Body under Constant Acceleration

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

The following equations apply:

$(3):$ Velocity after Distance
where:
 * $\mathbf u$ is the velocity at time $t = 0$
 * $\mathbf v$ is the velocity at time $t$
 * $\mathbf s$ is the displacement of $B$ from its initial position at time $t$
 * $\cdot$ denotes the dot product.

Proof
$B$ has acceleration $\mathbf a$.

Let $\mathbf x$ be the vector corresponding to the position of $B$ at time $t$.

Then:
 * $\dfrac {\d^2 \mathbf x} {\d t^2} = \mathbf a$

Solving this differential equation:
 * $\mathbf x = \mathbf c_0 + \mathbf c_1 t + \frac 1 2 \mathbf a t^2$

with $\mathbf c_0$ and $\mathbf c_1$ constant vectors.

Evaluating $\mathbf x$ at $t = 0$ shows that $\mathbf c_0$ is the value $\mathbf x_0$ of $\mathbf x$ at time $t=0$.

Taking the derivative of $\mathbf x$ at $t = 0$ shows that $\mathbf c_1$ corresponds to $\mathbf u$.

Therefore, since $\mathbf s = \mathbf x - \mathbf x_0$, we have:
 * $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

and by taking the derivative of $\mathbf x$, we have:
 * $\mathbf v = \mathbf u + \mathbf a t$

Next, we dot $\mathbf v$ into itself using the previous statement.

From the linearity and commutativity of the dot product:

The expression in parentheses is $2 \mathbf s$, so:
 * $\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$

and the proof is complete.

Also known as
These equations can often be seen collectively referred to, particularly at elementary and high school levels, as SUVAT, for the usual symbols used to represent the quantities involved.