Jnaanangkur The Learning Hub

JNAANANGKUR Overview Distance & Displacement Speed & Velocity Acceleration Equations of Motion Graphs NCERT Solutions MCQ Bank Question Bank Previous Year Qs Common Mistakes Revision Notes FAQs

● JNAANANGKUR — THE LEARNING HUB

Class 9 Science Chapter 8: Motion

A complete, exam-ready guide to distance, displacement, speed, velocity, acceleration and the equations of motion — built for NCERT, CBSE, and State Board learners (including SEBA/Assam Board), with graphs, derivations, 75+ MCQs and previous-year questions.

📘 NCERT Chapter 8 🎯 CBSE 2026 Pattern 🧭 SEBA / Assam Board Aligned ⏱ ~28 min read ✅ 75+ Solved MCQs

0.0

Velocity (m/s)

0.0

Time (s)

0.0

Distance (m)

2.0

Acceleration (m/s²)

👋 Welcome Note

Why does Chapter 8 matter so much?

Hello and welcome, young scientist! Every time you walk to school, watch a cricket ball fly across the field, or feel a bus jerk forward, you are witnessing motion — one of the oldest and most fundamental ideas in physics. This chapter gives you the mathematical language to describe movement precisely: not just "the car is fast," but exactly how fast, in what direction, and how that speed is changing. Once you master this chapter, every future physics topic — force, energy, gravitation, even the motion of planets — becomes far easier to understand.

🧠 Chapter Snapshot

Motion is the change in position of an object with respect to its surroundings, over time. This chapter (NCERT Class 9, Physics Unit) builds a complete toolkit: scalar and vector quantities of motion, uniform and non-uniform motion, the three equations of motion, and graphical representation using distance–time and velocity–time graphs.

NCERT ALIGNED CBSE 2025-26 SYLLABUS SEBA / ASSAM BOARD

🎯 Learning Objectives

By the end of this chapter, you will be able to:

📏 Describe Position

Explain why a reference point (origin) is essential to describe any motion.

↔️ Differentiate Quantities

Distinguish clearly between distance & displacement, and speed & velocity.

📈 Classify Motion

Identify uniform and non-uniform motion from real situations and data.

🧮 Apply Equations

Derive and use the three equations of motion to solve numericals.

📊 Read Graphs

Interpret and draw distance-time and velocity-time graphs confidently.

🌀 Understand Circular Motion

Calculate speed in uniform circular motion with real examples.

🌍 Real-Life Applications of Motion

🚗 Speedometers & Traffic Rules

Speed limits, stopping distances, and safe braking all rely on the physics of motion and acceleration.

🚀 Space Missions

ISRO scientists use equations of motion to calculate a rocket's velocity and the exact moment of stage separation.

🏃 Sports Science

Sprinters' acceleration off the blocks and a bowler's release speed are analysed using these very formulas.

🌦 Weather & Satellites

Satellites in circular orbit (uniform circular motion) help track cyclones over the Bay of Bengal.

📚 Table of Contents

1. Distance & Displacement
2. Speed & Velocity
3. Acceleration
4. Uniform vs Non-Uniform Motion
5. Equations of Motion
6. Uniform Circular Motion
7. Motion Graphs
8. NCERT In-Text & Exercise Solutions
9. 75+ MCQs (Topic-wise)
10. VSA / SA / LA / HOTS / Assertion-Reason / Case-Based
11. Previous Year Questions
12. Common Mistakes & Tips
13. Quick Revision Notes
14. FAQs

01 · Foundations

Distance and Displacement

Before we can measure motion, we must fix a reference point (origin). An object is said to be in motion if its position changes with time relative to this reference point.

📏 Distance

The total path length covered by an object during its motion, irrespective of direction. Distance is a scalar quantity — it has magnitude only, never negative, and never decreases with time.

➡️ Displacement

The shortest straight-line distance between the initial and final position of an object, along with its direction. Displacement is a vector quantity — it can be positive, negative, or even zero (when the object returns to its starting point).

Feature	Distance	Displacement
Type of quantity	Scalar	Vector
Depends on	Path taken	Only initial & final position
Value	Always positive	Positive, negative, or zero
Can it be zero while object moves?	No	Yes (if it returns to start)
Magnitude comparison	Distance ≥ Displacement	Displacement ≤ Distance
SI Unit	metre (m)	metre (m)

✏️ Worked Example

Rahim walks 4 m East and then 3 m North. Find the distance and displacement travelled.

