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Derivation of Equation of Motion

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Last updated date: 23rd May 2024
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Types of the Equation of the Motion :

There are three equations of Motion which are given below:

  1. \[\ v_{final} - u_{initial}= a\Delta t\]

  2. \[\ S = u_{initial} (\Delta t)+\frac {1}{2}a(\Delta t)^2\]

  3. \[\ v^2_{final} - u^2_{initial} = 2as\]

 

The above three equations represent the Motion of a particle in a three-dimensional space. One can write the equation in 1D or 2 D with their respective components. 

 

Derivation of First Equation of Motion     

The first equation of Motion is written as below:

 

\[\ v_{final} - u_{initial}= a\Delta t\]

 

This equation involves the initial and final velocity, constant acceleration, and time. Let an object of mass ‘m’ moves with an initial speed \[u_{initial}\]. The speed of the object changes due to constant acceleration ‘a’ which results in the final speed. \[v_{final}\] after a time interval \[\Delta t\]

  1. Derivation of First Equation of Motion by Algebraic Method

As it is well known that the rate of change in speed is called acceleration.

\[a = \frac {\Delta v}{t} = \frac {v_{final} - u_{initial}}{t} \]

On rearranging the above equation, one can obtain the first equation of Motion: 

\[v_{final} - u_{initial} = at\]

  1. Derivation of First Equation of Motion by Graphical Method 

In Figure 1 at time t = 0 at point E, an object has the initial speed \[u_{initial} = OE\] and the final speed is \[v_{final} = AC\]. From the graph: \[v_{final} = u_{initial} + BC\] which results in

\[BC = v_{final} - u_{initial} …..(1)\]

Acceleration ‘a’ is the slope of the velocity versus time graph and hence, can be written as-

\[a = \frac {BC}{BE} …….(2)\]

From the horizontal axis, \[BE = OA = \Delta t …….(3)\]

Putting the equation (3) in equation (2), the following equation is obtained: 

\[a = \frac {BC}{BE} = \frac {v_{final} - u_{initial}}{\Delta t}\]

Hence, we get the first equation of Motion \[v_{final} - u_{initial} = at\]

C. Derivation of First Equation of Motion by Calculus Method

It is already known that acceleration ‘a’ is the rate of change of the velocity, v. Mathematically, we can write

\[\frac {dv}{dt} = a\].

\[dv = adt\]

\[\int_{u_{initial}}^{v_{final}} dv = a \int_{0}^{1} dt\]

\[v_{final} - u_{initial} = a \Delta t\]

 

Derivation of Second Equation of Motion

The second equation of Motion is

\[S = S_{0} + u_{initial} (\Delta t) + \frac{1}{2} a(\Delta t)^2\]

which represents the total distances travelled by an object in a time interval of \[\Delta t\]  with an initial speed of \[u_{initial}\] and acceleration ‘a’.

  1. Derivation of Second Equation of Motion by Algebraic Method

The second equation of Motion gives the relationship between the positions of the object with time.

Consider an object that moves with an initial speed of \[u_{initial}\] which is under the influence of constant acceleration ‘a’. After time \[\Delta t\] travelling distance s, the speed of the object becomes \[v_{final}\]. The average speed is given as below  

\[v_{avg} = \frac {v_{final} + u_{initial}}{2}\]

It is also known that Displacement = Speed x time

\[S = v_{avg} \times \Delta t\]

Putting the value of \[v_{avg}\] in the above equation we obtain

\[S = \frac {v_{final} + u_{initial}}{2} \times \Delta t\]

Using the first equation of Motion in the above equation

\[S = \frac {(u_{initial}+ a\Delta t)+ u_{initial}}{2} \times \Delta t\]

On rearranging the above equation we get,       

\[S = \frac {(u_{initial}+ a\Delta t)+ u_{initial}}{2} \times \Delta t\]

         which is the second equation of Motion.

  1. Derivation of Second Equation of Motion by Graphical Method?

Consider Figure 2; the total distance travelled by the object in a time interval of \[\Delta t\] is equal to the area of the geometrical figure OECA.

Area of OECA = Area of triangle CEB + Area of rectangle OEBA

Area of triangle CEB = \[\frac {1}{2} \times base \times altitude\]

= \[\frac {1}{2} \times EB \times BC\]

= \[\frac {1}{2} \times \Delta t \times (AC-AB) = \frac {1}{2} \times \Delta t \times (v_{final} - u_{initial})\]

= \[\frac {1}{2} \times \Delta t \times a(\Delta t)\]

= \[\frac {1}{2} \times \Delta t \times a(\Delta t)\]

= \[\frac{1}{2} \times a(\Delta t)^2\]

Area of rectangle OEBA = length x breadth

      = OA x OB

      = \[\Delta t \times u_{initial}\]

                Hence, the total distance traveled, ‘s’,  is equal to 

Area of OECA = Area of triangle CEB + Area of rectangle OEBA, which gives the equation

\[S = u_{initial}(\Delta t) + \frac {1}{2} a (\Delta t)^2\]

  1. Derivation of Second Equation of Motion by Calculus Method

Velocity is the rate of change of displacement. Mathematically, this can be written as:

\[\frac {ds}{dt} = v_{final}\]

Here ‘ds’ is the small change in the displacement in a given small interval of time ‘dt’.

\[ds = v_{final} dt\]

Putting the first equation of Motion and eliminating the value of the final speed

\[v_{final} - u_{initial} = a\Delta t\] in the above equation, we get:

\[\int_{u_{initial}}^{v_{final}} dv = a \int_{0}^{t} dt\]

\[S =  \frac {(u_{initial} + a\Delta t) + u_{initial}}{2} \times \Delta t\]

\[S= u_{initial} \Delta t + \frac {1}{2} a \Delta t^2\]


Derivation of third Equation of Motion

The third equation of Motion is given as \[v^{2}_{final} - u^{2}_{initial} =2as\]. This shows the relation between the distance and speeds.

