Body under Constant Acceleration

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

Then the following equations apply:

$(1):$ Velocity after Time

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

$(2):$ Distance after Time

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

$(3):$ Velocity after Distance

$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$


$\mathbf u$ 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 scalar product.


$B$ has acceleration $\mathbf a$.

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


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

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:

\(\displaystyle \mathbf v \cdot \mathbf v\) \(=\) \(\displaystyle \paren {\mathbf u + \mathbf a t} \cdot \paren {\mathbf u + \mathbf a t}\)
\(\displaystyle \) \(=\) \(\displaystyle \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + t^2 \mathbf a \cdot \mathbf a\)
\(\displaystyle \) \(=\) \(\displaystyle \mathbf u \cdot \mathbf u + \mathbf a \cdot \paren {2 \mathbf u t + t^2 \mathbf a}\)

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.