Derivation of the 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 the graphical method and by the normal method are explained in an easily understandable way for the first, second and third equation of motion.

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First Equation by Algebraic MethodFirst Equation by Graphical MethodFirst Equation by Calculus MethodSecond Equation by Algebraic MethodSecond Equation by Graphical MethodSecond Equation by Calculus MethodThird Equation by Algebraic MethodThird Equation by Graphical MethodThird Equation by Calculus Method

*It is first important to understand what is motion and its laws to understand the derivations better.*

*Following is the table explaining more concepts related to motion*

## Derivation of Equation of Motion

There are mainly three equations of motion which describe the relationship between velocity, time, acceleration and displacement.

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 travelled be**

*v,Â*

*S.*The equation of motions derivation can be done in three 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 three 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 First 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 = v âˆ’ ut

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

v = u + at

### Derivation of First Equation of Motion by Graphical Method

Consider the diagram of the 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*Following details are obtained from the graph above:

The initial velocity of the body, u = OA

The 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 First 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 + Â½ a ^{2}*

### Derivation of Second Equation of Motion by Algebraic Method

Consider the same notations for the derivation of the second equation of motion by simple algebraic method.

### Derivation of Second Equation of Motion by Graphical Method

Taking the same diagram used in first law derivation:

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

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

And, Area of triangle ABD = (1/2) Ã— 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 Second Equation of Motion by Calculus Method

Velocity is the rate of change of displacement.

Mathematically, this is expressed as

\(v=\frac{ds}{dt}\)Rearranging the equation, we get

\(ds=vdt\)Substituting the first equation of motion in the above equation, we get

\(ds=(u+at)dt\) \(=(udt+at\,dt)\) \(\int_{0}^{s}ds=\int_{0}^{t}u\,dt+\int_{0}^{t}at\,dt\) \(s=ut+\frac{1}{2}at^2\)## Derivation of Third Equation of Motion

The third equation of motion is:

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

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

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

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

So, S= 1/2(SumofParallelSides)Ã—Height

S=(OA+CB)Ã—OC

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

The above equation becomes

S= 1/2(u+v)Ã—t

Now, since *t = (v â€“ u)/ a*

The above equation can be written as:

S= 1/2(u+v)Ã—(v-u)/a

Rearranging the equation, we get

S= 1/2(v+u)Ã—(v-u)/a

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

Third equation of motion is obtained by solving the above equation:

v^{2Â }= u^{2}+2aS

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

It is known that,

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

### Equations of Motion Formula

Equations of motion |
Formula |

First equation of motion | v=u+at |

Second equation of motion | \(s=ut+\frac{1}{2}at^{2}\) |

Third equation of motion | v^{2Â }= u^{2}+2as |

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