Proving $s = ut + \frac{1}{2} at^2 $ I have been asked to prove on a graph that $s = ut + \dfrac{1}{2} at^2 $ ![enter image description here]( I know that the area of the rectangle is $ut$ but the area under the tri...--prophetes.ai

Proving $s = ut + \frac{1}{2} at^2 $ I have been asked to prove on a graph that $s = ut + \dfrac{1}{2} at^2 $ ![enter image description here]( I know that the area of the rectangle is $ut$ but the area under the triangle is $\frac{1}{2}\times t \times (v-u)$ So total displacement is $s = ut + \frac{1}{2}\times t \times (v-u) $ so how do I get the equation above? Thanks.

By area under graph we find (add your rectangular and trianguar components not multiply) that:

$$s=ut+\frac{t(v-u)}{2}$$

now $a=\frac{v-u}{t}$ so multiplying both sides by $t^2/2$ gives $\frac{at^2}{2}=\frac{t(v-u)}{2}$ and the answer follows.