# Show by means of

The body has an initial velocity u at 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 that the slope of a velocity – time graph is equal to acceleration, a.

Thus, Acceleration, a = slope of line AB

But AD = OC = t,

so putting t in place of AD in the above relation, we get:

a = BD/t

or BD = at

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

v = at + u

This equation can be rearranged to give:

v = u + at

And this is the first equation of motion. It has been derived here by the graphical method.

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