Because the particles are situated on an n-sided regular polygon at $t=0$ they admit a radial symmetry at an angle of $\alpha = \frac{2\pi}{n}$. The velocity vectors preserve this symmetry, so the particles will stay on an n-sided regular polygon.
Therefore it suffices to consider one point $(x, y)$, or in the complex plane $z=x+yi$, if we take $0$ as the center of the polygon, the next point is at $e^{-\alpha i} z$. So the differential equation to figure out the path of $z$ is $$\frac{dz}{dt} = z - e^{-\alpha i} z = (1 - e^{-\alpha i}) z.$$
We won't care about the velocity of the point for now, because this differential equation is easy to solve. The solutions are $$z(t) = C e^{(1-e^{-\alpha i})t}$$ for some $C$ in $\mathbb{C}$ which is the equation for a spiral. We can rewrite this by noticing $e^{-\alpha i} = \cos(\alpha) - i\sin(\alpha)$ so that
$$z(t) = C e^{(1-\cos(\alpha))t}(e^{i\sin(\alpha)t}).$$
In polar coordinates $$r(\theta) = R e^{\frac{1-\cos(\alpha)}{\sin(\alpha)}\theta}.$$ Let's introduce the variable $\beta = \frac{1-\cos(\alpha)}{\sin(\alpha)} = \tan(\frac{\alpha}{2})$, then we see that the trajectory of the point is a classic logarithmic spiral $r(\theta) = R e^{\beta \theta}$.
The arc length parametrization of this function is known (check Wikipedia for a full explanation). This also explains the way to calculate the arc length to the origin, it requires the following calculation.
We know that $\beta = \tan(\frac{\alpha}{2}) = \tan(\frac{\pi}{n})$ so that the formula for the arc length between a point on the curve $(r, \theta)$ and the origin is equal to $\frac{r}{\sin(\frac{\pi}{n})}$. Meaning that the time it takes to reach the origin from a point at distance $r$ is equal to $\frac{r}{v\sin(\frac{\pi}{n})}$
Some values of $\frac{1}{\sin(\frac{\pi}{n})}$ for specific values of $n$.
- For $n=3$ the value is $\frac{2}{\sqrt 3} = 1.1547...$
- For $n=4$ the value is $\sqrt 2 = 1.4142...$
- For $n=5$ the value is around 1.7013,
- For $n=6$ the value is exactly 2, etc...
