Second equation of motion is used to calculate the position of an object after time t when it is undergoing constant acceleration a.
Suppose that at t = 0, x1 = x0; v1 = v0 and at t = t, x2 = x; v2 = v.
The distance traveled = area under v – t graph = Area of trapezium OABC
= ½(CB + OA) × OC
x – x0 = ½(v + v0)t
Since v = v0 + at,
x – x0 = ½(v0 + at + v0)t
x – x0 = v0t + ½at2
x = x0 + v0t + ½at2