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^(2alpha)`

Answer

A

Explanation

Given, `a_(c)prop t^(alpha)`
or `" "a_(c)=kt^(alpha)`
Now, `" "a_(c)=(v_(t)^(2))/(r)`
(where, `v_(t)` = tangential component of velocity)
`implies" "v_(t)=sqrt("kr.t"^(alpha))`
and `a_(t)=(dv_(t))/(dt)=sqrt(("kr")).(alpha)/(2).t^((alpha)/(2)-1)`
Now, power developed is
`P=Fv=ma_(1)v_(1)`
`=m sqrt("kr").(alpha)/(2).t^((alpha)/(2)-1)`
`implies" "P prop t^(alpha-1)`