Consider the velocity – time graph of a body shown in the below Figure.

The body has an initial velocityuat pointAand then its velocity changes at a uniform rate fromAtoBin timet.

In other words, there is a uniform acceleration 'a' fromAtoB, and after timetits final velocity becomes 'v' which is equal toBCin the graph. The timetis represented byOC.

To complete the figure, we draw the perpendicularCBfrom pointC, and drawADparallel toOC.

BEis the perpendicular from pointBtoOE.

Now, Initial velocity of the body,u=OA....(1)And, Final velocity of the body,v=BC......(2)But from the graphBC=BD +DC.Therefore,v=BD + DC......... (3)AgainDC=OASo,v=BD + OANow, From equation(1),OA=uSo,v=BD+u........... (4)We should find out the value ofBDnow. We know that the slope of a velocity – time graph is equal to acceleration,a.

Thus, Acceleration,a=slope of lineAB ora=BD/ADButAD=OC = t,so puttingtin place ofADin the above relation, we get:a=BD/torBD=atNow, putting this value ofBDin equation (4) we get :v=at+uThis equation can be rearranged to give:

v=u + atAnd this is the first equation of motion. It has been derived here by the graphical method.

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