**Derivation of Equations of motion with the help of graphs**

(a) Graph in case of the equation, **v = v _{0} + at**

In this equation, there are two variables— one is t and the other one is v. A graph is drawn taking t on the X-axis and v on the Y-axis [Fig]. In the figure, normal PN is drawn from point P to Y-axis. Let consider that the final velocity at time t = v = OY. Now OY = OA + AY i.e., v = v_{0} + at.

Here, slope, a = PM/AM = AY/AM = AY/t

The equation v = v_{0} + at is obtained from the graph.

In case of motion of a body from rest, v_{0} = 0, a = constant.

So, v = 0 + constant x t

so,** v ∞ t**

**That means, velocity is proportional to time.**

(b) Graph for the equation, **s = v _{0}t + ½ at^{2}**

In figure, graph of ‘v’ versus ‘t’ is a straight line.

In the figure PN ┴ OX; AM ┴ PN.

Let initial velocity = v_{0}, uniform acceleration = a, ON = t and distance travelled in time t = s.

Now, s = Area of OAPN

= area of the rectangle OAMN + area of the triangle AMP

= OA x ON + ½ AM PM

= v_{0}t + ½ AM PM

Again, slope ‘a’ PM/AM = PM/t

so, PM = at

and, s = v_{0}t + ½ t.at = v_{0}t + ½ at^{2}

So, the equation can be presented in graph.

In case of a body moving with uniform acceleration from rest, then,

s = 0 x t + ½ constant x t^{2}

or, s = constant x t^{2}

or, **s ∞ t ^{2}**

**That means, displacement is proportional to time.**