  1. Derivation of Third Equation of Motion by Algebraic Method

Let’s assume an object starts moving with an initial speed of \[u_{initial}\] and is subject to acceleration ‘a’. The second equation of Motion is written as \[S = u_{initial} (\Delta t) + \frac {1}{2} a (\Delta t)^2\] Putting the value of \[\Delta t\] from the first equation of Motion, which is \[\Delta t = \frac {v_{final} - u_{initial}}{a}\], we get the following equation

\[S = u_{initial} \lgroup \frac {v_{final} - u_{initial}}{a} \rgroup + \frac {1}{2} a \lgroup \frac {v_{final} - u_{initial}}{a} \rgroup^2\]

On solving the above equation, we obtain the third equation of Motion which is

\[v^{2}_{final} - u^{2}_{initial} = 2as\]

  1. Derivation of third Equation of Motion by Graphical Method?

Total distance travelled by an object is equal to the area of trapezium OECA, consider Figure 3.

\[\text {Area of a trapezium =} \frac{\text{(sum of the parallel sides ) x altitude}}{\text{2}}\]

\[\text{Area of a trapezium OECA} = \frac {(OE+CA) \times EB}{2} = \frac {(u_{initial}  + v_{final} \times \Delta t)}{2}\]

Eliminating time interval \[\Delta t\] from the above equation by using the first equation of Motion, such as \[\Delta t = \frac {v_{final} - u_{initial}}{a}\]

Hence, we can write;

\[\text{Area of a trapezium OECA} = s = \frac {(u_{initial} + v_{final})}{2} \times \frac {v_{final} - u_{initial}}{a}\]

\[S = \frac {(u_{initial} + v_{initial})}{2} \times \frac {v_{final} - u_{initial}}{a}\]

\[2as = v^2_{final} - u^2_{initial} \] 

  1. Derivation of third Equation of Motion by Calculus Method

As discussed in previous sections we can term acceleration and velocity in a mathematical form as below: 

\[\frac {dv}{dt} = a ……(A)\]

\[\frac {ds}{dt} = v_{final} ……(B)\]

Multiply \[\frac {ds}{dt}\] on both sides of equation A,

\[\frac {ds}{dt} \lgroup \frac {dv}{dt} \rgroup = a \lgroup \frac {ds}{dt} \rgroup …… (C)\]

\[\frac {ds}{dt}\] from equation (B) in equation (C),

\[v \lgroup \frac {dv}{dt} \rgroup = a \lgroup \lgroup \frac {ds}{dt} \rgroup \]

We can write the above equation in integral form as below-

\[\int_{u_{initial}}^{v_{final}} vdv = a \int_{0}^{s} ds \]

On solving the above equation, we obtain the third equation of Motion.


Equations of Motion – Explained along with its Derivation.

Explanation of Derivation of Equations of Motions is Available at Vedantu.

For the students, it is important to learn the topic of equations of Motion and their Derivation. There are basically three equations of Motion. The first one is the equation for velocity-time relation, the second is the equation for position-time relation, and the last one is the equation for the position velocity relation. It is important for the students to learn all of these three laws in a detail, and hence Vedantu provides the students, a complete explanation of the Equations of Motion.

FAQs on Derivation of Equation of Motion

1. What are Equations of Motions and what do they describe?

The equations of Motions are the kinetic Motions, and the basic Motion of the object can be understood by it. It explains the position of the object, as well as its acceleration and object. In other words, equations of Motion describe the behaviour, that is to say, the Motion of the physical object in the given time. There are three equations of the Motions and each of the equations describes the Motions of an object in different dimensions, such as 1D, 2D, and 3D.

2. Is the topic of Equations of Motions difficult to understand?

No, the equations of Motions are not difficult to understand at all. There are only three equations which you have to learn only once, and if you have understood the said three equations then the topic of equations of Motions is going to be extremely easy for you. Also, if you find the topic difficult then you must solve a few questions using the equations of Motions, because if you lack the practice of the topic, then it is really difficult to solve understand the equation of Motion.

3. Is it necessary for me to learn the Equation of Motion and its Derivation, or can I skip it?

No, you must not skip the topic of the Equation of Motion and its Derivation, because it is absolutely necessary for you to master this topic. As said earlier, there are three equations of motion that describe the behavior of the object in each of the dimensions, and each of these equations is important for the students of physics. Also, the equations of Motions are included in the syllabus for a reason, which in itself implies that it is important for the students to learn the topic.

4. Where can I find the complete explanation of the Equations of Motion Derivations?

If you are in search of the explanation of the equation of Motion and its Derivations, then it is time to end your search because you have already arrived at the right place. Yes, Vedantu provides the complete explanation of the equations of Motion Derivations, with the formulas of all the three equations, that is to say, equation of motion in one dimension, two dimensions, and three dimensions. And also, Vedantu provides examples of the same, so that it becomes easy for you to understand the equations of Motions.

5. Why should I choose Vedantu for the Explanation of Equations of Motion Derivation?

You must choose Vedantu for the explanation of Equations of Motion Derivation, because Vedantu does not only provide the equations of the Motions, but it also explains the way in which they are derived. And this explanation of Derivation is provided in a simple and lucid language so that you can understand it without any hassle. And since it explains how each of the equations of the Motions is derived, it becomes extremely easy for you to remember the equations as well.