time →
🏃 Chapter 7 · NCERT Science IX

MOTION

Speed, velocity, acceleration — and the three immortal equations that describe every moving object in the universe.

v = u + at s = ut + ½at² v² = u² + 2as
3Equations
2Graph Types
★★★★★Exam Weight
📍 Chapter Snapshot — Types of Graphs
time dist
Distance-Time
Slope = Speed
time vel
Velocity-Time
Slope = Acceleration; Area = Displacement
time
Non-Uniform
Variable acceleration
🎯 Why This Chapter Matters for Exams
📌 Equations of motion numericals are the most-asked questions in board exams (3–5 marks).
📌 Drawing and interpreting distance-time and velocity-time graphs is a guaranteed section.
📌 Uniform circular motion and centripetal acceleration — common MCQ in NTSE/Olympiad.
📌 Derivation of 3 equations of motion by graphical method — 5-mark theory staple.
⚗️ Important Formula Capsules
1st Equationv = u + atm/s
2nd Equations = ut + ½at²m
3rd Equationv² = u² + 2asm²/s²
Speedv = d / tm/s
Accelerationa = (v − u) / tm/s²
💡 Key Concept Highlights
Distance vs DisplacementSpeed vs VelocityUniform MotionNon-uniform MotionAccelerationRetardationD-T GraphV-T GraphCircular MotionScalar vs Vector
📚 What You Will Learn
  • 1Difference between distance and displacement; speed and velocity
  • 2Graphical representation of uniform and non-uniform motion
  • 3Deriving the three equations of motion graphically
  • 4Acceleration, retardation, and free fall under gravity
  • 5Uniform circular motion and concept of centripetal acceleration
🔗 Navigate to Chapter Resources
📄
Full Notes
🧩
MCQs Practice
✅
True / False
📝
NCERT Exercises
🏆 Exam Strategy & Preparation Tips
01
Know All 3 Derivations
Derive all 3 equations using the V-T graph method — these are 5-mark theory questions.
02
Read Graphs Carefully
Area under V-T graph = displacement. This is tested in almost every objective round.
03
Direction Matters
Always specify direction when writing velocity or displacement answers — it's a vector.
04
Do 20 Numericals
Practice at least 20 varied numericals — identify which of the 3 equations to apply.
Chapter 7 · CBSE · Class IX
🚀

Motion

Physics Kinematics Rest and Motion Scalar Quantities Vector Quantities Distance Displacement Uniform Motion Non-uniform Motion Speed Velocity Average Speed Acceleration Distance-Time Graphs Velocity-Time Graphs Equations of Motion Uniform Circular Motion NCERT Class 9 Science
📋
Table of Contents
15 sections
🗺️ Overview

Motion is the phenomenon in which an object changes its position with time with respect to a fixed reference point or an observer. Almost every event occurring around us involves some kind of motion, from the movement of vehicles on roads to the revolution of planets around the Sun.

The study of motion forms the foundation of Mechanics, which is an important branch of Physics dealing with the behaviour of moving objects and the forces acting on them. Understanding motion helps us explain numerous natural and technological processes such as transportation systems, satellite communication, sports, and planetary movements.

📘 Definition
💡 Rest and Motion are Relative Concepts
ℹ️ Reference Point
A reference point is a fixed point or object with respect to which the position of another object is determined.
In everyday life, buildings, trees, electric poles, milestones, or the surface of the Earth are often used as reference points.

Example:
If a car moves from one milestone to another, the milestones act as reference points to determine the change in position of the car.
📌 Essential Conditions for Motion
✏️ Examples of Motion in Daily Life
  • Fans rotating inside a room.
  • Clouds drifting across the sky.
  • Water flowing in rivers.
  • Players running on a field.
  • Hands of a clock moving continuously.
  • The Moon revolving around the Earth.
  • The Earth rotating about its axis.
ℹ️ Why Do We Study Motion?
Understanding motion helps us:
  • Design automobiles, trains, aircraft and spacecraft.
  • Predict planetary and satellite movements.
  • Determine travel time and distances.
  • Develop traffic management systems.
  • Understand sports movements and machine operations.
  • Study earthquakes, wind movement and ocean currents.
📘 Uniform Motion
📘 Non-Uniform Motion
⚖️ Comparison
Difference Between Uniform and Non-Uniform Motion
Uniform Motion Non-Uniform Motion
Equal distances in equal intervals of time. Unequal distances in equal intervals of time.
Speed remains constant. Speed changes continuously.
Acceleration is zero. Acceleration is non-zero or variable.
Distance-time graph is a straight line. Distance-time graph is curved.
Example: Train moving steadily. Example: Car moving in traffic.
✏️ Example
Solved Example
A bus travels 120 km in 2 hours at constant speed. Determine its speed.
\[\text{Speed}=\frac{\text{Distance}}{\text{Time}}\]
Given Distance → Given Time → Apply Formula → Calculate Speed
Given, \[ d=120\;km \] and, \[ t=2\;h \] Therefore, \[ v=\frac{120}{2} \] \[ v=60\;km/h \]
he speed of the bus is 60 km/h.
⚡ Exam Tip
❌ Common Mistakes
  • Writing "motion means movement" without mentioning change in position and time.
  • Assuming a moving train passenger is always in motion.
  • Thinking that constant speed automatically means constant velocity.
  • Ignoring the importance of direction while discussing velocity.
  • Confusing uniform motion with motion at high speed.
📋 CBSE Case Study (HOTS)

A student sits inside a bus moving at constant speed. The student places a book on the seat beside him. Another person stands on the roadside and observes the bus.

Questions:

  1. Is the book at rest or in motion for the student?
  2. Is the book at rest or in motion for the roadside observer?
  3. Which concept of motion is illustrated here?

Answers:

  1. The book is at rest with respect to the student.
  2. The book is in motion with respect to the roadside observer.
  3. This illustrates that rest and motion are relative concepts.
🎨 SVG Diagram
Illustrative Diagram of Motion and Reference Point
-40 m -30 m -20 m -10 m 0 m +10 m +20 m x Origin (x = 0) Reference Object (Stationary Frame) Object of Interest Position: x(t) Velocity Vector v 0 ONE-DIMENSIONAL KINEMATICS FRAME OF REFERENCE & MOTION DIAGRAM
🌟 Important Terms to Remember
🚀

Physical Quantities

📋
Table of Contents
13 sections
📘 Definition
🤔 Did You Know?
Why Do We Measure Physical Quantities?
Measurements are essential because they allow scientists and engineers to describe physical events accurately and communicate observations in a universally accepted manner.

Measurement helps us to:
  • Compare different objects and events.
  • Study motion scientifically.
  • Perform calculations and predictions.
  • Develop technology and engineering applications.
  • Maintain uniform scientific standards throughout the world.
🔷 Characteristics of a Physical Quantity
🔷 Characteristics
  • It must be measurable.
  • It must have a numerical value.
  • It must possess an appropriate unit.
  • Its measurement should be reproducible under similar conditions.
🗒️ Representation Of Physical Quantities
A physical quantity is represented as: \[ \text{Physical Quantity} = \text{Numerical Value} \times \text{Unit} \] Example:
If the length of a pencil is 15 centimetres, then \[\text{Length}=15\;cm\] Here,
  • 15 is the numerical value.
  • cm is the unit.
🗂️ Types / Category
Classification of Physical Quantities
Based on the requirement of direction, physical quantities are broadly classified into:
Scalar Quantities
A scalar quantity is a physical quantity that is completely described by its magnitude (numerical value and unit) only and does not require direction.

In other words, a scalar quantity answers only the question "How much?"

Examples of Scalar Quantities:

  • Mass
  • Distance
  • Time
  • Temperature
  • Area
  • Volume
  • Density
  • Speed
  • Energy
  • Work
Examples:
  • A person runs at a speed of 5 m/s.
  • The mass of a book is 500 g.
  • The distance between two cities is 250 km.
Notice that no direction is necessary to describe these quantities completely.
Vector Quantities
A vector quantity is a physical quantity that requires both magnitude and direction for its complete description.

  • How much?
  • In which direction?
Examples of Vector Quantities:
  • Displacement
  • Velocity
  • Acceleration
  • Force
  • Momentum
  • Weight
Examples:
  • A car moves with a velocity of 20 m/s towards the east.
  • A force of 50 N acts towards the north.
  • A particle has an acceleration of 3 m/s² downward.
If direction is not specified, a vector quantity remains incomplete.
🌟 Importance of Scalars and Vectors in Motion
⚖️ Difference Between Scalar and Vector Quantities
Scalar Quantity Vector Quantity
Has only magnitude. Has both magnitude and direction.
Can be completely described by a number and unit. Requires both numerical value and direction.
Simple arithmetic rules are usually sufficient. Special rules of vector addition are required.
Examples: Mass, Time, Distance, Speed. Examples: Displacement, Velocity, Force.
💡 Concept Builder: Distance and Displacement
✏️ Example
Solved Example
A cyclist travels with a velocity of 12 m/s towards the north. Is this quantity scalar or vector?
A quantity possessing both magnitude and direction is a vector quantity.
Identify Magnitude → Identify Direction → Classify Quantity
  • Magnitude = 12 m/s
  • Direction = North
Since both magnitude and direction are present, the given quantity is a vector quantity.
Velocity is a vector quantity.
⚡ Exam Tip
❌ Common Mistakes
  • Considering speed and velocity to be identical.
  • Writing displacement as a scalar quantity.
  • Ignoring units while writing physical quantities.
  • Assuming every moving object possesses vector quantities only.
  • Forgetting that mass and time are scalar quantities.
📋 CBSE Case Study (HOTS)

During a race, a runner covers a total path length of 400 m around a circular track and reaches the starting point again.

