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.

Now, Initial velocity of the body,

u = OA .........(1)

And, Final velocity of the body, v = BC .........(2)

But from the graph BC = BD + DC

Therefore, v=BD + DC ........(3)

Again DC = OA

So, v = BD + OA

Now, from equation (1), OA = u

So, v = BD + u ..........(4)

We should find out the value of BD now. We know the slope of a velocity-time graph is equal to the acceleration, a.

Thus, Acceleration, a = slope of line AB

or a = BD/AD

But AD = OC = t, so putting t in place of AD in the above relation, we get,

or BD = at

Now, putting this value of BD in equation(4), we get,

v = u+ at