The acceleration, $a$ m/s$^2$, of a particle is inversely proportional to the square of its distance $x$ metres from a fixed point.
The distance of the particle is reduced to $0.5x$. Find the ratio of the acceleration to the original acceleration. Answer ... : ...

1. Express the inverse square relationship using a constant of proportionality.

Since $a$ is inversely proportional to the square of the distance $x$,
$a \propto \frac{1}{x^2}$
so
$a=\frac{k}{x^2}$
where $k$ is a constant.

2. Find the new acceleration when the distance is reduced to $0.5x$.

Let the new acceleration be $A$.

When the distance is $0.5x$,
$A=\frac{k}{(0.5x)^2}$
$A=\frac{k}{0.25x^2}=\frac{4k}{x^2}$

3. Compare the new acceleration with the original acceleration as a ratio.

Original acceleration:
$a=\frac{k}{x^2}$

New acceleration:
$A=\frac{4k}{x^2}=4a$

Therefore,
$A:a=4a:a=4:1$