motion with constant acceleration

1. Displacement, \[s\] = area under graph
   

\begin{align*} s&=\frac{t\cdot (v-u)}{2}+tu \\ &=\frac{1}{2}(tv-tu)+tu \\ &=\frac{1}{2}(tv+tu) \\ \tag{1} &=\frac{1}{2}(u+v)t \\ \end{align*}

1. Acceleration, \[a\] = gradient of graph
   

\begin{align*} a&=\frac{v-u}{t-0} \\ \tag {2} &=\frac{v-u}{t} \end{align*}

1. By rearranging (2), we get
   

\begin{align*} at&=v-u \\ \tag{3} v&=u+at \end{align*}

1. By substituting (1) into (3), we get
   

\begin{align*} s&=\frac{1}{2}(u+(u+at))t \\ &=\frac{1}{2}(2u+at)t \\ \tag{4} &=ut+\frac{1}{2}at^{2} \end{align*}

1. By rearranging (3) then substituting into (1), we get
   

\begin{align*} s&=\frac{1}{2}((v-at)+v)t \\ \tag{5} &=vt-\frac{1}{2}at^{2} \end{align*}

1. By rearranging (2) then substituting into (1), we get
   

\begin{align*} s&=\frac{1}{2}(u+v)(\frac{v-u}{a}) \\ 2as&=v^{2}-u^{2} \\ \tag{6} v^{2}&=u^{2}+2as \end{align*}