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}

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