compute the acceleration of four dogs pursuit problem Four dogs pursuit problem is a classical puzzle.

At each of the four corners of a square room (the length of a wall
being L metres) is a dog. At a specified moment, each dog sets off at
the same speed chasing its neighbour. At any time during the ensuing
pursuit, each dog is always travelling directly towards the dog it is
chasing. How far does each dog travel before they all meet?

See: https://activityworkshop.net/puzzlesgames/fourdogs/solution.html
or
https://www.cut-the-knot.org/Curriculum/Geometry/FourTurtles.shtml
It's not to hard to compute the original problem, by computing the speed towards the center of the square. But what if I want to compute the acceleration (i.e.) the change of the dog's velocity vector at any given time?
I followed this solution to the original problem, but I am not sure how I can compute the acceleration out of it.
 A: The dogs are running at some constant speed $v =\dfrac{ds}{dt}$, where $s$ is the arclength along their path from the position at $t = 0$. For polar coordinates $$ds^2 = dr^2 + r^2d\theta^2$$
That is,
$$\left(\frac{ds}{dt}\right)^2 = \left(\frac{dr}{dt}\right)^2+ r^2\left(\frac{d\theta}{dt}\right)^2$$
Or just $\dot s^2 = \dot r^2 + r^2\dot \theta^2$
Since $\frac {dr}{d\theta} = -r$, we have $\dot r = -r\dot \theta$ by the chain rule. So
$$v^2 = 2r^2\dot \theta^2$$
$v$ is positive, as is $r$, and we can assume the dogs choose to run towards the next dog in the positive $\theta$ direction, so we can take $\dot\theta$ as positive as well. So $$\frac v{\sqrt 2} = r\dot\theta$$
Differentiating with respect to $t$ again:
$$0 = \dot r\dot\theta + r\ddot\theta = -r\dot\theta^2 + r\ddot\theta$$
$$\ddot\theta = \dot\theta^2$$
which has solution $$\dot\theta = -\frac 1{t+C_1}$$ and so $$\theta =C_2 - \log(t+C_1)$$
You can now plug this into the solution of the equation $\frac{dr}{d\theta} = -r$ to get a very surprising relationship between $r$ and $t$. With these two relationships you can calculate velocity (not just speed) and acceleration.