Questions:

  1. What is the distance travelled?
  2. What is the displacement?
  3. Which quantity is scalar?
  4. Which quantity is vector?

Answers:

  1. Distance = 400 m
  2. Displacement = 0 m
  3. Distance is a scalar quantity.
  4. Displacement is a vector quantity.
🎨 Illustrative Diagram: Scalar and Vector Quantities
Comparison: Scalar vs. Vector Quantities Scalar 20 m Defined by Magnitude only Vector 20 m Magnitude + Direction ( v 0)
🌟 Important Terms to Remember
🚀

Distance and Displacement

📋
Table of Contents
14 sections
📘 Definition
🤔 Did You Know?
Why Do We Need Two Different Quantities?
Suppose a person walks around a park and returns to the starting point. The person has travelled a certain path length, but the initial and final positions are the same.

In such situations, merely knowing the path length is insufficient. We also need to know the change in position of the object. Therefore, Physics uses:
  • Distance to describe the actual path travelled.
  • Displacement to describe the shortest change in position.
📘 Definition

Distance

📘 Definition
Distance is the total length of the actual path travelled by an object during its motion, irrespective of direction.

Distance depends upon the entire path followed by the object and not merely on its initial and final positions.
🔷 Characteristics
  • Distance is a scalar quantity.
  • It possesses only magnitude and no direction.
  • It is always positive.
  • Its SI unit is metre (m).
  • Its value can never be zero if the object has moved.
  • Distance travelled is always greater than or equal to the magnitude of displacement.
📘 Definition

Displacement

Displacement is the shortest straight-line distance between the initial and final positions of an object along with the direction of motion.

Displacement represents the overall change in position of the object and depends only on the starting and ending points.
🔷 Characteristics
  • Displacement is a vector quantity.
  • It has both magnitude and direction.
  • Its SI unit is metre (m).
  • It may be positive, negative or zero depending on the chosen direction.
  • It is independent of the actual path followed.
  • The magnitude of displacement can never exceed the distance travelled.
💡 Concept of Path Length and Shortest Distance
🔗 Mathematical Relation Between Distance and Displacement
The following relation always holds: \[\text{Distance} \geq \text{Magnitude of Displacement}\] Equality occurs only when the object moves along a straight line without changing its direction. \[\text{Distance} = |\text{Displacement}|\] Condition:
Straight-line motion in a single direction.
📌 Motion Along a Straight Line
📌 Distance and Displacement in Circular Motion
⚖️ Comparison
Distance Displacement
Total length of actual path travelled. Shortest straight-line distance between initial and final positions.
Scalar quantity. Vector quantity.
Has magnitude only. Has magnitude and direction.
Depends upon the path followed. Depends only on initial and final positions.
Always positive. Can be positive, negative or zero.
Can never be less than displacement. Can never exceed distance.
✏️ Example
Solved Example
A boy walks 12 m north and then 5 m south. Find the distance and displacement.
  • Distance is the total path length.
  • Displacement is the net change in position.
Calculate Total Path → Determine Final Position → Find Net Change
Distance: \[ \text{Distance}=12+5=17\;m \] Displacement: \[ \text{Displacement}=12-5=7\;m\;north \]
Distance travelled = 17 m
Displacement = 7 m north
A runner completes one full lap of a circular track of radius 35 m. Find the distance and displacement.
Distance: \[ \text{Distance}=2\pi r \] \[ =2\times\frac{22}{7}\times35 \] \[ =220\;m \] Displacement: \[ \text{Displacement}=0\;m \]
Distance travelled = 220 m
Displacement = 0 m
⚡ Exam Tip
❌ Common Mistakes
  • Writing distance and displacement as identical quantities.
  • Forgetting to specify direction in displacement.
  • Assuming displacement cannot be zero when an object has moved.
  • Using the total path length to calculate displacement.
  • Confusing the shortest distance with the actual path length.
📋 CBSE Case Study (HOTS)

A student starts from point A, walks 30 m east, then 40 m north, and finally stops at point C.

Questions:

  1. Calculate the total distance travelled.
  2. Calculate the magnitude of displacement.
  3. Which quantity depends on the path followed?

Solution:

Distance: \[ \text{Distance}=30+40=70\;m \] Displacement: Using Pythagoras theorem, \[ \text{Displacement} = \sqrt{30^2+40^2} \] \[ = \sqrt{900+1600} \] \[ = \sqrt{2500} = 50\;m \]

Answers:

  1. Total distance = 70 m.
  2. Displacement = 50 m.
  3. Distance depends on the path followed.
🎨 SVG Diagram
Illustrative Diagram: Distance and Displacement
Distance vs. Displacement A B Actual Path (Scalar: Distance) Shortest Path (Vector: Displacement s disp)
🌟 Important Terms to Remember
🚀

Speed and Velocity

📋
Table of Contents
15 sections
🗺️ Overview
While studying motion, merely knowing that an object changes its position is not sufficient. We also need to know how fast the object moves and in which direction it moves. These aspects are described by two important physical quantities: Speed and Velocity.

Speed and velocity are among the most important concepts of mechanics. They form the basis of numerical problems, graphical interpretation of motion, equations of motion, and applications in transportation, engineering, sports, and astronomy.
🤔 Did You Know?
Why Do We Need Speed and Velocity?
Consider two students travelling from home to school. Both cover the same distance, but one reaches earlier than the other.

The difference lies in how rapidly they travelled. Physics therefore introduces the concept of speed. Furthermore, two objects moving at the same speed may move in different directions. To completely describe such motion, we require velocity.
📘 Speed
🗂️ Types of Speed
Uniform Speed
An object is said to move with uniform speed if it covers equal distances in equal intervals of time, however small these intervals may be.
Example:
A train covers 60 km every hour throughout its journey.
Non-Uniform Speed
An object is said to move with non-uniform speed if it covers unequal distances in equal intervals of time or equal distances in unequal intervals of time.
Example:
A car moving through city traffic.
Average Speed
During non-uniform motion, speed changes continuously. In such situations, we define average speed.
Average speed is defined as the total distance travelled divided by the total time taken. \[ \text{Average Speed} = \frac{\text{Total Distance Travelled}} {\text{Total Time Taken}} \] Symbolically, \[ v_{avg} = \frac{d_{total}} {t_{total}} \]
Average speed is calculated using total distance, not displacement.
📘 Definition

Velocity

📘 Definition

Average Velocity

🔗 Relationship Between Speed and Velocity
Speed and velocity have the same units and dimensions but differ in their physical meaning.

Since distance is always greater than or equal to displacement, \[ \text{Average Speed} \geq |\text{Average Velocity}| \] Equality occurs only when an object moves along a straight line without changing its direction.
⚖️ Difference Between Speed and Velocity
Speed Velocity
Distance travelled per unit time. Displacement per unit time.
Scalar quantity. Vector quantity.
Magnitude only. Magnitude and direction.
Cannot be negative. May be positive, negative or zero.
Depends on total path length. Depends on displacement.
Average speed is always greater than or equal to average velocity. Average velocity is always less than or equal to average speed.
🔗 Conversion of Units
Since, \[ 1\;km=1000\;m \] and, \[ 1\;h=3600\;s \] Therefore, \[ 1\;km/h = \frac{1000}{3600}\;m/s \] \[ 1\;km/h = \frac{5}{18}\;m/s \] Hence,
Conversion from km/h to m/s
\[ m/s = \frac{5}{18} \times km/h \]
Conversion from m/s to km/h
\[ km/h = \frac{18}{5} \times m/s \]
📐 Derivation
Derivation of the Formula for Speed
Suppose an object travels a distance \(d\) in time \(t\).
By definition, speed is the distance travelled per unit time.

Therefore, \[ \text{Speed} = \frac{\text{Distance}} {\text{Time}} \] Hence, \[ v = \frac{d}{t} \] Rearranging, \[ d=vt \] and, \[ t=\frac{d}{v} \]
✏️ Example
Solved Examples
A car travels 180 km in 3 hours. Find its speed.
Identify Distance → Identify Time → Apply Formula
Given, \[ d=180\;km \] \[ t=3\;h \] Therefore, \[ v=\frac{180}{3} \] \[ v=60\;km/h \]
Speed of the car = 60 km/h.
A person walks 40 m east in 8 s. Find the velocity.
Given, \[ \text{Displacement}=40\;m\;east \] \[ t=8\;s \] Therefore, \[ \vec{v} = \frac{40}{8} = 5\;m/s\;east \]
Velocity = 5 m/s east.
A runner completes one lap of a circular track and returns to the starting point in 200 s. Find the average velocity.
Since, \[ \text{Net Displacement}=0 \] Therefore, \[ \text{Average Velocity} = \frac{0}{200} = 0\;m/s \]
Average velocity = 0 m/s.
⚡ Exam Tip
❌ Common Mistakes
  • Using distance instead of displacement while calculating velocity.
  • Forgetting to write the direction of velocity.
  • Assuming speed and velocity are always numerically equal.
  • Using incorrect conversion factors between km/h and m/s.
  • Assuming average velocity cannot be zero when an object has moved.
📋 CBSE Case Study (HOTS)

A student walks around a circular park of circumference 300 m and returns to the starting point in 150 s.

