[Physics Class Notes] on Derivation of Equation of Motion Pdf for Exam

There are three equations of Motion which are given below:

1. [ v_{final} – u_{initial}= aDelta 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}= aDelta 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]

2. 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}+ aDelta t)+ u_{initial}}{2} times Delta t]

On rearranging the above equation we get,       

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

         which is the second equation of Motion.

2. 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]

3. 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} = aDelta t] in the above equation, we get:

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

[S =  frac {(u_{initial} + aDelta t) + u_{initia
l}}{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]

2. 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} ] 

3. 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 .

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 provides the students, a complete explanation of the Equations of Motion.