As stated in the title, I want to find an expression or a way to calculate the minimum time to go from one point of a path to another when the path is given and acceleration is restricted.
Thus far, I have tried to consider a simplified situation where the 2D path can be represented as a differentiable function $f: [0, l]\to\mathbb{R}$. Here $(0,f(0))$ and $(l, f(l))$ represent the ending points of the 2D path. Also, for now I consider the situation where the initial speed is $v_0=0$.
Let the maximum allowed acceleration be $a_{\max}$.
I have tried to construct a suitable expression to calculate the total time and then minimise it (most likely with Euler-Lagrange) but so far I haven't been able to construct an expression I can manipulate.
We could consider the time to go through an infinitesimal arc length and integrate that: \begin{align*} dt&=\frac{ds}{v(x)}\\ dt&=\frac{\sqrt{1+f'(x)^2}}{v(x)}dx\\ T&=\int_0^l\frac{\sqrt{1+f'(x)^2}}{v(x)}dx. \end{align*} However, minimising this proves to be difficult, or at least I don't seem to get a grasp on it. First of all we'd have to somehow tie the acceleration here: intuitively it should always be at its maximum value so $a(x)=a_{\max}$. Secondly, you'd have to somehow include the limiting factor of $v(x)$ as its maximum value at a specific point is $$v_{\max}=a_{\max}\left|\frac{\left(1+f'(x)^2\right)^{3/2}}{f''(x)}\right|$$ (i.e. considering that the instantaneous axis of rotation comes from the radius of curvature).
Anyways, any idea on how to proceed, or is there a better way to approach this? Of course, when this simplified case is solved, the next step would be to change the initial speed. Then to give a restriction at the end i.e. $v(l)=v_l$. Furthermore, to make this result useful in some way we'd have to generalise this to 2D parametric curves and it would also be interesting to consider non-differentiable curves as well, but this goes way beyond the initial question.
