# Newton’s Law of Motion Part-3

**Newton’s Third Law of Motion**

** ****According to this law, ****if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.**

## Newton’s Third Law of Motion

Force is a push or pull acting on an object resulting in its interaction with another object. Force is a result of an interaction.

Forces can be classified into two categories: contact forces, such as frictional forces, and non-contact forces, such as gravitational forces.

According to Newton, when two bodies interact, they exert a force on each other, and these forces are known as action and reaction pairs, which are explained in Newton's third law of motion.

Newton’s third law of motion states that

**“When one body exerts a force on the other body, the first body experiences a force which is equal in magnitude in the opposite direction of the force which is exerted”.**

The above statement means that in every interaction, there is a pair of forces acting on the interacting objects. The magnitude of the forces are equal and the direction of the force on the first object is opposite to the direction of the force on the second object.

## Examples of Interaction Force Pairs

A variety of action-reaction pairs are evident in nature. We have listed a few below, and they are as follows:

- Pushing fish through water is an example of an action-reaction pair. The fish pushes the water back with its fins. This thrust accelerates the fish forward. The magnitude of the force in the water is equal to the magnitude of the force acting on the fish; the direction of the force on the water (backwards) is opposite to the direction of the force on the fish (forwards)..
- The flight of a bird is an example of an action-reaction pair. A bird's wings push the air down. Air pushes air up.
- A swimmer pushes against the water, while the water pushes back on the swimmer.
- Lift is created by helicopters by pushing the air down, thereby creating an upward reaction force.
- Rock climbers pull their vertical rope downwards to push themselves upwards.

**DERIVATION OF EQUATION OF MOTION**

The equations of motion can be derived using the following methods:

**Derivation of equations of motion by Simple Algebraic Method****Derivation of equations of Motion by Graphical Method****Derivation of equations of Motion by Calculus Method**

In the next few sections, the equations of motion are derived by all the three methods in a simple and easy to understand way.

DERIVATION OF THIRD EQUATION OF MOTION

For the derivation of the third equation of motion, consider the same variables that were used for the derivation of the first and second equations of motion.

**Derivation of Third Equation of Motion by Algebraic Method**

We know that displacement is the product of average velocity and time. Mathematically, this can be represented as:

Rearranging the above formula, we get

v^{2} = u^{2 }+ 2as

**Derivation of Third Equation of Motion by Graphical Method**

From the graph, we can say that

The total distance travelled, *s* is given by the **Area of trapezium OABC**.

Hence,

**s = ½ × (Sum of Parallel Sides) × Height**

**s = 1/2 x (OA + CB) x OC**

Since, **OA = u**, **CB = v**, and **OC = t**

The above equation becomes

**s = 1/2 x (u+v) x t**

Now, since **t = (v – u)/ a**

The above equation can be written as:

**s = ½ x ((u+v) × (v-u))/a**

Rearranging the equation, we get

**s = ½ x (v+u) × (v-u)/a**

**s = (v ^{2}-u^{2})/2a**

Third equation of motion is obtained by solving the above equation

Rearranging the above formula, we get

v^{2} = u^{2 }+ 2as

**Derivation of Third Equation of Motion by Calculus Method**

We know that acceleration is the rate of change of velocity and can be represented as:

Rearranging the above formula, we get

v^{2} = u^{2 }+ 2as