1. Assume uniform acceleration

Since the body starts from rest and moves on a frictionless plane, the standard interpretation is motion with constant acceleration $a$.

So, $s(t)=\frac{1}{2}at^2$ where $s(t)$ is displacement from the start at time $t$.

2. Distance moved in the $n$th second

The distance moved between $t=n-1$ and $t=n$ is $S_n=s(n)-s(n-1)$

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

Similarly, $S_{n-1}=s(n-1)-s(n-2)$ $=\frac{1}{2}a\left[(n-1)^2-(n-2)^2\right]$ $=\frac{1}{2}a(2n-3)$

3. Form the ratio

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

For $n=10$, $\frac{S_9}{S_{10}}=\frac{2(10)-3}{2(10)-1}=\frac{17}{19}$

4. Match with the given form

Given, $\frac{S_{n-1}}{S_n}=1-\frac{2}{x}$

So, $1-\frac{2}{x}=\frac{17}{19}$

Then, $\frac{2}{x}=1-\frac{17}{19}=\frac{2}{19}$

Hence, $x=19$

5. Final answer

$\boxed{19}$