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 $\phi$.
 A: The major and minor axes cut an ellipse into four congruent segments. It will suffice to find the optimal parametrization one of these segments, for the parametrization can be extended by symmetry. I will assume the ellipse has equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
$$
with $a\geq b$, and describe a parametrization from $(a,0)$ to $(0,b)$.
Because the curvature of the ellipse decreases as we move from $(a,0)$ to $(0,b)$, the optimal parametrization has a concrete description: start out as fast as possible, and accelerate as much as possible. More precisely, start out at a velocity such that the entire unit of acceleration must be directed inward to stay on the track, and at each point, choose a tangential component of acceleration to be as large as possible (given the bound on total acceleration).  I will give an argument this parametrization is optimal at the end of this post.
I will use a parameter $0\leq w\leq \pi/2$, with a point on the ellipse written as $(a\cos w,b\sin w)$. Write $t$ for time (starting at $t=0$), $v$ for speed, and $K$ for the curvature of the ellipse. I will also write $s$ for arc length along the ellipse. Each of these quantities is a function of $w$.
First some auxiliary calculations. It is convenient to introduce the function $C(w):=\sqrt{a^2\sin^2 w+b^2\cos^2 w}$. A formula for $K$ is given by
$
K(w) = {ab}/{C(w)^3}
$. We have $ds/dw=C(w)$. And of course $ds/dt=v$. The inward acceleration needed to stay on the track is $K v^2$, so our tangential acceleration will be
$$
\frac{dv}{dt}=\sqrt{1-K^2v^4}.
$$
We can combine these equation to get a differential equation for $v$:
\begin{align*}
\frac{dv}{dw}&=\frac{dv}{dt}\left(\frac{ds}{dt}\right)^{-1}\frac{ds}{dw}\\
&=\frac{\sqrt{1-\frac{a^2b^2}{C(w)^6}v^4}}{v}C(w).
\end{align*}
This is a non-linear first order ordinary differential equation. The initial condition will have $K(0)v(0)^2=1$, so $v(0)=b/\sqrt{a}$. I have no reason to believe there is a closed form solution. But if a solution $v(w)$ is found numerically, then the period $T$ can be computed as
$$
T=4\int_0^{\pi/2}\frac{ds}{v}=4\int_0^{\pi/2}\frac{C(w)}{v(w)}dw.
$$
Proof of optimality: Write $s_1$ for our speed function, as suppose $s_2$ is the speed function for some other parametrization. By the way we chose acceleration, we have $s_1'>s_2'$ at every point where $s_2>s_1$. So in particular, at the point where $s_2-s_1$ is maximized, we have $s_1'=s_2'$, and so $s_2\leq s_1$ at that point. Thus there is no parametrization that is faster at any point on the ellipse than the one we describe, so ours gives the minimal period.
A: 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.
