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 $v=u+at$: $$ \begin{equation}\begin{aligned} v^2&=(u+at)^2\\ v^2&=u^2+2uat+a^2t^2\\ v^2&=u^2+2a(ut+\frac12 at^2)\\ v^2&=u^2+2as\\ \end{aligned}\end{equation} $$