Derivation of Equation of Motion: Methods and Solved Examples

Collegedunia Team logo

Collegedunia Team Content Curator

Content Curator

Derivation of the equation of motion is the mathematical method through which the three equations of motion are derived. Algebraic, Graphic, and Calculus methods are used to calculate various parameters such as time, velocity, acceleration, or distance in a given problem of motion related to daily life. Of these parameters, u, v, a, and s are vector quantities. Vectors going in one direction are positive and vectors in the opposite direction are negative. Equations of motion apply to uniformly accelerated motion. Moving objects have momentum, and forces cause variations in them. The total momentum in an explosion or collision is conserved and stays the same.

Read More: Potential Energy

Key Terms: Equations of Motion, Equations of Motion Derivation, Moment Inertia, Motion, Distance, Acceleration, Velocity, Time


Equations of Motion

[Click Here for Sample Questions]

There are three equations of motions in physics used to calculate various parameters such as time, acceleration, Velocity and Distance. These equations are as follows.

  1. First Equation of Motion: v = u + at
  2. Second Equation of Motion: s = ut + ½ at2
  3. Third Equation of Motion: v2 = u2 + 2as

Where

  • v is final Velocity 
  • u is initial velocity 
  • t is the time taken 
  • a is acceleration
  • s is the distance travelled

Equations of Motion

Equations of Motion


Methods to Derive Equations of Motion

[Click Here for Previous Year Questions]

There are three methods used to derive the equations of motion, these methods are as follows: 

  1. Derivation using Algebraic Method
  2. Derivation using Graphical Method
  3. Derivation using Calculus Method

Motion in a Straight Line Video Explanation

Also Read:


Derivation of First Equation of Motion

[Click Here for Sample Questions]

Let’s imagine a moving body, travelling with uniform acceleration in a straight line. So, let the acceleration be a, initial velocity be u, the time period be t, distance travelled be s and final velocity be v.

Algebraic Method

The acceleration of a body is said to be the rate of change of velocity.

So, acceleration is represented mathematically as: a = (v-u)/t

Here v is the final velocity and u is the initial velocity.

By Rearranging the equation, we get the first equation of motion as: v = u + at

Graphical Method

In the below graph,

  • The velocity changes in time t from A to B at a uniform rate.
  • BC is the v and OC is t.
  • A perpendicular is drawn from B to OE, A parallel line is drawn from A to D and A perpendicular is drawn from B to OC.

Derivation of First Equation of Motion

Derivation of First Equation of Motion

So, we know from the graph that

BC = BD + DC

Hence, v = BD + DC

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

Therefore, v = BD + u (since OA = u) (Equation 1)

Now,

a = slope of line AB

a = BD/AD

Since AD = AC = t,

BD = at (Equation 2)

Equation 1 + equation 2, we get: v = u + at

Read More: Angular Momentum

Calculus Method

Since acceleration is the rate of change of velocity,

a=dv/dt

Rearranging the above equation, we get

adt=dv

Integrating both sides, we get

tadt = ∫vudv

at = v−u

Rearranging, we get v=u+at

Read More: Equations of Motion


Derivation of Second Equation of Motion

[Click Here for Previous Year Questions]

The second equation of motion can be derived via the following methods:

Algebraic Method

The rate of change of displacement is known as velocity.

Velocity = Displacement/Time

Displacement = Velocity×Time

Use average velocity in the place of velocity, if the velocity is not constant.:

Displacement={(Initial Velocity + Final Velocity)/2} × Time

s={(u+v)/2} × t

s=[{u+ (u + at)}/2] × t

s=[{2u + at/2}]×t

s=({2u/2} + {at/2})×t

s=(u + ½ at)×t

On further simplification, the equation becomes s=ut+½ at2

Read More: Application and Uses of Convex Mirror, Examples

Graphical Method

From the graph above, 

Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD

s=(½ AB×BD)+(OA×OC)

Derivation of Second Equation of Motion

Derivation of Second Equation of Motion

Since BD = EA,

s=(½ AB×EA) + (u × t)

As EA = at, 

s=½ × at × t+ ut

So, the equation becomes s= ut+ ½ at2

Calculus Method

The rate of change of displacement is known as velocity.

Mathematically, this is:

v=ds/dt

ds=vdt

ds=(u + at) dt

ds=(u + at) dt = (udt + atdt)

On further simplification, the equation becomes:

0sds=∫0tudt+∫0tatdt

s = ut + ½ at2


Derivation of Third Equation of Motion

[Click Here for Sample Questions]

The third equation of motion can be derived by: 

Algebraic Method

Displacement is an object’s rate of change of position. Mathematically, this is:

Displacement=(Initial Velocity+Final Velocity/2) × t

s=(u+v/2) × t

From the first equation of motion, we know that

v= u + at

t = (v−u)/a

s=(v+u/2) (v−u/a)

s=(v2−u2/2a)

2as=v2−u2

That results in v2 = u2 + 2as

Read More: Types of Motors and its Application

Graphical Method

s is the total distance travelled given by the Area of trapezium OABC.

