Three Equations of Motion

The three equations of motion v = u + at; s = ut + (1/2) at² and v² = u² + 2as can be derived with the help of graphs as described below.

Q1

Derive v = u + at by Graphical Method

Sol:

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

Velocity – Time graph to derive the equations of motion.

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.

Q2

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

Sol:

Velocity-Time graph to derive the second equation of motion

Velocity–Time graph to derive the equations of motion.

Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus

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 ABD