Entry ($0 \le s \le L$): $k(s)=s/(LR)$. Exit ($L \le s \le 2L$): $k(s)=(2L-s)/(LR)$. Peak curvature at $s=L$: $k=1/R$.

$$\sqrt{V^2 k^2 + V^2 \left(\frac{dV}{ds}\right)^2} = V\sqrt{k^2 + \left(\frac{dV}{ds}\right)^2} = f\,g$$

Equivalent ODE on $[0,L]$: $\displaystyle\left(\frac{dV}{ds}\right)^2=\left(\frac{f g}{V}\right)^2-(V k)^2$. Terminal $V(L)=\sqrt{f\,g\,R}$; with $\sigma=L-s$, $\displaystyle\frac{dV}{d\sigma}=\sqrt{\cdots}$ — stable IVP from $\sigma=0$ (no shooting).

$\rho=1.25\ \mathrm{kg/m^3}$, $S C_z=3.2\ \mathrm{m^2}$, mass $m$. Apex speed:

$$V_R = \sqrt{\frac{m\,f\,g}{\,m/R - \tfrac{1}{2} f \rho S C_z\,}}$$ $$m\,V\sqrt{k^2 + \left(\frac{dV}{ds}\right)^2} = f\,m\,g + \frac{f}{2}\rho S C_z\, V^2$$

On $[0,L]$ solve for $V(s)$; on $[L,2L]$ use symmetry $V(s)=V(2L-s)$ for both cars.

$$a_x = V\frac{dV}{ds},\qquad a_y = V^2 k,\qquad j_y = \frac{d a_y}{dt} = V\left(2V\frac{dV}{ds}\,k + V^2\frac{dk}{ds}\right)$$

Motion: $ds/dt = V(s)$.