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 dot product.

Proof

From Body under Constant Acceleration: Velocity after Time

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

Then:

\(\ds \mathbf v \cdot \mathbf v\)

\(=\)

\(\ds \paren {\mathbf u + \mathbf a t} \cdot \paren {\mathbf u + \mathbf a t}\)

\(\ds \)

\(=\)

\(\ds \mathbf u \cdot \mathbf u + \mathbf u \cdot \mathbf a t + \mathbf a t \cdot \mathbf u + \paren {\mathbf a t} \cdot \paren {\mathbf a t}\)

Dot Product Distributes over Addition

\(\ds \)

\(=\)

\(\ds \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \paren {\mathbf a t} \cdot \paren {\mathbf a t}\)

Dot Product Operator is Commutative

\(\ds \)

\(=\)

\(\ds \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \mathbf a \cdot \mathbf a t^2\)

Dot Product Associates with Scalar Multiplication

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds \mathbf u \cdot \mathbf u + \mathbf a \cdot \paren {2 \mathbf u t + \mathbf a t^2}\)

Dot Product Distributes over Addition

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.

$\blacksquare$