Coyote VS Road Runner

Let $T_{n}$ denote the distance of the $n^{th}$ track.

Then we can set up a geometric progression of the series as the following:

$T_{1} = 90$

$T_{2} = 90\times (\dfrac{80}{100}) = 72$

$T_{3} = 90\times (\dfrac{80}{100})^2$

$vdots$

$T_{n} = 90\times (\dfrac{80}{100})^{n-1}$

However, since the distance is also classified as eastward and westward, we can mark "eastward" direction as "+" and "westward" as "-" from point $A$ .

In other words, $T_{n} = (-1)^{n-1} (90\times (\dfrac{80}{100})^{n-1})$

Now for eastward track, we can see from $T_{1}$ to $T_{3}$ that the distance is reduced to $(\dfrac{80}{100})^2 = \dfrac{64}{100}$ , which also applies for the next $n+2$ term in the eastward series, and the progression is the same as in the westward series.

Therefore, the total distance traveled eastward = $\dfrac{T_{1}}{1 - r}$ = $\dfrac{90}{1 - \dfrac{64}{100}}$ = $250$ meters.

Then, the total distance traveled westward = $\dfrac{T_{2}}{1 - r}$ = $\dfrac{72}{1 - \dfrac{64}{100}}$ = $200$ meters.

As a result, as the number of tracks reach infinity, the Coyote's distance from the starting point $A$ = $250 - 200 = \boxed{50}$ meters eastward.