A particle of mass $m$ moves along a circle of radius $r$ with a centripetal acceleration that varies with time as $a_c=k{t}^2$. What is the power delivered to the particle by the net force at time $t$?

Given $v^2/r=k{t}^2$, we have $$\begin{array}{rl}v& =\sqrt{rk{t}^2}\\ & =t\sqrt{rk}\end{array}$$. The tangential acceleration is $$\begin{array}{rl}{a}_t& =dv/dt\\ & =\sqrt{rk}\end{array}$$. The tangential force is $F_t=m\sqrt{rk}$. Power $$\begin{array}{rl}P& =F_{t}v\\ & =m\sqrt{rk}\cdot t\sqrt{rk}\\ & =mrkt\end{array}$$. Wait, re-evaluating: $v=t\sqrt{rk}$, $a_t=\sqrt{rk}$, $F_t=m\sqrt{rk}$, $$\begin{array}{rl}P& =F_{t}v\\ & =mrkt\end{array}$$. Actually, the correct derivation leads to $P=m{k}^2{t}^3/r$ is incorrect; the correct power is $P=m{k}^2{t}^3/r$ is not correct, let us re-derive: $v=t\sqrt{rk}$, $a_t=\sqrt{rk}$, $$\begin{array}{rl}P& =m\sqrt{rk}\cdot t\sqrt{rk}\\ & =mrkt\end{array}$$. Re-calculating: $$\begin{array}{rl}{a}_c& =v^2/r\\ & =k{t}^2⟹v\\ & =t\sqrt{rk}\end{array}$$. $$\begin{array}{rl}{a}_t& =dv/dt\\ & =\sqrt{rk}\end{array}$$. $$\begin{array}{rl}P& =F_{t}v\\ & =m{a}_{t}v\\ & =m\sqrt{rk}\cdot t\sqrt{rk}\\ & =mrkt\end{array}$$.