How fast can one move around an ellipse with bounded acceleration? Given a smooth closed planar curve $\Gamma$, I'm looking for its periodic parametrization $\phi : \mathbb{R}\to\Gamma$ such that 


*

*the second derivative $\phi''$ is bounded by $1$ in the norm: $|\phi''|\le 1$

*the period $T$ of the parameterization is as small as possible.


The "real world" motivation is auto racing: set the best lap time $T$ given that the car's acceleration is limited. 
Question: what is the smallest $T$ for an ellipse with semiaxes $a,b$? (Auto racing tracks often look like an ellipse, although they are not really.)
Special cases
The special case $a=b$, a circle, is easy. Assume it centered at the origin. Then $|\phi|^2\equiv a^2$. Differentiate twice: $|\phi'|^2+\phi\cdot \phi'' =0$. This yields $|\phi'|\le \sqrt{|\phi||\phi''|} \le \sqrt{a}$, hence $$T\ge 2\pi a/\sqrt{a} = 2\pi\sqrt{a}$$ 
This lower bound is attained by $\phi(t) = a(\cos (t/\sqrt{a}), \sin (t/\sqrt{a}))$.
The degenerate ellipse, $b=0$, is also easy to handle: the car has to stop at the endpoints of the interval $[-a,a]$, and then have full acceleration up to the midpoint $0$. This yields $T=4\sqrt{2a}$, because one loop consists of $4$ segments of constant acceleration or deceleration. 
In general, the smallest period $T(a,b)$ scales like $T(\lambda a,\lambda b)=\sqrt{\lambda}T(a,b)$ because the function $\lambda \phi(t/\sqrt{\lambda})$ has the same top acceleration as $\phi$.

I see the problem as follows. For an ellipse with semi-major/minor axes $(A,B)$, the trajectory is parameterized in the complex plane as
$$z=A\cos(t/\sqrt{a})+iB\sin(t/\sqrt{a})$$
Here, $a$ must have the dimensions [$\text{t}^2$], and, of course $(A,B)$ are [L].
The velocity and acceleration are given by
$$
v=\dot z=\frac{1}{\sqrt{a}}\left[-A\sin(t/\sqrt{a})+B\cos(t/\sqrt{a})\right],\quad \text{[L/t]}\\
\dot v=\ddot z=\frac{1}{a}\left[-A\cos(t/\sqrt{a})-B\sin(t/\sqrt{a})\right]=-\frac{z}{a},\quad \text{[L/t$^2$]}\\
$$
And finally, the instantaneous arc length, if needed, is given by
$$s=\int_0^t |\dot z|~dt=\frac{B}{\sqrt{a}}\int_0^t \sqrt{C\sin^2(t/\sqrt{a})+1}~dt,\quad C=\left(\frac{A^2-B^2}{B^2}\right)\\$$
The integral is an incomplete elliptic function of the second kind.
This should be sufficient to complete your task.
