Derivation of Three Equations of Motion – Step-by-Step Explanation

Yashwant Parihar,

The derivation of the three equations of motion is a basic and important topic in physics and kinematics. These equations explain the relationship between distance, displacement, velocity, time, and acceleration for an object moving in a straight line with uniform acceleration. Students of Class 9 and Class 10 often learn these formulas to solve numerical problems and exam questions. In this article, you will find a step-by-step derivation of equations of motion, along with simple explanations and examples, to help you clearly understand the motion formulas in physics.

Basic Terms in Motion

1. Distance

Distance is the total length travelled by an object for a particular path. It is a path-dependent quantity. Distance cannot be negative, but can be zero.

Key Points:

  • It is a scalar quantity (has only magnitude, no direction)
  • It is always positive or zero
  • It depends on the path taken, not just the starting and ending points
  • The SI unit of distance is the meter (m)

2. Displacement

Displacement is the length covered by the object from its initial position to its final position. It cannot depend on the path.

Key Points:

  • It is a vector quantity (has both magnitude and direction)
  • It can be positive, negative, or zero
  • It depends only on the start and end points, not the path taken
  • The SI unit of displacement is the meter (m)

Difference Between Distance and Displacement

DistanceDisplacement
It is a scalar quantity It is a vector quantity
It is always positive or zeroIt can be positive, negative, or zero
It depends on the path takenIt depends only on the start and end points
SI unit – MeterSI unit – Meter

3. Velocity

Velocity is the speed of an object in a given direction. Velocity tells us how fast an object is moving and in which direction it is moving.

Key Points:

  • It is a vector quantity (has both magnitude and direction)
  • It can be positive, negative, or zero, depending on the direction of motion
  • It depends on both distance/displacement and time
  • The SI unit of velocity is meters per second (m/s)

Formula of Velocity: Displacement/time

4. Acceleration

The rate of change in velocity.

Key Points:

  • It is a vector quantity (has both magnitude and direction)
  • It can be positive (speeding up), negative (slowing down, also called retardation), or zero (constant velocity)
  • The SI unit of acceleration is meters per second squared (m/s²)

Derivation of the three equations of motion

First Equation of Motion (v = u + at)

Let an object move with initial velocity (u) and with (t) time, the velocity becomes (v) then:-

Acceleration: The rate of change in velocity.

a=v−uta =\frac{v – u}{t}
at=v−uat = v – u
v=u+atv = u + at
Handwritten derivation of first equation of motion

Second Equation of Motion (s = ut + ½at²)

Displacement:- Average velocity x time

S=u+v2×tS = \frac{u+v}{2} \times t

From Equation 1

u+at+u2×t\frac {u+at+u}{2} \times t

Multiply t in the bracket

S=2ut+at22S = \frac {2ut + at^2}{2}
S=2ut2+at22S = \frac {2ut}{2} + \frac {at^2}{2}
S=Ut+12at2S = Ut + \frac {1}{2}at^2
Handwritten derivation of second equation of motion

Third Equation of Motion (v² = u² + 2as)

Squaring equation 1

v2=(u+at)2v^2 = (u +at)^2
v2=u2+a2t2+2vatv^2 = u^2 + a^2t^2 + 2vat
v2=u2+2a[Ut+12at2]v^2 = u^2 + 2a [Ut + \frac {1}{2}at^2]
v2=u2+2asv^2 = u^2 + 2as
Handwritten derivation of third equation of motion

Question Related to Equation of Motions

A car is moving at 20/sec, and suddenly it applies the brakes and stops the car after 20m distance. Then, find the time taken by car to stop it?

From equation third

v2=u2+2asv^2 = u^2 + 2as
02=202+2×a×200^2 = 20^2 + 2 \times a \times20
0=400+40a0 = 400 + 40a
a=−40040a = \frac {-400}{40}
𝐚 = -𝟏𝟎\textbf{a = -10}
v=u+atv = u + at
0=20−10t0 = 20 -10t
10t=2010t = 20
t=2010t = \frac {20}{10}
 𝐭 = 𝟐 𝐬𝐞𝐜\textbf{ t = 2 sec}

A stone thrown upward with a velocity of 10m/sec and after 6 seconds it reaches the hands of a man. Find the height covered by the stone?

v = ut + 1/2 at^2
10×3 + 1/2 (-10)x3^2
30x-45
-15 ( height cannot be negative )
15

NCERT Class 11 Ncert Solutions Physics