Questions:

  1. Calculate the average speed.
  2. Calculate the average velocity.
  3. Explain why both answers are different.

Solution:

Average Speed: \[ \text{Average Speed} = \frac{300}{150} = 2\;m/s \] Average Velocity: \[ \text{Average Velocity} = \frac{0}{150} = 0\;m/s \]

Explanation:

The runner has travelled a distance of 300 m but the net displacement is zero because the initial and final positions are identical.

🎨 SVG Diagram
Illustrative Diagram: Speed and Velocity
Illustrative Diagram: Speed vs. Velocity Speed (Scalar) Magnitude only (e.g., 50 km/h) Velocity (Vector) Magnitude + Direction (e.g., 50 km/h North)
🌟 Important Terms to Remember
🚀

Acceleration and Retardation

📋
Table of Contents
16 sections
📘 Definition
🤔 Did You Know?
Why Do We Need Acceleration?
Consider the following situations:
  • A car increases its speed from 20 km/h to 60 km/h.
  • A train slows down before reaching a station.
  • A ball thrown upward gradually slows down.
  • An aeroplane gains speed before take-off.
In all these cases, the velocity of the object changes with time. Physics uses acceleration to measure how rapidly this change occurs.
📘 Definition

Acceleration

📐 Derivation of the Formula of Acceleration
Suppose an object initially moves with velocity \(u\).
After time \(t\), its velocity becomes \(v\).
Therefore,
Change in velocity: \[ =v-u \] By definition, \[ \text{Acceleration} = \frac{\text{Change in Velocity}} {\text{Time}} \] Therefore, \[ a = \frac{v-u}{t} \] Rearranging, \[ at=v-u \] Hence, \[ v=u+at \] This equation becomes the first equation of motion studied later in this chapter.
📌 SI Unit of Acceleration
📌 Dimensional Formula of Acceleration
🗂️ Types of Acceleration
Positive Acceleration
When the velocity of an object increases with time, the acceleration is called positive acceleration.
Examples:
  • A car increasing its speed from 20 km/h to 50 km/h.
  • A train accelerating after leaving a station.
Negative Acceleration
When the velocity decreases with time, the acceleration becomes negative and is called negative acceleration or retardation.
Zero Acceleration
When velocity remains constant, there is no change in velocity.
Therefore, \[ v=u \] Hence, \[ a=\frac{u-u}{t}=0 \] Thus, an object moving with constant velocity has zero acceleration.
📘 Definition

Retardation (Deceleration)

📌 Physical Meaning of Acceleration
⚖️ Difference Between Acceleration and Retardation
Acceleration Retardation
Velocity increases with time. Velocity decreases with time.
Usually positive. Usually negative.
Acts in the direction of motion. Acts opposite to the direction of motion.
Object speeds up. Object slows down.
Example: Accelerating car. Example: Braking train.
✏️ Example
Solved Examples
A car increases its velocity from 10 m/s to 30 m/s in 5 s. Find its acceleration.
Identify \(u\) → Identify \(v\) → Identify \(t\) → Apply Formula
Given, \[ u=10\;m/s \] \[ v=30\;m/s \] \[ t=5\;s \] Therefore, \[ a = \frac{30-10}{5} \] \[ a = \frac{20}{5} \] \[ a=4\;m/s^2 \]
Acceleration of the car = 4 m/s².
A train moving at 25 m/s comes to rest in 5 s. Find its retardation.
Given, \[ u=25\;m/s \] \[ v=0\;m/s \] \[ t=5\;s \] Therefore, \[ a = \frac{0-25}{5} \] \[ a=-5\;m/s^2 \]
Retardation of the train = 5 m/s². The negative sign indicates that the train is slowing down.
🛠️ Application
Applications of Acceleration in Daily Life
  • Design of automobiles and braking systems.
  • Rocket launches and spacecraft missions.
  • Performance analysis in sports.
  • Traffic management and road safety.
  • Manufacturing and industrial machinery.
  • Study of planetary and satellite motion.
⚡ Exam Tip
❌ Common Mistakes
  • Writing acceleration as a scalar quantity.
  • Confusing speed with velocity.
  • Ignoring the negative sign in retardation problems.
  • Using incorrect units such as m/s instead of m/s².
  • Assuming acceleration means only increase in speed.
📋 CBSE Case Study (HOTS)

A motorcycle starts from rest and reaches a velocity of 18 m/s in 6 s.

Questions:

  1. What is the initial velocity?
  2. Calculate the acceleration.
  3. Is the acceleration positive or negative?
  4. What does the obtained value physically represent?

Solution:

Given, \[ u=0\;m/s \] \[ v=18\;m/s \] \[ t=6\;s \] Therefore, \[ a = \frac{18-0}{6} \] \[ a=3\;m/s^2 \] Interpretation: The velocity of the motorcycle increases by \(3\;m/s\) every second.
🎨 SVG Diagram
Illustrative Diagram: Positive Acceleration and Retardation
POSITIVE ACCELERATION & RETARDATION 1. Positive Acceleration (Speeding Up) Condition: Applied Force (F) is in the SAME direction as Velocity (v) v(0) = 10 m/s Constant Applied Force → (Positive Accel.) v(3) = 25 m/s 2. Retardation / Deceleration (Slowing Down) Condition: Applied Force (F) is in the OPPOSITE direction to Velocity (v) v(0) = 25 m/s ← Constant Opposing Force (Retardation) v(3) = 10 m/s
🌟 Important Terms to Remember
🚀

Equations of Motion

📋
Table of Contents
12 sections
📘 Definition
⚙️ Conditions For Applying Equations Of Motion
The equations of motion are valid only under the following conditions:
  • The object must move along a straight line.
  • The acceleration should remain constant throughout the motion.
  • The motion should be uniformly accelerated or uniformly retarded.
  • The same sign convention should be followed throughout the problem.
If acceleration changes continuously or the motion is not in a straight line, these equations cannot be directly applied.
📌 Symbols Used in Equations of Motion
🗂️ Types / Category

