Body under Constant Acceleration/Velocity after Distance

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

Then:
 * $\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$

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

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

Then:

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

Substituting for $\mathbf s$ in $(1)$ gives:
 * $\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$

and the proof is complete.