The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE. The distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

Distance travelled, s = Area of trapezium OABC

s = \(\frac{(sum\;of\;parallel\;sides) \times Height}{2}\)

s = \(\frac{(OA + CB) \times OC}{2}\)

Now, OA + CB = u + v and OC = t

Putting these values in the above relation, we get,

s = \((\frac{u + v}{2})\times 2\)  .....(1)

Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.

Thus, v = u + at (first equation of motion)

And, at = v