Distance = 4 m + 3 m = 7 m (total path covered)

Displacement = straight line from start to end = √(4² + 3²) = √25 = 5 m, directed north-east of the starting point.

💡 Memory Trick

"Distance is the Diary, Displacement is the Direction." A diary records every step of the journey (path-dependent), while displacement only cares about where you began and where you ended — like a compass arrow drawn straight from start to finish.

02 · Rate of Motion

Speed and Velocity

To compare how fast different objects move, we need a quantity that combines distance (or displacement) with time.

⚡ Speed

Speed (v) = Distance travelled / Time taken

SI Unit: metre per second (m/s or m s⁻¹). Speed is a scalar quantity.

🧭 Velocity

Velocity = Displacement / Time taken

SI Unit: m/s. Velocity is a vector quantity — it changes if either speed or direction changes.

Types of Speed & Velocity

Uniform Speed

Equal distances covered in equal intervals of time (e.g., a car on cruise control on a straight highway).

Non-Uniform (Variable) Speed

Unequal distances in equal time intervals — most real-world motion, like a car in city traffic.

Average Speed

Total distance ÷ total time, even if speed varies throughout the journey.

Instantaneous Speed

Speed of an object at one particular moment (what a speedometer shows).

📐 Average Velocity (for uniform acceleration)

v_avg = (u + v) / 2

where u = initial velocity and v = final velocity — valid only when acceleration is uniform.

✏️ Worked Example

An athlete completes one round of a circular track of diameter 200 m in 40 s. Find the distance, displacement, speed and velocity at the end of 2 minutes 20 s (140 s).

Circumference = πD = 3.14 × 200 = 628 m. Time for 1 round = 40 s.

140 s = 3 full rounds (120 s) + 20 s extra. In 20 s (half a round, since 40 s = 1 round), the athlete covers half the circumference and ends up diametrically opposite the start.

Distance = 3 × 628 + 314 = 2198 m. Displacement = diameter = 200 m.

Speed = 2198/140 ≈ 15.7 m/s. Velocity = 200/140 ≈ 1.43 m/s.

03 · Change in Velocity

Acceleration

🚀 Definition

Acceleration is the rate of change of velocity with time. It is a vector quantity.

Formula

a = (v − u) / t

u = initial velocity, v = final velocity, t = time taken. SI Unit: m/s² (m s⁻²).

➕ Positive Acceleration

Velocity increases with time — object is speeding up (e.g., a bus leaving a stop).

➖ Negative Acceleration (Retardation/Deceleration)

Velocity decreases with time — object is slowing down (e.g., a car braking before a signal).

⚖️ Uniform Acceleration

Velocity changes by equal amounts in equal time intervals (e.g., a freely falling body).

🔀 Non-Uniform Acceleration

Velocity changes by unequal amounts in equal time intervals (e.g., a bus in city traffic).

✏️ Worked Example

A bus increases its velocity from 10 m/s to 30 m/s in 5 s. Find its acceleration.

a = (v − u)/t = (30 − 10)/5 = 4 m/s²

04 · Classifying Motion

Uniform and Non-Uniform Motion

Uniform Motion	Non-Uniform Motion
Equal distances covered in equal time intervals	Unequal distances covered in equal time intervals
Speed remains constant	Speed keeps changing
Distance-time graph is a straight line	Distance-time graph is a curve
Acceleration = 0	Acceleration ≠ 0 (generally)
Example: a train on a straight track at constant speed	Example: a two-wheeler in city traffic

Motion

→

Uniform (constant speed)

→

a = 0

Motion

→

Non-Uniform (changing speed)

→

a ≠ 0

05 · The Core Toolkit

Equations of Motion (for Uniform Acceleration)

These three equations connect initial velocity (u), final velocity (v), acceleration (a), time (t) and displacement (s). They apply only when acceleration is uniform (constant).

1️⃣ v = u + at

2️⃣ s = ut + ½at²

3️⃣ v² = u² + 2as

📐 Derivation 1 — By Algebraic (Graphical) Method

First Equation: v = u + at

1. Consider an object moving with initial velocity u and uniform acceleration a. On a velocity-time graph, plot velocity on the y-axis and time on the x-axis.
2. The slope of this straight line graph equals acceleration: a = (v − u) / (t − 0).
3. Rearranging gives at = v − u.
4. Therefore, v = u + at — the first equation of motion.

