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 

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


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