# Body under Constant Acceleration/Distance after Time

Jump to navigation
Jump to search

## Theorem

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

Then:

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

where:

- $\mathbf s$ is the displacement of $B$ from its initial position at time $t$
- $\mathbf u$ is the velocity at time $t = 0$.

## Proof

From Body under Constant Acceleration: Velocity after Time:

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

By definition of velocity, this can be expressed as:

- $\dfrac {\mathrm d \mathrm s}{\mathrm d t} = \mathbf u + \mathbf a t$

where both $\mathbf u$ and $\mathbf a$ are constant.

By Solution to Linear First Order Ordinary Differential Equation:

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

where $\mathbf c$ is a constant vector.

We are (implicitly) given the initial condition:

- $\big.{\mathbf s}\big\rvert_{\, t \mathop = 0} = \mathbf 0$

from which it follows immediately that:

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

$\blacksquare$