The Three Equations of Motion

First Equation of Motion
The first equation of motion gives the relation among initial velocity, final velocity, acceleration, and time.
Formula
\[\boxed{\bbox[2pt]{v=u+at}}\]
Derivation
By definition, \[ a=\frac{v-u}{t} \] Multiplying both sides by \(t\), \[ at=v-u \] Adding \(u\) to both sides, \[ v=u+at \]
Special Case
  • If \(a=0\), then \(v=u\). The object moves with constant velocity.
  • If \(u=0\), then \(v=at\). The object starts from rest.
Second Equation of Motion
The second equation of motion gives the relation among displacement, initial velocity, acceleration, and time.
Formula
\[\boxed{\bbox[2pt]{s=ut+\frac{1}{2}at^2}}\]
Derivation
Average velocity, \[ \text{Average Velocity} = \frac{u+v}{2} \] Since, \[ \text{Displacement} = \text{Average Velocity} \times \text{Time} \] Therefore, \[ s=\frac{u+v}{2}\times t \] Using, \[ v=u+at \] we obtain, \[ s=\frac{u+(u+at)}{2}\times t \] \[ s=\frac{2u+at}{2}\times t \] Therefore, \[ s=ut+\frac{1}{2}at^2 \]
Special Cases
  • If \(a=0\), then \(s=ut\).
  • If \(u=0\), then
\[ s=\frac{1}{2}at^2 \] This equation is particularly useful for bodies starting from rest.
Third Equation of Motion
The third equation of motion establishes a relation among displacement, initial velocity, final velocity, and acceleration without involving time.
Formula
\[\boxed{\bbox[2pt]{v^2-u^2=2as}}\]
Derivation
From the first equation, \[ t=\frac{v-u}{a} \] Substituting this value of \(t\) into \[ s=\frac{u+v}{2}\times t \] we obtain, \[ s = \frac{u+v}{2} \times \frac{v-u}{a} \] \[ s = \frac{v^2-u^2}{2a} \] Therefore, \[ v^2-u^2=2as \]
Importance
This equation is extremely useful when time is not given in the problem.
ℹ️ Sign Convention
Before applying equations of motion, a positive direction should be chosen.
  • Quantities in the chosen direction are taken as positive.
  • Quantities opposite to the chosen direction are taken as negative.
  • Retardation is represented by negative acceleration.
Example:
A car moving east slows down.
  • East direction: Positive
  • Acceleration: Negative
🤔 Did You Know?
How to Select the Correct Equation?
Known Quantities Required Quantity Equation Used
\(u,a,t\) \(v\) \(v=u+at\)
\(u,a,t\) \(s\) \(s=ut+\frac{1}{2}at^2\)
\(u,v,a\) \(s\) \(v^2-u^2=2as\)
\(u,v,t\) \(s\) \(s=\frac{u+v}{2}t\)
🎨 SVG Diagram
Graphical Interpretation of Equations of Motion
Velocity-Time Graph: Equations of Motion Time (t) Velocity (v) 0 1 2 3 4 5 0 10 20 30 40 Δt Δv AREA = Displacement Key Concepts 1 Slope of v-t graph = Acceleration (a) a = Δv / Δt = (v₂ - v₁) / (t₂ - t₁) 2 Area under v-t graph = Displacement (s) s = ½ × base × height = ½ × t × v (for triangle) 3 Uniform acceleration → Straight line on v-t graph v = u + at (linear equation) EQUATIONS v = u + at s = ut + ½at² v² = u² + 2as
✏️ Example
Solved Example
A car starts from rest and accelerates uniformly at \(3\;m/s^2\) for 6 s. Find its final velocity.
Identify \(u\), \(a\), and \(t\) → Apply First Equation
Given, \[ u=0\;m/s \] \[ a=3\;m/s^2 \] \[ t=6\;s \] Therefore, \[ v=u+at \] \[ v=0+(3\times6) \] \[ v=18\;m/s \]
Final velocity = 18 m/s.
A train moving at \(10\;m/s\) accelerates uniformly at \(2\;m/s^2\) for 5 s. Find the displacement.
Given, \[ u=10\;m/s \] \[ a=2\;m/s^2 \] \[ t=5\;s \] Using, \[ s=ut+\frac{1}{2}at^2 \] Therefore, \[ s = (10)(5) + \frac{1}{2}(2)(5)^2 \] \[ s=50+25 \] \[ s=75\;m \]
Displacement = 75 m.
A car increases its speed from \(5\;m/s\) to \(15\;m/s\) with acceleration \(2\;m/s^2\). Find the displacement.
Given, \[ u=5\;m/s \] \[ v=15\;m/s \] \[ a=2\;m/s^2 \] Using, \[ v^2-u^2=2as \] Therefore, \[ 15^2-5^2=2(2)s \] \[ 225-25=4s \] \[ 200=4s \] \[ s=50\;m \]
Displacement = 50 m.
🛠️ Applications of Equations of Motion
  • Designing braking systems of vehicles.
  • Determining stopping distances of trains and cars.
  • Studying freely falling bodies.
  • Planning aircraft take-off and landing distances.
  • Predicting the motion of satellites and rockets.
  • Calculating travel time and displacement in transportation systems.
⚡ Exam Tip
❌ Common Mistakes
  • Using equations of motion for non-uniform acceleration.
  • Confusing distance with displacement.
  • Forgetting to convert km/h into m/s.
  • Ignoring negative acceleration during retardation.
  • Using an equation containing more than one unknown quantity.
  • Substituting values without mentioning units.
📋 CBSE Case Study (HOTS)

A metro train starts from rest and accelerates uniformly at \(1.5\;m/s^2\) for 20 s.

Questions:

  1. Find the final velocity.
  2. Find the displacement covered during this period.
  3. Which equations of motion are used?

Solution:

Final velocity: \[ v=u+at \] \[ v=0+(1.5)(20) \] \[ v=30\;m/s \] Displacement: \[ s = ut+\frac{1}{2}at^2 \] \[ s = 0+ \frac{1}{2}(1.5)(20)^2 \] \[ s=300\;m \]

Answer:

Final velocity = 30 m/s
Displacement = 300 m

🌟 Important Terms to Remember
🚀

First Equation of Motion

📋
Table of Contents
15 sections
📘 Definition
🗒️ Statement of the First Equation of Motion
If an object moves with uniform acceleration, then its final velocity is equal to its initial velocity plus the product of acceleration and time. \[ \boxed{v=u+at} \] where,
  • \(u\) = Initial velocity of the object
  • \(v\) = Final velocity of the object
  • \(a\) = Uniform acceleration
  • \(t\) = Time interval
🤔 Did You Know?
Why Do We Need the First Equation of Motion?
In real life, objects often move with changing velocities. For example:
  • A car starts from rest and gradually gains speed.
  • A train accelerates after leaving a station.
  • An aeroplane increases its speed during take-off.
  • A cyclist pedals faster and increases velocity.
In all such situations, we need a mathematical relation to determine the final velocity after a given time interval. The first equation of motion provides this relation.
🔎 Conditions for Applying the First Equation of Motion
📐 Derivation

Derivation of the First Equation of Motion

Consider an object moving along a straight line.
  • Initial velocity = \(u\)
  • Final velocity after time \(t\) = \(v\)
  • Uniform acceleration = \(a\)
By definition,
Acceleration = Change in Velocity / Time Taken
herefore, \[ a = \frac{v-u}{t} \] Multiplying both sides by \(t\), \[ at=v-u \] Rearranging, \[ v-u=at \] Adding \(u\) to both sides, \[ v=u+at \] Therefore, \[ \boxed{\bbox[2pt]{v=u+at}} \] This equation is known as the First Equation of Motion.
🔍 Physical Interpretation of the Equation
The equation \(v=u+at\) shows that,
  • \(u\) represents the velocity already possessed by the object.
  • \(at\) represents the additional velocity gained due to acceleration during time \(t\).
  • \(v\) is the final velocity after acceleration acts for time \(t\).
Therefore, the final velocity is simply:
Final Velocity = Initial Velocity + Gain in Velocity
⭐ Special Cases of the First Equation of Motion
📌 First Equation During Retardation
✏️ Example
Solved Example
A car starts from rest and accelerates uniformly at \(2\;m/s^2\) for 10 seconds. Find its final velocity.
\[v=u+at\]
Identify \(u\) → Identify \(a\) → Identify \(t\) → Apply Formula
Given, \[ u=0\;m/s \] \[ a=2\;m/s^2 \] \[ t=10\;s \] Therefore, \[ v=u+at \] \[ v=0+(2)(10) \] \[ v=20\;m/s \]
Final velocity = 20 m/s.
A train moving at \(15\;m/s\) accelerates uniformly at \(3\;m/s^2\) for 5 seconds. Find the final velocity.
Given, \[ u=15\;m/s \] \[ a=3\;m/s^2 \] \[ t=5\;s \] Therefore, \[ v=u+at \] \[ v=15+(3)(5) \] \[ v=30\;m/s \]
Final velocity = 30 m/s.
A car moving at \(25\;m/s\) slows down uniformly at \(5\;m/s^2\). Find its velocity after 3 seconds.
Given, \[ u=25\;m/s \] \[ a=-5\;m/s^2 \] \[ t=3\;s \] Therefore, \[ v=u+at \] \[ v=25+(-5)(3) \] \[ v=25-15 \] \[ v=10\;m/s \]
Velocity after 3 seconds = 10 m/s.
🛠️ Applications of the First Equation of Motion
  • Calculating the speed of vehicles after acceleration.
  • Determining take-off speed of aircraft.
  • Studying braking systems of automobiles.
  • Predicting the velocity of freely falling objects.
  • Analysing motion in sports and athletics.
  • Designing elevators and transportation systems.
⚡ Exam Tip
❌ Common Mistakes
  • Using km/h and m/s together without conversion.
  • Ignoring the negative sign in retardation problems.
  • Applying the equation to non-uniform acceleration.
  • Interchanging initial and final velocities.
  • Substituting values without writing units.
📋 CBSE Case Study (HOTS)

An electric car starts from rest and accelerates uniformly at \(4\;m/s^2\). Determine its velocity after 8 seconds.

Solution:

Given, \[ u=0\;m/s \] \[ a=4\;m/s^2 \] \[ t=8\;s \] Therefore, \[ v=u+at \] \[ v=0+(4)(8) \] \[ v=32\;m/s \]
🗒️ Interpretation
The car gains a velocity of \(32\;m/s\) in 8 seconds because its speed increases uniformly by \(4\;m/s\) every second.
🌟 Important Terms to Remember
🚀

Second Equation of Motion

📋
Table of Contents
13 sections
📘 Definition
🔎 Statement of the Second Equation of Motion
🤔 Did You Know?
Why Do We Need the Second Equation of Motion?
In many practical situations, we need to determine how far an object has travelled after a given time. Examples:
  • Distance covered by a car after accelerating for 10 seconds.
  • Height fallen by a freely falling body.
  • Runway length required for an aeroplane to take off.
  • Distance travelled by a train while accelerating.
The second equation of motion enables us to calculate displacement directly without first finding the final velocity.
⚙️ Conditions For Applying The Second Equation
  • The motion must occur along a straight line.
  • The acceleration must remain constant.
  • The same sign convention should be used throughout the problem.
  • The equation is applicable only for uniformly accelerated or uniformly retarded motion.
📐 Derivation
Derivation of the Second Equation of Motion
Consider an object moving with:
  • Initial velocity = \(u\)
  • Uniform acceleration = \(a\)
  • Time interval = \(t\)
  • Final velocity = \(v\)
  • Displacement = \(s\)
The average velocity of the object is: \[ v_{av} = \frac{u+v}{2} \] Since,

