# 2.2.2 Derivation of Equations of Motion

You will be using the equations of motion all the time. Surely you want to see their derivation/proof?

v-t

The v-t equation basically comes directly from the definition of acceleration. $\displaystyle a=\frac{{v-u}}{t}\quad \Rightarrow \quad v=u+at\quad \cdots (1)$

s-t

For displacement, we can start from $s={{v}_{{ave}}}t$, where vave refers to the average velocity during the duration t. Since $\displaystyle {{v}_{{ave}}}=\frac{{u+v}}{2}$ for uniform acceleration motion, $\displaystyle s={{v}_{{ave}}}\times t\Rightarrow s=\frac{1}{2}(u+v)\times t\quad \cdots (2)$

Another way to derive equation (2) is by considering the trapezium area under the v-t graph.

From equation (2), we just need to get rid of v by substituting $v=u+at$ into equation (2). \displaystyle \begin{aligned}s&=\frac{1}{2}(u+v)\times t\\&=\frac{1}{2}(u+(u+at))t\text{ }\Rightarrow \text{ }s=ut+\frac{1}{2}a{{t}^{2}}\quad \cdots (3)\end{aligned}

v-s

To derive the v-s equation, we just need to get rid of t by substituting $t=\frac{{v-u}}{a}$ into equation (2). \displaystyle \begin{aligned}s&=\frac{1}{2}(u+v)\times t\\&=\frac{1}{2}(u+v)\times \frac{{v-u}}{a}\\&=\frac{{{{v}^{2}}-{{u}^{2}}}}{{2a}}\text{ }\Rightarrow \text{ }{{v}^{2}}={{u}^{2}}+2as\quad \cdots (4)\end{aligned}