t^(alpha-1))`
`therefore P alpha t^(alpha - 1)`