Displacement = Average Velocity × Time

Therefore, \[ s=v_{av}\times t \] Substituting the value of average velocity, \[ s = \frac{u+v}{2} \times t \tag{i} \] From the first equation of motion, \[ v=u+at \tag{ii} \] Substituting equation (ii) into equation (i), \[ s = \frac{u+(u+at)}{2} \times t \] \[ s = \frac{2u+at}{2} \times t \] \[ s = \frac{2ut+at^2}{2} \] Therefore, \[ s = \frac{2ut}{2} + \frac{at^2}{2} \] Hence, \[ \boxed{s=ut+\frac{1}{2}at^2} \] This equation is known as the Second Equation of Motion.
⭐ Special Cases of the Second Equation
🔗 Important Proportionality
For an object starting from rest, \[ s=\frac{1}{2}at^2 \] Therefore, \[ s\propto t^2 \] Thus, displacement is directly proportional to the square of time. Example:
  • In 1 s → Distance = 1 unit
  • In 2 s → Distance = 4 units
  • In 3 s → Distance = 9 units
  • In 4 s → Distance = 16 units
✏️ Example
Solved Examples
A car starts from rest and accelerates uniformly at \(2\;m/s^2\) for 8 seconds. Find the displacement.
Identify \(u\) → Identify \(a\) → Identify \(t\) → Apply Formula
Given, \[ u=0\;m/s \] \[ a=2\;m/s^2 \] \[ t=8\;s \] Using, \[ s=ut+\frac{1}{2}at^2 \] Therefore, \[ s = 0+ \frac{1}{2}(2)(8)^2 \] \[ s=64\;m \]
Displacement = 64 m.
A train moves with an initial velocity of \(10\;m/s\) and accelerates uniformly at \(3\;m/s^2\) for 5 seconds. Find the displacement.
Given, \[ u=10\;m/s \] \[ a=3\;m/s^2 \] \[ t=5\;s \] Therefore, \[ s = (10)(5) + \frac{1}{2}(3)(5)^2 \] \[ s = 50+37.5 \] \[ s=87.5\;m \]
Displacement = 87.5 m.
A car moving at \(20\;m/s\) slows down uniformly at \(2\;m/s^2\) for 5 seconds. Find the displacement.
Given, \[ u=20\;m/s \] \[ a=-2\;m/s^2 \] \[ t=5\;s \] Therefore, \[ s = (20)(5) + \frac{1}{2}(-2)(5)^2 \] \[ s = 100-25 \] \[ s=75\;m \]
Displacement = 75 m.
🛠️ Applications of the Second Equation of Motion
  • Determining the distance travelled by moving vehicles.
  • Calculating the height covered by freely falling objects.
  • Estimating braking distances of automobiles.
  • Designing airport runways.
  • Predicting displacement in transportation systems.
  • Studying motions of elevators and amusement rides.
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(s=ut+at^2\) and forgetting the factor \(\frac{1}{2}\).
  • Using positive acceleration during retardation problems.
  • Using the equation for non-uniform acceleration.
  • Substituting velocities without converting units.
  • Confusing displacement with total distance travelled.
📋 CBSE Case Study (HOTS)

An electric train starts from rest and accelerates uniformly at \(1.5\;m/s^2\) for 20 seconds.

Questions:

  1. Find the displacement covered.
  2. What type of motion is represented?
  3. How does displacement depend on time in this case?

Solution:

Given, \[ u=0\;m/s \] \[ a=1.5\;m/s^2 \] \[ t=20\;s \] Therefore, \[ s = 0+ \frac{1}{2}(1.5)(20)^2 \] \[ s = 300\;m \]

Interpretation:

The train covers a displacement of 300 m and exhibits uniformly accelerated motion. Since it starts from rest, its displacement is proportional to the square of time.

🌟 Important Terms to Remember
🚀

Third Equation of Motion

📋
Table of Contents
12 sections
📘 Definition
📌 Statement of the Third Equation of Motion
🤔 Why Do We Need the Third Equation?
Consider the following situations:
  • A car comes to rest after applying brakes.
  • A train accelerates over a known distance.
  • A stone falls from a certain height.
  • An aircraft requires a specific runway length for take-off.
In these cases, displacement and velocities are often known, but time may not be given. The third equation directly relates these quantities without involving time.
⚙️ Conditions For Applying The Third Equation
  • The motion must occur along a straight line.
  • Acceleration should remain constant.
  • Proper sign convention must be followed.
  • The equation is applicable only for uniformly accelerated or uniformly retarded motion.
📐 Derivation of the Third Equation of Motion
We know that:
From the second equation of motion, \[ s=ut+\frac{1}{2}at^2 \tag{i} \]

From the first equation of motion,

\[ v=u+at \] Therefore, \[ t=\frac{v-u}{a} \tag{ii} \]

Substituting the value of \(t\) from equation (ii) into equation (i):

\[ s= u\left(\frac{v-u}{a}\right) + \frac{1}{2} a \left( \frac{v-u}{a} \right)^2 \] \[ s= \frac{u(v-u)}{a} + \frac{1}{2} a \left( \frac{(v-u)^2}{a^2} \right) \] \[ s= \frac{u(v-u)}{a} + \frac{(v-u)^2}{2a} \] Taking the LCM, \[ s= \frac{ 2u(v-u)+(v-u)^2 } {2a} \] Expanding, \[ s= \frac{ 2uv-2u^2+ v^2+u^2-2uv } {2a} \] \[ s= \frac{ v^2-u^2 } {2a} \] Multiplying both sides by \(2a\), \[ 2as=v^2-u^2 \] Therefore, \[ \boxed{v^2=u^2+2as} \] This equation is known as the Third Equation of Motion.

Alternative Derivation Using Average Velocity
Since, \[ s= \frac{u+v}{2} \times t \] and, \[ t= \frac{v-u}{a} \] Therefore, \[ s= \frac{u+v}{2} \times \frac{v-u}{a} \] \[ s= \frac{ (u+v)(v-u) } {2a} \] Using the identity, \[ (a+b)(a-b)=a^2-b^2 \] we obtain, \[ s= \frac{v^2-u^2}{2a} \] Hence, \[ \boxed{v^2=u^2+2as} \]
⭐ Special Cases of the Third Equation
✏️ Solved Example
A car starts from rest and accelerates uniformly at \(4\;m/s^2\) through a displacement of 50 m. Find its final velocity.
Identify \(u\), \(a\), and \(s\) → Apply Third Equation
Given, \[ u=0\;m/s \] \[ a=4\;m/s^2 \] \[ s=50\;m \] Using, \[ v^2=u^2+2as \] Therefore, \[ v^2= 0+ 2(4)(50) \] \[ v^2=400 \] \[ v=20\;m/s \]
Final velocity = 20 m/s.
A train moving at \(30\;m/s\) comes to rest after travelling 90 m. Find its retardation.
Given, \[ u=30\;m/s \] \[ v=0\;m/s \] \[ s=90\;m \] Using, \[ v^2=u^2+2as \] Therefore, \[ 0= (30)^2+ 2(a)(90) \] \[ 0= 900+180a \] \[ 180a=-900 \] \[ a=-5\;m/s^2 \]
Retardation = 5 m/s².
🛠️ Applications of the Third Equation of Motion
  • Calculating stopping distances of vehicles.
  • Determining the speed acquired by freely falling bodies.
  • Designing railway braking systems.
  • Estimating runway lengths of aircraft.
  • Studying vehicle safety and accident analysis.
  • Predicting displacement under uniform acceleration.
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(v=u^2+2as\) instead of \(v^2=u^2+2as\).
  • Ignoring negative acceleration in braking problems.
  • Substituting velocities without unit conversion.
  • Using the equation for non-uniform acceleration.
  • Forgetting to take the square root while calculating velocity.
📋 CBSE Case Study (HOTS)

A motorcycle moving at \(15\;m/s\) accelerates uniformly at \(3\;m/s^2\) and covers a distance of 40 m.

Questions:

  1. Calculate the final velocity.
  2. Which equation of motion is most suitable?
  3. Why is this equation preferred?

Solution:

Given, \[ u=15\;m/s \] \[ a=3\;m/s^2 \] \[ s=40\;m \] Therefore, \[ v^2= (15)^2+ 2(3)(40) \] \[ v^2= 225+240 = 465 \] \[ v= \sqrt{465} \approx 21.56\;m/s \]

Interpretation:

The third equation is preferred because time is not given and the equation directly relates velocity, acceleration, and displacement.

🌟 Important Terms to Remember
🚀

Graphical Representation of Motion

📋
Table of Contents
10 sections
🗺️ Overview
Graphs provide a visual and pictorial representation of motion. They help us understand the relationship between physical quantities such as distance, displacement, velocity, acceleration, and time more easily than numerical tables.

