Types of the Equation of the Motion :
There are three equations of Motion which are given below:
\[\ v_{final} - u_{initial}= a\Delta t\]
\[\ S = u_{initial} (\Delta t)+\frac {1}{2}a(\Delta t)^2\]
\[\ v^2_{final} - u^2_{initial} = 2as\]
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 \[\Delta t\]
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:
\[v_{final} - u_{initial} = at\]
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’.
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.
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
\[S = u_{initial}(\Delta t) + \frac {1}{2} a (\Delta t)^2\]
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.
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
\[v^{2}_{final} - u^{2}_{initial} = 2as\]
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;
\[\text{Area of a trapezium OECA} = s = \frac {(u_{initial} + v_{final})}{2} \times \frac {v_{final} - u_{initial}}{a}\]
\[S = \frac {(u_{initial} + v_{initial})}{2} \times \frac {v_{final} - u_{initial}}{a}\]
\[2as = v^2_{final} - u^2_{initial} \]
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 the three fundamental equations of motion for an object moving with uniform acceleration?
The three equations of motion describe the relationship between displacement, velocity, acceleration, and time for an object in motion. For the 2025-26 CBSE syllabus, they are:
- First Equation (Velocity-Time Relation): v = u + at
- Second Equation (Position-Time Relation): s = ut + (1/2)at²
- Third Equation (Position-Velocity Relation): v² = u² + 2as
Here, 'u' is the initial velocity, 'v' is the final velocity, 'a' is the constant acceleration, 't' is the time interval, and 's' is the displacement.
2. What is the single most important condition that must be met to apply these three equations of motion?
The most critical condition is that the acceleration ('a') of the object must be constant. If the acceleration is changing, these equations are not valid. For example, they can be used for a ball in freefall (where acceleration 'g' is constant) but not for a car accelerating and braking unpredictably in traffic.
3. What are the different methods used to derive the equations of motion?
There are three primary methods taught in the NCERT curriculum to derive the equations of motion:
- Algebraic Method: Uses basic algebraic manipulation of the definitions of acceleration and average velocity.
- Graphical Method: Involves analysing the slope and area of a velocity-time graph. This method is highly visual and a key topic in Class 9.
- Calculus Method: Uses integration and differentiation, starting from the fundamental definitions a = dv/dt and v = ds/dt. This is typically introduced in Class 11.
4. How is the second equation of motion, s = ut + ½at², derived using the graphical method?
Using the graphical method, the displacement (s) is represented by the area under the velocity-time graph. For an object with initial velocity 'u' and final velocity 'v' over time 't', this area forms a trapezium. This area can be split into:
- A rectangle with area = u × t.
- A triangle with area = ½ × base × height = ½ × t × (v-u).
Since we know from the first equation that (v-u) = at, the triangle's area becomes ½ × t × (at) = ½at². Adding the areas gives the total displacement: s = ut + ½at².
5. Why is it important to understand the derivation of these equations, not just memorise the formulas?
Understanding the derivation is crucial for deep conceptual clarity. It helps you to:
- Grasp the fundamental relationship between displacement, velocity, and acceleration.
- Recognise the assumptions made, such as the requirement of constant acceleration.
- Apply the concepts to solve complex problems, rather than just plugging numbers into a formula.
- Develop a stronger foundation for more advanced topics in Physics, like projectile motion and work-energy theorems.
6. How does the calculus method of derivation differ from the graphical method?
The key difference lies in their approach. The graphical method is a visual representation that relies on geometric shapes (area of a trapezium) to find total displacement. The calculus method is more fundamental; it treats motion as a series of infinitesimal (extremely small) changes. By integrating the instantaneous velocity (v = ds/dt) with respect to time, we can precisely sum up these tiny displacements to find the total displacement, providing a more powerful tool that forms the basis of modern physics.
7. Can the standard equations of motion be used for an object in uniform circular motion?
No, the standard linear equations of motion cannot be directly applied to uniform circular motion. While the object's speed might be constant, its velocity is continuously changing because its direction of motion is always changing. This change in velocity means there is an acceleration (centripetal acceleration), but it is not constant in direction. Therefore, the conditions for using v = u + at, etc., are not met. Rotational motion has its own set of kinematic equations.
