## 'SUVAT' Equations Proofs

There are 5 'SUVAT' equations:

$①:v=u+at$

$②:s=\frac{1}{2} (u+v)t$

$③:v^2=u^2+2as$

$④:s=ut+\frac{1}{2} at^2$

$⑤:s=vt- \frac{1}{2} at^2$

## Proof using a velocity-time graph

Consider the following general velocity-time graph for constant acceleration:

1. The gradient of the line is given by \frac{v-u}{t}, and by definition, the gradient is the acceleration (since both are the change in velocity over time). Therefore: $a=\frac{v-u}{t}$ $at=v-u$ $\therefore v=u+at$
2. By definition, the area under the graph is displacement (since displacement = velocity $$\times$$ time). Therefore (using the area of a trapezium): $s=\left(\frac{u+v}{2}\right) t$ $\therefore s=\frac{1}{2} (u+v)t$
3. Rearranging $$①$$ gives: $t=\frac{v-u}{a}$ Substituting this into $$②$$: $s=\left(\frac{u+v}{2}\right)\left(\frac{v-u}{a}\right)$ $2as=(u+v)(v-u)$ $2as=v^2-u^2$ $\therefore v^2=u^2+2as$
4. Substituting $$①$$ into $$②$$ gives: $s=\left(\frac{u+u+at}{2}\right) t$ $s=\left( \frac{2u}{2}+\frac{at}{2} \right) t$ $\therefore s=ut+\frac{1}{2} at^2$
5. Substituting $$①$$ into $$④$$ gives: $s=(v-at)t+\frac{1}{2}at^2$ $s=vt-at^2+\frac{1}{2} at^2$ $\therefore s=vt-\frac{1}{2} at^2$

## Proof using integration

We can derive equations $$①$$ and $$④$$ using integration.

1. Acceleration is the rate of change of velocity with respect to time, so: $v=\int a \hspace{1 mm} dt$ $=at+c$ when $$t=0$$ ,  $$v=u$$ so substituting this in gives: $u=0+c$ $\therefore c=u$ $\therefore v=u+at$
2. Velocitiy is the rate of change of displacement with respect to time, so: $s=\int v \hspace{1mm} dt$ $=\int (u+at)\hspace{1mm}dt$ $=ut+\frac{1}{2} at^2+c$ $$s=0$$ when $$t=0$$ meaning that $$c=0$$ $\therefore s=ut+\frac{1}{2} at^2$
(1 Vote)