Second Equation: s = ut + ½at²

1. On the same velocity-time graph, displacement equals the area under the line between t = 0 and t = t.
2. This area is a trapezium: a rectangle of height u and width t, plus a triangle of base t and height (v − u).
3. Area of rectangle = ut. Area of triangle = ½ × t × (v − u).
4. Since (v − u) = at (from equation 1), area of triangle = ½ × t × at = ½at².
5. Adding both areas: s = ut + ½at² — the second equation of motion.

Third Equation: v² = u² + 2as

1. Displacement is also the area of the trapezium using parallel sides u and v with height t: s = ½(u + v)t.
2. From equation 1, t = (v − u)/a.
3. Substitute t into the area formula: s = ½(u + v) × (v − u)/a.
4. This simplifies to s = (v² − u²) / 2a.
5. Rearranging: v² = u² + 2as — the third equation of motion.

💡 Memory Trick

"VAT, SUAT, V²U²2AS" — say it like a rhythm: v = u + at → s = ut + ½at² → v² = u² + 2as. Notice each equation drops one variable: the 1st has no 's', the 2nd has no 'v', the 3rd has no 't'.

✏️ Numerical 1

A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Find (a) the velocity attained and (b) the distance travelled.

u = 0, a = 2 m/s², t = 10 s

(a) v = u + at = 0 + 2×10 = 20 m/s

(b) s = ut + ½at² = 0 + ½×2×100 = 100 m

✏️ Numerical 2

A train travelling at 90 km/h is brought to rest in 10 s by applying brakes. Find the retardation and distance travelled before it stops.

u = 90 km/h = 25 m/s, v = 0, t = 10 s

a = (v−u)/t = (0−25)/10 = −2.5 m/s² (retardation = 2.5 m/s²)

s = ut + ½at² = 25×10 + ½×(−2.5)×100 = 250 − 125 = 125 m

✏️ Numerical 3

A stone is dropped from a cliff and hits the ground with a velocity of 40 m/s. Taking g = 10 m/s², find the height of the cliff.

u = 0, v = 40 m/s, a = 10 m/s²

v² = u² + 2as → 1600 = 0 + 2×10×s → s = 80 m

06 · Special Case

Uniform Circular Motion

🌀 Definition

When an object moves in a circular path at constant speed, its motion is called uniform circular motion. Even though speed is constant, velocity keeps changing because direction changes continuously — so the motion is always accelerated.

Speed in Circular Motion

v = 2πr / t

r = radius of the circular path, t = time for one complete revolution.

🌍 Examples

Earth revolving around the Sun, a satellite orbiting Earth, the tip of a clock's second hand, a stone tied to a string whirled in a circle.

❓ Why is it accelerated?

Because velocity is a vector — a change in direction alone (even with constant speed) counts as a change in velocity, hence acceleration exists (centripetal acceleration).

07 · Visualising Motion

Distance–Time and Velocity–Time Graphs

📊 Distance–Time Graphs

Straight sloped line → uniform speed (constant slope = speed)

Curved line → non-uniform speed (slope changes = speed changes)

📈 Velocity–Time Graphs

Horizontal line → uniform velocity (a = 0)

Rising straight line → uniform acceleration; area under graph = displacement

💡 Golden Rule for Graphs

In a distance-time graph, slope = speed. In a velocity-time graph, slope = acceleration and area under the graph = distance/displacement. A straight line always means "uniform"; a curve always means "non-uniform" or "accelerated."

08 · Textbook Support

NCERT In-Text & Exercise Solutions

Yes. If an object returns to its starting point after moving along any path, its displacement is zero even though the distance covered is not. Example: an athlete running one full round of a circular track returns to the starting point — distance = circumference, displacement = 0.

Perimeter = 40 m, time for one round = 40 s. In 140 s, the farmer completes 3 full rounds (120 s) + 20 s more, i.e., half of the 4th round (2 sides = 20 m). Starting from one corner, after covering 2 sides he reaches the diagonally opposite corner. Displacement = diagonal of the square = 10√2 ≈ 14.14 m.

Both statements are false. (a) Displacement can be zero when the object returns to its starting point. (b) The magnitude of displacement can never be greater than the distance travelled; it is always less than or equal to the distance.

Speed is the distance travelled per unit time and is a scalar quantity; velocity is the displacement per unit time and is a vector quantity, since it also specifies direction. (See the comparison table in the Speed & Velocity section above for full detail.)

When the object moves along a straight line in one direction only (no change in direction), the distance travelled equals the magnitude of displacement, so average speed = magnitude of average velocity.

The odometer measures the total distance travelled by the vehicle, not displacement.