S = Height × ½ (Sum of Parallel Sides)

S=(OA+CB)×OC

Derivation of Third Equation of Motion

Derivation of Third Equation of Motion

Here, CB = v, OA = u and OC = t, 

so,

s= ½ (u+v) × t

t = (v – u)/ a

s= ½ ((u+v) × (v-u))/a

s= ½ (v+u) × (v-u)/a

s = (v2-u2)/2a

Third equation of motion is: v2 = u2 + 2as

Read More: Unit of Frequency

Calculus Method

The rate of change of velocity is Acceleration:

a= dv/dt (1)

The rate of change of displacement is velocity.

v= ds/dt (2)

Cross multiplying (1) and (2), we get

ads/dt=vdv/dt

0sads=∫uvvds

as = (v2−u2)/2

This results in v2 = u2 + 2as

Read More: Capacitor Types


Things to Remember

  • Derivation of the equation of motion is the mathematical method through which the three equations of motion are derived using different methods, these equations are used to calculate various parameters such as time, velocity, acceleration or distance in a given problem of motion related to daily life. 
  • When you know the derivation of a given equation, you tend to understand the basic concept behind that equation, and that in turn helps to solve problems in a better way.
  • let the acceleration be a, initial velocity be u, the time period be t, distance traveled be s and final velocity be v.
  • There are three methods used to derive the equations of motion, these methods are as follows: Derivation using Algebraic Method, Derivation using Graphical Method, Derivation using Calculus Method.

Read More:


Previous Year Questions

  1. A charged particle enters a uniform magnetic field with a certain speed at… [JKCET 2006]
  2. A person walked up a stalled escalator in 90s… [NEET 2006]
  3. An automobile in travelling at 50 km/h, can be stopped at a distance of… [KEAM 2003]
  4. An object is thrown vertically upward with a speed of… [TS EAMCET 2017]
  5. An automobile travelling at 50 km/h, can be stopped at a distance of… [JCECE 2005]
  6. A body of mass 2kg has an initial velocity of… [BHU UET]
  7. A body dropped from a height h with initial velocity zero… [NEET 1996]
  8. A ball is dropped from a high rise platform at… [NEET 2010]
  9. A ball is thrown vertically downward with a velocity of 20m/s from the… [NEET 2020]
  10. A ball is thrown vertically upward. It has a speed of… [NEET 2005]

Sample Questions

Ques: What are the three equations of motion? (3 marks)

Ans: The three equations of motion are as follows:

  1. v = u+at
  1. s = ut+ (½ ) at2
  1. v2 = u2+2as

Ques: In what three ways can an equation be derived? (3 marks)

Ans: There are three ways through which an equation can be derived, these are: 

  • By Calculus Method
  • By Algebraic Method
  • By Graphical Method

Ques: What is the derivation of a motion? (2 marks)

Ans: Derivation of the equation of motion is the mathematical method through which the three equations of motion are derived using different methods.

Ques: Why is a derivation of an equation necessary to understand? (2 marks)

Ans: When you know the derivation of a given equation, you tend to understand the basic concept behind that equation, and that in turn helps to solve problems in a better way.

Ques: What parameters are found out by using the equations of motion? (2 marks)

Ans: The equations of motion are used to calculate various parameters such as time, velocity, acceleration or distance in a given problem of motion related to daily life. 

Read more: Electrical Resistance

Ques: What to do when the acceleration in a given situation is not constant? (2 marks)

Ans: When deriving an equation and the acceleration given is not constant, one can use average acceleration in its place.

Ques: Use the calculus method to derive the first equation of motion. (3 marks)

Ans: Since acceleration is the rate of change of velocity, it can be mathematically written as:

a=dv/dt

Rearranging the above equation, we get

adt=dv

Integrating both sides, we get

t0adt=∫vudv

at=v−u

Rearranging, we get v=u+at

Ques: Derive the third equation of motion using the calculus method. (3 marks)

Ans: We know that acceleration is the rate of change of velocity and can be represented as:

a= dv/dt (1)

We also know that velocity is the rate of change of displacement and can be represented as:

v= ds/dt (2)

Cross multiplying (1) and (2), we get

ads/dt=vdv/dt

0sads=∫uvvds

as = (v2−u2)/2

v2 = u+ 2as

Read More: Rolling Friction

Ques: Derive the second equation of motion using the calculus method. (3 marks)

Ans: Velocity is the rate of change of displacement.

Mathematically, this is:

v=ds/dt

ds=vdt

ds=(u + at) dt

ds=(u + at) dt = (udt + atdt)

On further simplification, the equation becomes:

0sds=∫0tudt+∫0tatdt

s = ut + ½ at2


Read Also: 

CBSE CLASS XII Related Questions

  • 1.
    Answer the following giving reason:
    (a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
    (b) The saturation current in case (a) is different for different intensity.
    (c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.


      • 2.
        A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:
        vertically held bar magnet is dropped along the axis of a copper ring

          • zero
          • less than \( g \)
          • \( g \)
          • greater than \( g \)

        • 3.
          The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.


            • 4.
              In the circuit, three ideal cells of e.m.f. \( V \), \( V \), and \( 2V \) are connected to a resistor of resistance \( R \), a capacitor of capacitance \( C \), and another resistor of resistance \( 2R \) as shown in the figure. In the steady state, find (i) the potential difference between P and Q, (ii) the potential difference across capacitor C.
              potential difference across capacitor C


                • 5.
                  Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), \( r \) being the separation between the two charges \( (q_1, q_2) \) and \( (q_2, q_3) \). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1, q_2 \), and \( q_3 \) are such that:
                  variation of Coulomb force

                    • \( q_2<q_1<q_3 \)
                    • \( q_3<q_1<q_2 \)
                    • \( q_1<q_2<q_3 \)
                    • \( q_2<q_3<q_1 \)

                  • 6.
                    Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
                    Three batteries E1, E2, and E3 of emfs and internal resistances

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show