A particle of mass m moves on a circularpath ofradius r . Its centripetal acceleration is kt^(2), where k isa constant andt is time . Express power as function of t .

A particle of mass m moves on a circularpath ofradius r . Its centripe

1.	A particle of mass m moves on a circularpath ofradius r . Its centripetal acceleration is kt^(2), where k isa constant andt is time . Express power as function of t .
Answer» <P>kmt ktmr km/t kmr/t Solution :Here radial centripetal acceleration`a = (v^(2))/r` ` = kt^(2) ` is given Now , TAKING derivation w.r.ttime , t . `(2V (dv)/(dt)) 1/r = 2 kt` Multiplyingon both sides by .m. ` :.MV (dv)/(dt) = mktr` ` :. FV = ktmr"" ( :. F = ma = m (dv)/(dt))` ` :. P = ktmr"" (:. P = Fv)`