# Body under Constant Acceleration

## Theorem

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

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

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

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

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: The justification for the derivation of $(1)$ from $(2)$ is handwavey and clumsy. Better to do $(1)$ first and then derive $(2)$ from it, to save having to solve a second order DE whose solution we have not yet put on $\mathsf{Pr} \infty \mathsf{fWiki}$. As the separate proofs for the separate parts have now been extracted into their own pages, complete with justifying links throughout (and no handwvery), it is suggested that this proof may be deleted.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

$B$ 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:

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

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

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

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.

$\blacksquare$

## 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.

## Sources

- 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**constant acceleration**