Third Equation of Motion by Graphical Method

Here in this article, we will derive the third equation of motion by graphical method.

Formula for the third equation of motion:

Third equation of motion is given by the relation
\(v^2=u^2+2as\)
Where,
\(v=\) final velocity
\(u=\) initial velocity
\(a=\) acceleration
\(s=\) distance travelled
Note: - This equation along with other kinematics equations of motion are valid for objects moving with uniform acceleration.

Derivation of 3rd equation of motion by graphical method:

To derive 3rd equation of motion we will make 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 distance covered by 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 distance travelled by 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 the distance covered by the object moving with uniform acceleration is given by the area under the velocity-time graph. Here area under the velocity-time graph is equal to area of trapezium OPQS

∴ Area of trapezium OPQS \[= \frac{1}{2}\text{(Sum of Parallel Slides + Distance between Parallel Slides)}\]
Or, \[s=\frac{OP+SQ}{2}\times PR\]

We have knowledge about acceleration of the moving object. So, 
Acceleration  \[a=\frac{\text{Change in velocity}}{time}=\frac{QR}{PR}\]
Or,
\[PR=\frac{QR}{a}\]
From graph we can see that
 \[QR=v-u\]
\[t=PR=\frac{v-u}{a}\]
From above figure we can see that
\[OP=u\]
\[SQ=v\]
\[OP+SQ=u+v\]
Substituting these values, we get
\[s=\left(\frac{u+v}{2}\right)\times\left(\frac{v-u}{a}\right)=\frac{v^2-u^2}{2a}\]
Rearranging it we get
\[v^2=u^2+2as\]
Which is third equation of motion.

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