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» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>kmt <br/>ktmr<br/>km/t<br/>kmr/t </p>Solution :Here radial centripetal acceleration`a = (v^(2))/r`<br/> ` = kt^(2) ` is given<br/>Now , <a href="https://interviewquestions.tuteehub.com/tag/taking-1238585" style="font-weight:bold;" target="_blank" title="Click to know more about TAKING">TAKING</a> derivation w.r.ttime , t . <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a> (dv)/(dt)) 1/r = 2 kt` <br/> Multiplyingon both sides by .m.<br/> ` :.<a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a> (dv)/(dt) = mktr` <br/> ` :. <a href="https://interviewquestions.tuteehub.com/tag/fv-458643" style="font-weight:bold;" target="_blank" title="Click to know more about FV">FV</a> = ktmr"" ( :. F = ma = m (dv)/(dt))` <br/> ` :. P = ktmr"" (:. P = Fv)`</body></html>