The constant acceleration equations of motion are:
- $v=u+at$ (no $s$)
- $s=(\frac{u+v}{2})t$ (no $a$)
- $s=ut+\frac{1}{2}at^2$ (no $v$)
- $s=vt-\frac{1}{2}at^2$ (no $u$)
- $v^2=u^2+2as$ (no $t$)
Where $s$ is displacement, $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration and $t$ is time. Time is the only scalar thus we apply the sign convention to the other physical quantities.
Deriving the equations of motion
Starting with the defining equation for acceleration (the rate of change of velocity with respect to time): $$ \begin{equation}\begin{aligned} a=\frac{v-u}{t}\\ \end{aligned}\end{equation} $$
Making $v$ the subject of the formula: $$ \begin{equation}\begin{aligned} v=u+at\\ \end{aligned}\end{equation} $$
Starting with the defining equation for displacement being the product of average velocity and time: $$ \begin{equation}\begin{aligned} s=(\frac{u+v}{2})t\\ \end{aligned}\end{equation} $$
We substitute $v$ as $u+at$: $$ \begin{equation}\begin{aligned} s&=(\frac{u+u+at}{2})t\\ s&=(\frac{2u+at}{2})t\\ s&=(u+\frac{1}{2}at)t\\ s&=ut+\frac{1}{2}at^2\\ \end{aligned}\end{equation} $$
Using $v=u+at$, and finding for $u$: $$ \begin{equation}\begin{aligned} u=v-at\\ \end{aligned}\end{equation} $$
We substitute $u$ as $v-at$: $$ \begin{equation}\begin{aligned} s&=(\frac{v-at+v}{2})t\\ s&=(\frac{2v-at}{2})t\\ s&=(v-\frac{1}{2}at)t\\ s&=vt-\frac{1}{2}at^2\\ \end{aligned}\end{equation} $$
Squaring both sides of the equation $v=u+at$: $$ \begin{equation}\begin{aligned} v^2&=(u+at)^2\\ v^2&=u^2+2uat+a^2t^2\\ v^2&=u^2+2a(ut+\frac12 at^2)\\ v^2&=u^2+2as\\ \end{aligned}\end{equation} $$
Object in free fall
Consider an object released from rest and falling through the air without influence from forces other than gravity:
- $u_y=0ms^{-1}$
- $a_y=g$
Then we can find the relationship between the time taken ($t$) for the object to hit the ground and the displacement ($s_y$) of the object relative to the ground:
$$ \begin{equation}\begin{aligned} s_y&=u_yt+\frac12a_yt^2\\ s_y&=\frac12 gt^2\\ gt^2&=2s_y\\ t^2&=\frac{2s_y}{g}\\ \therefore t&=\sqrt{\frac{2s_y}{g}}\\ \end{aligned}\end{equation} $$