# Second Equation of motion by graphical Method

In this article, we will derive the second equation of motion by graphical method.

### Formula for the second equation of motion

First equation of motion is given by the relation

$s=ut+\frac{1}{2}at^2$

Where

$v$= final velocity

$u$= initial velocity

$a$= acceleration

$s$= displacement of the object

$t$= time taken

Note: - This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.

## Derivation of the 2nd equation of motion by graphical method:

To derive the 2nd equation of motion we will make the following assumptions

* Object under consideration is moving with acceleration $a\,\, m/s^2$

* At time $t=0$ object have some initial velocity. Let’s denote it by $u\,\, m/s$

* At time $ts$ object have some final velocity. Let’s denote it by $v\,\, m/s$

* Total displacement of the object in time $t$ seconds is $s$ meters.

Object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time $t = 0$ and $t = t\,\, s$ are $u$ and $v$ respectively. During time $t$, let $s$ be the total displacement of the object.

Figure given below show the velocity-time graph for the object whose initial velocity is $u$ at time $t=0$ and velocity $v$ at time $t$.

We know that area covered under velocity-time graph gives the displacement of the object in given time $t$ So,

Net Displacement = Area under velocity-time graph.
Or,

$s=$ Area of trapezium $OPQS$

$s=$ Area of rectangle $OPRS$ + Area of triangle $PQR$
Thus,

$s=OP\times PR + \frac{RQ\times PR}{2}$

Substituting various values we get

$s=u\times t +\frac{1}{2}(v-u)\times t$

Since $RQ=(v-u)$ and $PR=OS=t$

Or,

$s=u\times t+\frac{1}{2}at\times t$

Since $(v-u)=at$

So, our first equation of motion is

$s=ut+\frac{1}{2}at^2$