**Equation 3 **

As stated before the area under the graph represents displacement

We have already shown that s = (v+u)/2 x t

Another way of representing the area under the graph is to split it into two sections;

A rectangular section with area = base x height

and a triangular section with area = ^{1}/_{2} x base x height

This gives us the equation s = ut + ^{1}/_{2}(v-u)t

From equation 2 we know that a = (v-u)/t and therefore (v-u) = at

Therefore we can substitute the term (v-u) in the above equation for at

This gives;

**s = ut + ^{1}/_{2} at^{2} (eq. 3)**

[i.e. total displacement = displacement due to initial velocity plus addition displacement due to the objects acceleration]