In Physics, graphs are powerful analytical tools because they allow us to:
  • Study motion at a glance.
  • Compare the motions of different objects.
  • Determine speed and acceleration mathematically.
  • Predict future motion of an object.
  • Derive equations of motion graphically.
  • Represent large amounts of data in a simple manner.
🗂️ Types of Graphs Used in Motion
Distance-Time Graph
A distance-time graph represents the variation of distance travelled by an object with time.

In this graph:
  • Time is plotted along the x-axis (horizontal axis).
  • Distance is plotted along the y-axis (vertical axis).
Every point on the graph represents the distance travelled by the object at a particular instant of time.
Uniform Motion (Constant Speed)
  • The graph is a straight line with positive slope.
  • Equal distances are covered in equal intervals of time.
  • The slope remains constant.
  • The steeper the graph, the greater is the speed.
Non-Uniform Motion
  • The graph is a curve.
  • The slope changes continuously.
  • Unequal distances are covered in equal intervals of time.
Object at Rest
  • The graph is a horizontal straight line.
  • Distance remains constant with time.
  • The slope of the graph is zero.
Distance–Time Graph Three patterns of motion compared a) Uniform Motion Straight line, constant slope Distance Time Equal distances are covered in equal intervals of time → constant speed. b) Non-Uniform Motion Curved line, changing slope Distance Time Unequal distances in equal time intervals → changing (variable) speed. c) Object at Rest Horizontal line, zero slope Distance Time Distance stays constant as time passes → the object is stationary. NCERT Class IX Science — Motion · Academia Aeternum
Velocity-Time Graph
A velocity-time graph represents the variation of velocity with time. In this graph:
  • Time is plotted along the x-axis.
  • Velocity is plotted along the y-axis.
Velocity-time graphs are extremely important because they help us determine acceleration and displacement directly from the graph.
Uniform Velocity on a Velocity-Time Graph
If an object moves with constant velocity, the graph is a horizontal straight line parallel to the time axis.
Since the velocity remains constant,\[ \text{Acceleration}=0 \] Therefore, the slope of the graph is zero.
Uniformly Accelerated Motion
If the velocity increases uniformly with time:
  • The graph is a straight line with positive slope.
  • The slope remains constant.
  • The object possesses uniform acceleration.
A steeper line indicates greater acceleration.
Uniform Retardation
If the velocity decreases uniformly with time:
  • The graph is a straight line with negative slope.
  • The object possesses uniform retardation.
  • The magnitude of slope gives the retardation.
Velocity–Time Graph Three patterns of changing velocity Uniform Velocity Horizontal line, zero slope Velocity Time Velocity stays the same over time → acceleration is zero. Uniformly Accelerated Motion Rising straight line, constant slope Velocity Time Velocity increases by equal amounts in equal time → constant positive acceleration. Uniform Retardation Falling straight line, constant slope Velocity Time Velocity decreases by equal amounts in equal time → constant negative acceleration. NCERT Class IX Science — Motion · Academia Aeternum
📌 Slope of a Velocity-Time Graph
📌 Area Under a Velocity-Time Graph
🌟 Importance of Graphical Representation of Motion
✏️ Example
Solved Example
A car travels 120 m in 6 s with uniform motion. Find the slope of the distance-time graph.
\[ \text{Slope} = \frac{\Delta d}{\Delta t} \]
\[ \text{Slope} = \frac{120}{6} \] \[ \text{Slope} = 20\;m/s \]
Speed of the car = 20 m/s.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing speed with acceleration while interpreting slopes.
  • Interchanging distance-time and velocity-time graphs.
  • Assuming every straight-line graph represents acceleration.
  • Ignoring the units while calculating slopes.
  • Forgetting that area under a velocity-time graph gives displacement.
📋 CBSE Case Study (HOTS)

A velocity-time graph of a car is a straight line that rises uniformly from 0 m/s to 20 m/s in 10 s.

Questions:

  1. What type of motion does the graph represent?
  2. Calculate the acceleration.
  3. Calculate the displacement during the interval.

Solution:

Acceleration: \[ a= \frac{20-0}{10} = 2\;m/s^2 \] Displacement: \[ s= \frac{1}{2} \times 10 \times 20 \] \[ s=100\;m \]

Interpretation:

The car undergoes uniformly accelerated motion with an acceleration of 2 m/s² and covers a displacement of 100 m.

🌟 Important Terms to Remember
🚀

Uniform Circular Motion

📋
Table of Contents
17 sections
🗺️ Overview
Uniform Circular Motion (UCM) is the motion of an object along a circular path with constant speed. Although the speed remains constant, the direction of motion changes continuously at every point of the circular path. Therefore, the velocity of the object is not constant.

Uniform circular motion is one of the most important examples of non-uniform motion because velocity is a vector quantity and its direction changes continuously even though its magnitude remains unchanged.
📘 Definition
✏️ Examples of Uniform Circular Motion
  • The motion of the blades of a ceiling fan.
  • The hands of an analog clock.
  • A stone tied to a string and whirled in a circle.
  • The rotation of Earth about its own axis.
  • The revolution of satellites around Earth.
  • A cyclist moving around a circular track at constant speed.
  • The motion of amusement park rides such as merry-go-rounds.
🗒️ Why is Uniform Circular Motion Non-Uniform?
Why is Uniform Circular Motion Non-Uniform?
At first glance, it appears that the motion is uniform because the speed remains constant. However, velocity is a vector quantity and depends upon both magnitude and direction.
  • The magnitude of velocity remains constant.
  • The direction of velocity changes continuously.
  • Hence, velocity changes continuously.
Since acceleration is the rate of change of velocity, an object moving in a circle possesses acceleration.

Constant Speed + Changing Direction = Changing Velocity = Accelerated Motion
📘 Tangential Velocity
📘 Time Period (Period of Revolution)
📘 Frequency of Revolution
🔢 Formula
Formula for Tangential Speed
📐 Derivation of Tangential Speed Formula
Distance travelled in one complete revolution: \[ =2\pi r \] Time taken: \[ =T \] Therefore, \[ v= \frac{\text{Distance}}{\text{Time}} \] Hence, \[ v= \frac{2\pi r}{T} \] Thus, tangential speed is directly proportional to the radius and inversely proportional to the time period.
🔗 Relation Between Speed and Frequency
Since, \[ T=\frac{1}{f} \] Therefore, \[ v= \frac{2\pi r}{1/f} \] Hence, \[ \boxed{ v=2\pi rf } \] This form is particularly useful when frequency is given instead of time period.
📌 Acceleration in Uniform Circular Motion
⚖️ Difference Between Uniform Motion and Uniform Circular Motion
Uniform Motion Uniform Circular Motion
Occurs along a straight line. Occurs along a circular path.
Speed and velocity remain constant. Speed remains constant but velocity changes.
Acceleration is zero. Acceleration is non-zero.
No change in direction. Direction changes continuously.
✏️ Example
Solved Example
A stone tied to a string moves in a circle of radius \(0.5\;m\). It completes one revolution in \(2\;s\). Find its tangential speed.
v=\frac{2\pi r}{T}
Given, \[ r=0.5\;m \] \[ T=2\;s \] Therefore, \[ v= \frac{2\times3.14\times0.5}{2} \] \[ v=1.57\;m/s \]
Tangential speed = 1.57 m/s.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming velocity remains constant because speed is constant.
  • Calling uniform circular motion a uniform motion.
  • Confusing time period with frequency.
  • Using diameter instead of radius in formulas.
  • Ignoring the continuous change in direction.
📋 CBSE Case Study (HOTS)

The blade of a ceiling fan completes 120 revolutions in one minute. The radius of the blade is \(0.25\;m\).

Questions:

  1. Find the frequency of rotation.
  2. Calculate the time period.
  3. Why is the motion accelerated even though the speed remains constant?

Hints:

Frequency: \[ f=\frac{120}{60}=2\;Hz \] Time Period: \[ T=\frac{1}{f} =0.5\;s \]

The motion is accelerated because the direction of velocity changes continuously.

🎨 SVG Diagram
Uniform Circular Motion
Uniform Circular Motion Constant speed along a circular path The Motion Diagram O r v v v v velocity (tangent, ever-changing direction) centripetal acceleration (toward centre O) Key Concepts Definition When a body moves in a circle at a constant speed, its motion is called uniform circular motion. Speed is Constant The magnitude of velocity (|v|) stays the same at every point. Velocity Keeps Changing Direction of velocity changes continuously — it is always tangent to the circular path. Centripetal Acceleration Since velocity changes direction, the body accelerates. This acceleration always points toward the centre O. Period & Speed Time for one full revolution is the period, T. v = 2πr / T NCERT Class IX Science — Motion · Academia Aeternum
🌟 Important Terms to Remember
🚀

Centripetal Acceleration

📋
Table of Contents
15 sections
🗺️ Overview
Centripetal acceleration is the acceleration that keeps an object moving along a circular path. It is always directed towards the centre of the circular path and continuously changes the direction of velocity of the object.

