There are two athletes (say A & B) - starting from same fixed point-running with speed X m/s and Y m/s around a circle of radius R in anti-clockwise direction - infinitely. How can I find the maximum distance difference possible b/w the runners throughout their infinite run?
The reset of the distance happens when the athletes overlap (whenever one of them overtakes)
For e.g.:
A and B = 7m/s and 5 m/s with radius 0.79m - after 1 lap the distance difference would be dx1 - A is ahead by some amount - since its faster and in the next round it will increase....
What can be the maximum distance difference (along the circular arc) possible in an infinite run?
Note: The distance shall be measured in anti-clockwise direction - the direction in which runners are running
If the speeds are the same, then the runners distance will be 0 (or the starting value if they aren't).
If the speeds are different, then at some point one runner will overtake the next runner (since they run for an infinite amount of time): right before this point, the runners will be just a little bit under the full circumference away from each other (slightly less than $2\pi r$ for $r$ the radius, measured in the direction they're running).
Now if we think about the maximum possible value, we can see that it's also the full circumference ($2 \pi r$- anything larger would be more than a full lap).
So we never actually reach the maximum length, but we can get arbitrarily close to it: for $r$ the radius, and any small distance $\varepsilon > 0$, then there will be a moment (very shortly before overtaking) that the runners are $2 \pi r - \varepsilon$ distance apart.
Edit - Number of laps/amount of time before this happens:
Without loss of generality assume A is faster than B, A travelling at speed $a$, B at speed $b$. Then you can view this as A travelling at speed $a-b$ faster than B with B stationary (imagine that you are watching from B's perspective), so that A overtakes B at
$$ \textrm{time} = \dfrac{\textrm{distance}}{\textrm{speed}} = \dfrac{2 \pi r}{a-b} $$
time into the race, so that in that time A has travelled a total of $a \cdot\dfrac{2 \pi r}{a-b}$ distance, and then dividing by the length of a lap ($2\pi r$) to get number of laps (effectively changing unit of distance from meters to laps), we get that A overtakes B after exactly $$ \dfrac{a}{a-b} \ \quad \textrm{laps.} $$
(And similarly, B should have made one less lap, or $$ \dfrac{a}{a-b} - 1 = \dfrac{a-(a-b)}{a-b} = \dfrac{b}{a-b} $$ which can be found in the same way after calculating the amount of time and multiplying it by B's speed)