Body under Constant Acceleration/Distance after Time

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:

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

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:

$\bigvalueat {\mathbf s} {t \mathop = 0} = \mathbf 0$

from which it follows immediately that:

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

$\blacksquare$

Sources

• 1914: G.W. Caunt: Introduction to Infinitesimal Calculus ... (previous) ... (next): Chapter $\text I$: Functions and their Graphs: $1$. Constants and Variables