# How do you derive this? SImple question

1. Mar 12, 2012

### Alipepsi

Derive for time!
D=Vit+1/2at^2

THANKS for you help

2. Mar 12, 2012

### Dickfore

The displacement is given by the area under the velocity-time graph. Your velocity-time equation is:
$$v(t) = v_i + a t$$
and between the vertical lines $t_1 = 0$ and $t_2 = t$. The geometric figure is a right trapezoid (trapzium outside U.S.) and the area is:
$$D = \frac{1}{2} \left[ v_i + (v_i + a \, t) \right] \cdot t = v_i \, t + \frac{a \, t^2}{2}$$