Derive the following relation graphically:
s = ut + 1/2 `at^(2)`

Explanation

Let us consider that the object has travelling a distances in time t under uniform acceleration a. In the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distances travelled by the object is. given by
S = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABD
`=OAxxOC+1/2(ADxxBD)`....(4)
Substituting OA = u, OC = AD = t and BD = at, we get
`S = u xx t + 1/2 (I xx at) `
or `S=ut+1/2at^2`