Body under Constant Acceleration/Distance after Time

Theorem
Let $B$ be a body under constant acceleration $\mathbf a$.

Then:
 * $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

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

Proof
From Body under Constant Acceleration: Velocity after Time:
 * $\mathbf v = \mathbf u + \mathbf a t$

By definition of velocity, this can be expressed as:
 * $\dfrac {\mathrm d \mathrm s}{\mathrm d t} = \mathbf u + \mathbf a t$

where both $\mathbf u$ and $\mathbf a$ are constant.

By Solution to Linear First Order Ordinary Differential Equation:
 * $\mathbf s = \mathbf c + \mathbf u t + \dfrac {\mathbf a t^2} 2$

where $\mathbf c$ is a constant vector.

We are (implicitly) given the initial condition:
 * $\big.{\mathbf s}\big\rvert_{\, t \mathop = 0} = \mathbf 0$

from which it follows immediately that:
 * $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$