# 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

$\mathbf v = \mathbf u + \mathbf a t$

Then:

 $\displaystyle \mathbf v \cdot \mathbf v$ $=$ $\displaystyle \left({\mathbf u + \mathbf a t}\right) \cdot \left({\mathbf u + \mathbf a t}\right)$ $\displaystyle$ $=$ $\displaystyle \mathbf u \cdot \mathbf u + \mathbf u \cdot \mathbf a t + \mathbf a t \cdot \mathbf u + \left({\mathbf a t}\right) \cdot \left({\mathbf a t}\right)$ Dot Product Distributes over Addition $\displaystyle$ $=$ $\displaystyle \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \left({\mathbf a t}\right) \cdot \left({\mathbf a t}\right)$ Dot Product Operator is Commutative $\displaystyle$ $=$ $\displaystyle \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \mathbf a \cdot \mathbf a t^2$ Dot Product Associates with Scalar Multiplication $(1):\quad$ $\displaystyle$ $=$ $\displaystyle \mathbf u \cdot \mathbf u + \mathbf a \cdot \left({2 \mathbf u t + \mathbf a t^2}\right)$ Dot Product Distributes over Addition
$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

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.

$\blacksquare$