 Solution

Ignoring the small radiative back-reaction, conservation of energy gives $\dot{x}=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot{x}=-f^{\prime }(x)/m$. Integrating the Larmor formula with $dt=dx/\dot{x}$ yields, for $E>V_0$,If $E<V_0$, the particle turns at $x_{\ast }$ defined by $f(x_{\ast })=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_{\ast }$.