Although the speed of an object in uniform circular motion remains constant, its velocity changes continuously because its direction changes at every instant. This continuous change in velocity produces an acceleration known as centripetal acceleration.
🗒️ Centripetal Acceleration
Centripetal acceleration is the acceleration of an object moving in a circular path that always acts towards the centre of the circle.
The word centripetal is derived from two Latin words:
  • Centrum = Centre
  • Petere = To seek
Therefore, centripetal means "centre-seeking".
🤔 Did You Know?
Why is Centripetal Acceleration Needed?
According to Newton's First Law of Motion, every object tends to continue moving in a straight line unless acted upon by an external force.

However, in circular motion, the object continuously changes its direction and does not move in a straight line. Therefore, an inward acceleration is required to keep the object on the circular path.

Without centripetal acceleration, an object would move tangentially in a straight line due to inertia.
🔎 Direction of Centripetal Acceleration
🔢 Formula for Centripetal Acceleration
🗂️ Dependence of Centripetal Acceleration
Dependence on Radius
Since, \[ a_c=\frac{v^2}{r} \] Therefore, \[ a_c\propto v^2 \] Centripetal acceleration is directly proportional to the square of the speed.

Implication:
  • If speed becomes twice, acceleration becomes four times.
  • If speed becomes three times, acceleration becomes nine times.
Relation with Time Period
We know that, \[ v=\frac{2\pi r}{T} \] Substituting this value into \[ a_c=\frac{v^2}{r} \] we obtain, \[ a_c = \frac{ \left( \frac{2\pi r}{T} \right)^2 } {r} \] \[ a_c = \frac{ 4\pi^2r^2 } {T^2r} \] Therefore, \[ \boxed{ a_c= \frac{4\pi^2r}{T^2} } \] This form is useful when the time period of revolution is known.
Relation with Frequency
Since, \[ v=2\pi rf \] Therefore, \[ a_c = \frac{ (2\pi rf)^2 } {r} \] Hence, \[ \boxed{ a_c= 4\pi^2rf^2 } \] This relation is useful when frequency is given instead of time period.
🌟 Physical Significance of Centripetal Acceleration
Centripetal acceleration does not change the magnitude of velocity. Instead, it continuously changes the direction of velocity and keeps the object moving along a circular path.

Centripetal acceleration changes direction, not speed.
✏️ Examples of Centripetal Acceleration in Daily Life
  • A satellite revolving around Earth.
  • The blades of a ceiling fan.
  • A stone tied to a string and whirled in a circle.
  • A cyclist moving around a circular track.
  • A car turning on a curved road.
  • The Moon revolving around Earth.
  • Rotating amusement park rides.
🌟 Importance of Centripetal Acceleration
⚖️ Difference Between Linear and Centripetal Acceleration
Linear Acceleration Centripetal Acceleration
Changes the speed of an object. Changes only the direction of velocity.
Acts along the direction of motion. Acts towards the centre of the circle.
May increase or decrease speed. Maintains circular motion.
Associated with straight-line motion. Associated with circular motion.
✏️ Example
Solved Example
A stone moves in a circular path of radius \(2\;m\) with a speed of \(4\;m/s\). Calculate its centripetal acceleration.
\[a_c=\frac{v^2}{r}\]
Given, \[ v=4\;m/s \] \[ r=2\;m \] Therefore, \[ a_c = \frac{4^2}{2} \] \[ a_c = \frac{16}{2} \] \[ a_c=8\;m/s^2 \]
Centripetal acceleration = 8 m/s².
⚡ Exam Tip
❌ Common Mistakes
  • Assuming acceleration is zero because speed is constant.
  • Thinking centripetal acceleration acts tangentially.
  • Using diameter instead of radius in formulas.
  • Ignoring that acceleration changes direction continuously.
  • Confusing tangential velocity with centripetal acceleration.
📋 CBSE Case Study (HOTS)

A satellite moves around Earth in a circular orbit with constant speed.

Questions:

  1. Is the satellite accelerating?
  2. If yes, in which direction does the acceleration act?
  3. Why does the satellite not move in a straight line?

Answer:

  • Yes, the satellite is accelerating.
  • The acceleration acts towards the centre of Earth.
  • The inward centripetal acceleration continuously changes the direction of velocity and keeps the satellite in its orbit.
🌟 Importance
Important Terms to Remember
🚀

Important Points to Remember

📝 Summary
· Updated
NCERT Class IX · Science · Chapter 7

Motion

Distance, displacement, speed, velocity, acceleration, graphs, equations of motion, and uniform circular motion — explored through interactive tools, a step-by-step solver, and original practice problems.

☉Core Concepts

Three building blocks of Motion, explained with worked reasoning and a visual for each idea.

Concept 01

Describing Motion: Distance & Displacement

An object is said to be in motion if its position changes with time, relative to a chosen reference point (origin). Motion is always described relative to something — a tree appears stationary to you, but it is moving relative to the Sun.

Distance is the total path length covered by an object, irrespective of direction. It is a scalar quantity — it has magnitude only, and is always positive.

Displacement is the shortest straight-line distance between the initial and final position, along with direction. It is a vector quantity, and unlike distance, it can be zero even when distance is not.

  • If an object returns to its starting point, displacement = 0, but distance > 0.
  • Displacement can never be greater than distance travelled.
  • For straight-line motion in one direction, distance = displacement.
Why it matters: A runner completing one lap of a circular track covers a large distance, but their displacement is zero because they end where they began.
A B arc = distance travelled chord = displacement

Path A→B: the dashed arc is the distance travelled; the solid chord is the displacement.

Concept 02

Speed, Velocity, Acceleration & Motion Graphs

Speed is the distance covered per unit time (scalar): speed = distance ÷ time. Velocity is displacement per unit time (vector): velocity = displacement ÷ time, and includes direction.

Motion is uniform when equal distances are covered in equal intervals of time (constant speed); it is non-uniform when this is not the case. For non-uniform motion we use average speed = total distance ÷ total time.

Acceleration is the rate of change of velocity: a = (v − u) ÷ t, where u is initial velocity and v is final velocity. Acceleration is positive when velocity increases, and negative (sometimes called retardation) when velocity decreases.

On a distance–time graph, the slope at any point equals the speed; a straight line means uniform motion, a curve means the speed is changing. On a velocity–time graph, the slope equals acceleration, and the area enclosed between the line and the time axis equals the distance (or displacement) covered.

Time (s) Distance (m) constant slope = speed

Uniform motion: a straight distance–time line, constant slope.

Time (s) Distance (m) changing slope = changing speed

Non-uniform motion: the curve steepens as speed increases over time.

Time (s) Velocity (m/s) shaded area = displacement

Velocity–time graph: the shaded triangular area gives the displacement covered.

Concept 03

Equations of Motion & Uniform Circular Motion

For motion with uniform acceleration, three equations connect initial velocity (u), final velocity (v), acceleration (a), time (t), and distance (s). They can be derived directly by reading the slope and area of a velocity–time graph:

  • v = u + at — from the definition of acceleration (slope of the v–t graph).
  • s = ut + ½at² — from the area under the v–t graph (a rectangle plus a triangle).
  • v² = u² + 2as — obtained by eliminating t from the first two equations.

In uniform circular motion, an object moves on a circular path at a constant speed, but its velocity is continuously changing because direction changes at every point. Since velocity changes, the motion is accelerated even though the speed stays constant. For a circle of radius r completed in time period T, speed = 2πr ÷ T.

Key insight: "Constant speed" does not mean "no acceleration" — acceleration depends on the change of velocity (a vector), and direction change alone produces acceleration.

Tangential velocity vectors at four points — same length (speed), different direction.

ΣFormula Quick Reference

Every formula from Chapter 7, with units and a one-line reminder of when to use it.

Speed

speed = distance ÷ time

Scalar. Use when direction does not matter.

SI unit: m/s  |  also: km/h (×5/18 to convert to m/s)

Average Speed

vavg = total distance ÷ total time

For motion where speed is not constant throughout the journey.

SI unit: m/s

Velocity

velocity = displacement ÷ time

Vector. Direction must always be stated with the answer.

SI unit: m/s

Average Velocity (uniform acceleration)

vavg = (u + v) ÷ 2

Only valid when acceleration is uniform (constant).

SI unit: m/s

Acceleration

a = (v − u) ÷ t

Positive = speeding up. Negative = slowing down (retardation).

SI unit: m/s²

First Equation of Motion

v = u + at

Relates velocity and time. No distance term — use when s is not given/needed.

u, v in m/s · a in m/s² · t in s

Second Equation of Motion

s = ut + ½at²

Relates distance and time. No final velocity term.

s in m · u in m/s · a in m/s² · t in s

Third Equation of Motion

v² = u² + 2as

Relates velocity and distance. No time term — use when t is not given/needed.

u, v in m/s · a in m/s² · s in m

Speed in Uniform Circular Motion

v = 2πr ÷ T

r = radius of the circular path, T = time period (time for one full revolution).

r in m · T in s · v in m/s

Unit Conversion

km/h × (5/18) = m/s

Reverse: m/s × (18/5) = km/h. Memorise this single factor.

—

⚠Ticks, Tips & Common Mistakes

What sharp students do differently — and the slips that cost the most marks.

