Answer :

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*

or *a* = *BD/AD*

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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