Derivation of equation of motion is one of the most important topics in physics. Several important concepts in physics are based on the equation of motion. In this article, the equation of motion derivations by graphical method and by normal method are explained in an easily understandable way.

It is first important to understand what motion is and its laws to understand the derivations better. Visit introduction to motion to know everything about motion and its related terminologies.

## Derivation of Equations of Motion:

There are mainly 3 equations of motion which describe the relationship between velocity, time, acceleration and displacement. For more details about the equations, click Equation Of Motion And Its Application.

First, consider a body moving in a straight line with uniform acceleration. Then, let the initial velocity be ** u**, acceleration be

**, time period be**

*a***, velocity be**

*t***and the distance traveled be**

*v*

*S.*The equation of motions derivation can be done in 3 ways which are:

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

*Below, the equations of motion are derived by all the 3 methods in a simple and easy to understand way. *

**Derivation of First Equation of Motion:**

The first equation of motion is->

*v = u + at*

**Derivation of 1st Equation of Motion by Algebraic Method:**

It is known that the acceleration (a) of the body is defined as the rate of change of velocity. So, the acceleration can be written as->

*\(a = \frac{v-u}{t}\)*

From this, rearranging the terms, the first equation of motion is obtained which is:

*v = u + at*

**Derivation of 1st Equation of Motion by Graphical Method:**

Consider the diagram of velocity – time graph of a body below:

In this, the body is moving with an initial velocity of ** u **at point A. The velocity of the body then changes from A to B in time

**at a uniform rate. In the above diagram, BC is the final velocity i.e.**

*t***after the body travels from A to B at a uniform acceleration of**

*v***. In the graph, OC is the time**

*a***. Then, a perpendicular is drawn from B to OC, a parallel line is drawn from A to D and another perpendicular is drawn from B to OE (represented by dotted lines).**

*t*Compiling all the information from the graph, the following details can be obtained:

Initial velocity of the body, u = OA

Final velocity of the body, v = BC

From the graph,

BC = BD + DC

So, v = BD + DC

=> v = BD + OA (since DC = OA)

Finally, ** v = BD + u** (since OA = u)

**———–(Equation 1)**

Now, since the slope of a velocity – time graph is equal to acceleration ** a**,

So, a = slope of line AB

=> *a = BD/AD*

Since AD = AC = t, the above equation becomes->

*BD = at *———–(Equation 2)

Now, combining Equation 1 & 2, the following is obtained->

*v = at + u*

**Derivation of 1st Equation of Motion by Calculus Method:**

It is known that,

So,

### Derivation of Second Equation of Motion:

The second equation of motion is->

*S = ut + ½ a ^{2}*

**Derivation of 2nd Equation of Motion by Algebraic Method:**

Considering the same denotations for all the terms, the second equation of motion can be derived easily by simple algebraic method.

**Derivation of 2nd Equation of Motion by Graphical Method:**

Taking the same diagram used in 1st law derivation->

In this diagram, the distance traveled ** (S) **= Area of figure OABC = Area of rectangle OADC + Area of triangle ABD.

Now, area of the rectangle OADC = OA × OC = *ut*

And, Area of triangle ABD = (1/2) × Area of rectangle AEBD = ** (1/2) at^{2 }**(Since, AD = t and BD = at)

Thus, the total distance covered will be->

*S = ut + (1/2) at ^{2}*

**Derivation of 2nd Equation of Motion by Calculus Method:**

It is known that *v = dS/dt.*

So,

### Derivation of Third Equation of Motion:

The third equation of motion is->

*\(v^{2} = u^{2} + 2aS\)*

**Derivation of 3rd Equation of Motion by Algebraic Method:**

**Derivation of 3rd Equation of Motion by Graphical Method:**

The total distance traveled, *S = Area of trapezium OABC.*

So,

*\(S = \frac{\left ( Sum\, of\, Parallel \, Sides \right )\times Height}{2}\)*

*=> \(S = \frac{\left ( OA + CB \right )\times OC}{2}\)*

Since, OA = ** u**, CB =

**, and OC =**

*v***, the above equation becomes->**

*t**\(S = \frac{\left ( u + v \right )\times t}{2}\)*

Now, since ** t = (v – u)/ a**, the above equation can be written as->

*\(S = \frac{\left ( u + v \right )\times \left ( v – u \right )}{2a}\)*

Solving this equation, the third equation of motion is obtained which is->

*\(v^{2} = u^{2} + 2aS\)*

**Derivation of 3rd Equation of Motion by Calculus Method:**

It is known that,

*These were the detailed derivations for equations of motion in graphical method, algebraic method and calculus method.*

**Also Read: **Newton’s Laws of motion

*For more important derivations and other physics articles, stay tuned with BYJU’S.*