✓ Ticks & Tips

💡Always convert units first. If speed is in km/h and time is in seconds, convert km/h to m/s (×5/18) before substituting into any equation of motion.
💡Write what is given, then what is asked, before picking a formula. List u, v, a, t, s on one line — the missing one tells you which equation to use.
💡Use the "missing variable" rule. No s in the data → use v = u + at. No v → use s = ut + ½at². No t → use v² = u² + 2as.
💡Treat "starts from rest" as u = 0 and "comes to rest" or "stops" as v = 0. These phrases are the most common hidden data in word problems.
💡Retardation is just negative acceleration. Substitute it with a minus sign rather than treating it as a separate formula.
💡On a velocity–time graph, area = distance. Split the shape into a rectangle and a triangle, find each area separately, then add.
💡For average speed over two equal time intervals, simply average the two speeds. For two equal distances at different speeds, use total distance ÷ total time instead — do not average the speeds directly.

✗ Common Mistakes

⚠Confusing distance and displacement. Students often use the path length where direction-sensitive displacement was actually needed (or vice versa), especially in circular or to-and-fro motion.
⚠Forgetting to convert km/h to m/s before applying SI-unit formulas — this single slip invalidates an otherwise correct method.
⚠Dropping the sign of acceleration. When a body decelerates, writing "a = 5 m/s²" instead of "a = −5 m/s²" gives a wrong final answer even with the right formula.
⚠Averaging two speeds directly when the distances (not the times) covered at each speed are equal — this gives the wrong average speed; total distance ÷ total time must be used instead.
⚠Reading the wrong axis on a graph. Mixing up "slope" with "area" — slope of a v–t graph gives acceleration, area under it gives distance; these are often swapped in a hurry.
⚠Assuming uniform circular motion has zero acceleration because speed is constant — direction is changing, so velocity (and hence acceleration) is not zero.
⚠Squaring incorrectly in v² = u² + 2as. A frequent arithmetic slip is computing 2as before squaring u, or forgetting to take the square root at the final step to recover v.

⚙Step-by-Step AI Solver

A rule-based solver — pick a problem type, enter the known values, and it works through the solution one step at a time. Runs entirely in your browser.

Choose a problem type, fill in the known values on the left, and click Solve Step-by-Step to see the full worked solution here.

✎Concept-Building Practice Questions

Original questions, organised by concept, each with a complete worked solution. Click any question to expand it.

☉ Concept 1 — Distance & Displacement

Q1. A farmer walks along the boundary of a square field of side 15 m, starting at one corner and walking along two sides to reach the diagonally opposite corner. Find the distance travelled and the displacement.Easy▾
1
Identify the path

Walking along two adjacent sides of the square covers a path of length 15 m + 15 m.

distance = 15 + 15 = 30 m
2
Find the displacement

The diagonally opposite corner is connected to the start by the diagonal of the square, which is the shortest straight-line path.

displacement = √(15² + 15²) = 15√2 ≈ 21.2 m
Distance = 30 m  ·  Displacement = ≈ 21.2 m (along the diagonal)
Q2. An object moves 8 m east, then 8 m west, and finally 5 m east. Find the total distance covered and the net displacement.Easy▾
1
Add up every segment for distance

Distance ignores direction, so every leg of the journey is simply added.

distance = 8 + 8 + 5 = 21 m
2
Track direction for displacement

Take east as positive: +8 (east), then −8 (west), then +5 (east).

displacement = 8 − 8 + 5 = 5 m (east)
Distance = 21 m  ·  Displacement = 5 m east
Q3. A person jogs on a circular track of radius 35 m and completes 1.75 revolutions, starting and stopping on the track. Find the distance covered and the magnitude of displacement. (Use π = 22/7)Challenging▾
1
Find the circumference

Circumference = 2πr.

2 × (22/7) × 35 = 220 m
2
Multiply by number of revolutions for distance

Distance covered in 1.75 revolutions is 1.75 times the circumference.

distance = 1.75 × 220 = 385 m
3
Locate the final position

1 full revolution brings the jogger back to the start; the remaining 0.75 revolution = 270° of arc sets the final position 270° around the circle from the start.

4
Compute the chord (displacement)

For a central angle θ, the straight-line chord between start and end is 2r·sin(θ/2).

displacement = 2 × 35 × sin(135°) = 70 × (√2/2) = 35√2 ≈ 49.5 m
Distance = 385 m  ·  Displacement ≈ 49.5 m

☉ Concept 2 — Speed, Velocity, Acceleration & Graphs

Q4. A train covers the first 120 km of a journey in 2 hours, then the next 180 km in 3 hours. Find its average speed for the entire journey.Easy▾
1
Find total distance
total distance = 120 + 180 = 300 km
2
Find total time
total time = 2 + 3 = 5 h
3
Apply average speed = total distance ÷ total time

Note: this is not the same as averaging 60 km/h and 60 km/h from each leg individually — here it happens to match because both legs individually average to 60 km/h too, but in general you must use total distance ÷ total time, not an average of speeds.

average speed = 300 ÷ 5 = 60 km/h
Average speed = 60 km/h
Q5. A car moving at 30 m/s applies its brakes and comes to rest uniformly in 6 seconds. Find (a) its acceleration, and (b) the distance it travels while braking.Medium▾
1
Note the given values

"Comes to rest" means final velocity v = 0.

u = 30 m/s, v = 0, t = 6 s
2
Find acceleration
a = (v − u) ÷ t = (0 − 30) ÷ 6 = −5 m/s²

The negative sign shows this is retardation (deceleration).

3
Find distance using the velocity–time graph area

The v–t graph is a triangle with base = time, height = initial velocity.

distance = ½ × base × height = ½ × 6 × 30 = 90 m
Acceleration = −5 m/s² (retardation)  ·  Distance = 90 m
Q6. A scooter's velocity rises uniformly from 5 m/s to 25 m/s over 8 seconds, then stays constant at 25 m/s for the next 4 seconds. Find the acceleration during the first 8 seconds, and the total distance covered over the full 12 seconds.Challenging▾
1
Acceleration in phase 1
a = (25 − 5) ÷ 8 = 2.5 m/s²
2
Distance in phase 1 (trapezoid area under v–t graph)
s₁ = ½(5 + 25) × 8 = 120 m
3
Distance in phase 2 (constant velocity, rectangle area)
s₂ = 25 × 4 = 100 m
4
Add both phases
total distance = 120 + 100 = 220 m
Acceleration (phase 1) = 2.5 m/s²  ·  Total distance = 220 m

☉ Concept 3 — Equations of Motion & Circular Motion

Q7. A scooter starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. Find the velocity it attains and the distance it covers.Easy▾
1
Note given values

"Starts from rest" means u = 0.

u = 0, a = 2 m/s², t = 10 s
2
Find final velocity using v = u + at
v = 0 + 2 × 10 = 20 m/s
3
Find distance using s = ut + ½at²
s = 0 + ½ × 2 × 10² = 100 m
Velocity attained = 20 m/s  ·  Distance covered = 100 m
Q8. A ball is thrown vertically upward with an initial velocity of 14 m/s. Taking the deceleration due to gravity as 9.8 m/s², find the maximum height it reaches.Medium▾
1
Note given values

At the highest point, the ball's velocity momentarily becomes zero, so v = 0. Gravity acts downward, opposing the upward motion, so a = −9.8 m/s².

u = 14 m/s, v = 0, a = −9.8 m/s²
2
Apply v² = u² + 2as and solve for s
0 = 14² + 2(−9.8)s ⇒ s = 196 ÷ 19.6 = 10 m
Maximum height reached = 10 m
Q9. A cyclist rides on a circular track of radius 49 m and completes one full round in 22 seconds. Find her speed. (Use π = 22/7)Medium▾
1
Note the values
r = 49 m, T = 22 s
2
Apply v = 2πr ÷ T
v = 2 × (22/7) × 49 ÷ 22 = (2 × 49) ÷ 7 = 14 m/s
Speed of the cyclist = 14 m/s

❖Interactive Learning Modules

Six hands-on tools for building intuition about Motion — pick any one below.

Term

Definition

Click a term, then click its matching definition.

DISTANCE–TIME

VELOCITY–TIME

Initial velocity (u)0 m/s
Acceleration (a)1 m/s²
Time elapsed0.0 s
Current velocity0.0 m/s
Distance covered0.0 m

Question 1 of 5

Academia Aeternum · NCERT Class IX Science · Chapter 7: Motion

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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    Motion — Learning Resources

    🧠 Practice MCQs
    ✔️ True / False
    📌 Exercise
    📝 Exercises
    Motion-Exercise

    Frequently Asked Questions

    Motion is the change in position of an object with respect to time and a reference point.

    An object is said to be at rest if it does not change its position with respect to its surroundings.

    Distance is the total path length covered by an object during motion.

    Displacement is the shortest straight-line distance between the initial and final positions of an object.

    The SI unit for both distance and displacement is metre (m).

    Speed is the distance covered by an object per unit time.

    Speed = Distance / Time.

    Average speed is the total distance traveled divided by total time taken.

    Velocity is the rate of change of displacement with respect to time.

    The SI unit for both is metre per second (m/s).

    Motion of an object with respect to another moving or stationary object is relative motion.

    It means velocity is constant and the object is in uniform motion.

    It represents uniform motion.

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