A body moves on a frictionless plane starting from rest. If $$S_n$$ is distance moved between $$t = n - 1$$ and $$t = n$$ and $$S_{n-1}$$ is distance moved between $$t = n - 2$$ and $$t = n - 1$$, then the ratio $$\frac{S_{n-1}}{S_n}$$ is $$\left(1 - \frac{2}{x}\right)$$ for $$n = 10$$. The value of $$x$$ is ______.

A body starts from rest on a frictionless plane. We are given the formula for the ratio $$\frac{S_{n-1}}{S_n} = \left(1 - \frac{2}{x}\right)$$ for $$n = 10$$, and we need to find $$x$$.

We first recall the distance covered in the $$n$$-th second under uniform acceleration.

For a body starting from rest (initial velocity $$u = 0$$) with uniform acceleration $$a$$, the distance covered from $$t = 0$$ to $$t = n$$ is:

$$ S(n) = \frac{1}{2}an^2 $$

The distance covered in the $$n$$-th second (between $$t = n-1$$ and $$t = n$$) is:

$$ S_n = S(n) - S(n-1) = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2 $$

Expanding $$(n-1)^2 = n^2 - 2n + 1$$:

$$ S_n = \frac{a}{2}\left[n^2 - (n^2 - 2n + 1)\right] = \frac{a}{2}(2n - 1) $$

Similarly, replacing $$n$$ with $$n-1$$ in the formula gives:

$$ S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3) $$

Next, we compute the ratio $$\frac{S_{n-1}}{S_n}$$.

$$ \frac{S_{n-1}}{S_n} = \frac{\frac{a}{2}(2n-3)}{\frac{a}{2}(2n-1)} = \frac{2n - 3}{2n - 1} $$

This fraction can be rewritten as:

$$ \frac{2n - 3}{2n - 1} = \frac{(2n-1) - 2}{2n - 1} = 1 - \frac{2}{2n - 1} $$

Since $$\frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}$$, equating this to $$1 - \frac{2}{2n - 1}$$ gives:

$$ \frac{2}{x} = \frac{2}{2n - 1} $$

$$ x = 2n - 1 $$

Finally, substituting $$n = 10$$ gives:

$$ x = 2(10) - 1 = 20 - 1 = 19 $$

The value of $$x$$ is $$\boxed{19}$$.