You are driving a car with maximal acceleration (and deceleration) A on a road that’s been blocked off in both directions (or, if you prefer, on the landing strip of an aircraft carrier). Let L be the length of the road available to you.

What is the maximal speed you can reach?

Besides A and L, the answer also depends on your mood: do you want to live, or are you willing to go out in a blaze of glory? In the latter case the answer is obvious: position the car at one end of the interval, and put the pedal to the metal. The car will cover the distance L within the time , reaching the speed at the end. In the former scenario one has to switch to the brake pedal midway through the distance, so the maximal speed will be attained at half the length, .

Rephrased in mathematical terms: if is a twice differentiable function and for , then if is defined on a half-infinite interval, and if the domain of is the entire line. To connect the notation, just put and in the previous paragraph… and I guess some proof other than “this is obvious” is called for, but it’s not hard to find one: this is problem 5.15 in Rudin’s *Principles of Mathematical Analysis*.

Perhaps more interesting is to study the problem in higher dimensions: one could be driving in a parking lot of some shape, etc. Let’s normalize the maximal acceleration as 1, keeping in mind it’s a vector. Given a set E, let S(E) be the **square** of maximal speed attainable by a unit-acceleration vehicle which stays in E indefinitely. Also let U(E) be the square of maximal speed one can attain while crashing out of bounds after the record is set. Squaring makes these quantities scale linearly with the size of the set. Both are monotone with respect to set inclusion. And we know what they are for an interval of length L: namely, and , so that gives some lower bounds for sets that contain a line interval.

When E is a *circle* of radius 1, the best we can do is to drive along it with constant speed 1; then the centripetal acceleration is also 1. Any higher speed will exceed the allowable acceleration in the normal direction, never mind the tangential one. So, for a circle both S and U are equal to its radius.

On the other hand, if E is a *disk* of radius R, then driving along its diameter is better: it gives and .

Some questions:

- If E is a convex set of diameter D, is it true that and ?
- Is it true that in general?
- How to express S and U for a smooth closed curve in terms of its curvature? They are not necessarily equal (like they are for a circle): consider thin ellipses converging to a line segment, for which S and U approach the corresponding values for that segment.

The answer to Question 1 is yes. Consider the orthogonal projection of E, and of a trajectory it contains, onto some line L. This does not increase the diameter or the acceleration; thus, the one-dimensional result implies that the projection of velocity vector onto L does not exceed (or for the crashing-out version). Since L was arbitrary, it follows that and . These upper bounds hold for general sets, not only convex ones. But when E is convex, we get matching lower bounds by considering the longest segment contained in E.

I don’t have answers to questions 2 and 3.

Relevant math StackExchange post: “How fast can one move around an ellipse with bounded acceleration?” https://math.stackexchange.com/q/2384849