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.
Rate this question :