Acceleration of Point in Straight Line

Theorem

Let $\mathbf a$ be the acceleration of a particle $P$ in space.

Let $P$ be moving along a straight line $\LL$ whose positive direction has been established.

Then the motion of $P$ can be defined by:

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

where $\mathbf i$ denotes the unit vector in the positive direction of $\LL$.

We have by hypothesis that $P$ moves along a straight line $\LL$.

Let the $\LL$ be embedded in a Cartesian space $\CC$.

From the definition of the components of acceleration vector, we have:

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

where:

$\mathbf v$ is the velocity of $P$ at time $t$

$\mathbf r$ is the displacement of $P$ at time $t$ expressed as a position vector from a given origin.

Thus:

$\mathbf a = \dfrac {\d^2 x} {\d t^2} \mathbf i + \dfrac {\d^2 x} {\d t^2} \mathbf j + \dfrac {\d^2 x} {\d t^2} \mathbf k$

where:

$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$

Let us define $\CC$ such that $\LL$ is parallel to the $x$-axis of $\CC$.

Then we have:

$\ds \mathbf v$

$=$

$\ds \dfrac {\d x} {\d t} \mathbf i + \dfrac {\d y} {\d t} \mathbf j + \dfrac {\d z} {\d t} \mathbf k$

Definition of Velocity

$\ds $

$=$

$\ds \dfrac {\d x} {\d t} \mathbf i + 0 \mathbf j + 0 \mathbf k$

$\ds \leadsto \ \ $

$\ds \mathbf a$

$=$

$\ds \dfrac {\d^2 x} {\d t} \mathbf i$

Definition of Acceleration

$\blacksquare$