Question

Derive Eq. of motion by Graph Method.

Answer 1

1. Derive v = u + at by Graphical Method

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 + uThis 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.


2.Derive s = ut + (1/2) at2 by Graphical Method

Distance travelled=Area of figure OABC
=Area of rectangle OADC + Area of triangle ABD

We will now find out the area of the rectangle OADC and the area of the triangle ABD.
(i) Area of rectangle OADC=OA × OC
=u × t
=ut ...... (5)

(ii) Area of triangle ABD=(1/2) × Area of rectangle AEBD
=(1/2) × AD × BD
=(1/2) × t × at (because AD = t and BD = at)
=(1/2) at2...... (6)

So, Distance travelled, s=Area of rectangle OADC + Area of triangle ABDor s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method

Answer 2

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