Distance Moved by Body from Rest under Constant Acceleration

Theorem
Let a body $B$ be stationary.

Let $B$ be subject to a constant acceleration.

Then the distance travelled by $B$ is proportional to the square of the length of time $B$ is under the acceleration.

Proof
From Body under Constant Acceleration: Distance after Time:
 * $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

where:
 * $\mathbf s$ is the displacement of $B$ at time $t$ from its initial position at time $t$
 * $\mathbf u$ is the velocity at time $t = 0$
 * $\mathbf a$ is theconstant acceleration $t$

In this scenario, $\mathbf u = \mathbf 0$.

Thus:
 * $\mathbf s = \dfrac {\mathbf a} 2 t^2$

and so by taking the magnitudes of the vector quantities:
 * $s = \dfrac a 2 t^2$

Hence the result, by definition of proportional.