The distance-time graph for uniform motion is always a straight line, showing equal distances covered in equal intervals of time.

t = 5 min = 300 s, speed = 3×10⁸ m/s. Distance = speed × time = 3×10⁸ × 300 = 9×10¹⁰ m.

(i) Uniform acceleration: velocity changes by equal amounts in equal time intervals (e.g., a freely falling body). (ii) Non-uniform acceleration: velocity changes by unequal amounts in equal time intervals (e.g., a car moving through crowded traffic).

u = 80 km/h = 22.22 m/s, v = 60 km/h = 16.67 m/s, t = 5 s. a = (v−u)/t = (16.67−22.22)/5 ≈ −1.11 m/s² (retardation).

u = 0, v = 40 km/h = 11.11 m/s, t = 10 min = 600 s. a = (v−u)/t = 11.11/600 ≈ 0.0185 m/s².

09 · Practice Zone

75+ MCQs with Detailed Explanations

Tap any question to reveal the correct answer and explanation. Use the filters to practise topic-wise.

10 · All Question Types

VSA · SA · LA · HOTS · Assertion–Reason · Case-Based · Competency-Based

🔹 Very Short Answer (1 Mark)

The shortest distance between the initial and final positions of an object, along a specific direction; a vector quantity.

metre per second squared (m/s²).

Scalar — it has magnitude only.

An athlete completing one full round of a circular track.

A straight line parallel to the time axis.

🔹 Short Answer (2–3 Marks)

Because velocity is a vector; even though the magnitude (speed) stays constant, the direction of motion changes continuously at every point of the circle. A change in direction means a change in velocity, and any change in velocity implies acceleration.

Yes, if it continues covering equal distances (20 m) in every subsequent equal interval of 2 s, its speed is constant, so it is in uniform motion — provided the direction also does not change.

On a v-t graph, slope of the line = acceleration a = (v − u)/t. Rearranging gives v = u + at. (Full derivation given in the Equations of Motion section above.)

🔹 Long Answer (5 Marks)

Assumption: acceleration is uniform throughout the motion. Using a velocity-time graph with initial velocity u, final velocity v, time t and acceleration a: (i) slope gives v = u + at, (ii) area under the graph (trapezium) gives s = ut + ½at², (iii) eliminating t between the two gives v² = u² + 2as. (Full step-by-step derivation is provided in the Equations of Motion section above — refer to it for the complete answer.)

At maximum height, v = 0. Using v² = u² − 2gs: 0 = 900 − 20s → s = 45 m. Using v = u − gt: 0 = 30 − 10t → t = 3 s.

🔹 HOTS (Higher Order Thinking Skills)

Relative speed (opposite directions) = 40 + 60 = 100 km/h. Time = distance/speed = 200/100 = 2 hours.

Yes. When a ball thrown upward reaches its highest point, its velocity momentarily becomes zero, but gravity still acts on it, so acceleration (g, downward) is non-zero at that instant.

By counting/estimating the area under the irregular curve using graphical methods such as dividing it into small trapezium/rectangle strips and summing their areas (approximation method), since the simple equations of motion do not directly apply to non-uniform acceleration.

🔹 Assertion–Reason Questions

Choose: (A) Both A and R are true, R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Answer: (A). Both statements are true and R correctly explains A, since displacement being solely dependent on initial and final position is exactly why it can be zero for a closed path.

Answer: (D). A is false (uniform velocity means zero acceleration); R is true in a different sense but incompletely — acceleration measures rate of change of velocity, not just speed, so R as stated is actually false too. (Note: strictly, both A and R are false here, guiding students to option D/​"neither correct" per board conventions — teachers may award D as A is false and R's claim about "speed only" is imprecise.)

Answer: (D). A is false because velocity (a vector) changes direction continuously, so it isn't constant. R is true — speed does remain constant.

🔹 Case-Based / Competency-Based Question

📖 Case Study: The Guwahati–Silchar Bus Journey

A bus travels from Guwahati towards Silchar. For the first 2 hours, it maintains a steady speed of 60 km/h on the highway. It then slows down through hilly terrain, covering the next 40 km in 1 hour, and finally covers the last 30 minutes at 50 km/h.

Uniform motion, since it covered equal distances (60 km each hour) in equal time intervals.

First part: 60×2 = 120 km. Second part: 40 km. Third part: 50×0.5 = 25 km. Total = 120+40+25 = 185 km.

Total time = 2 + 1 + 0.5 = 3.5 h. Average speed = 185/3.5 ≈ 52.86 km/h.

Non-uniform, because the speed changed across the three segments of the journey (60 km/h → 40 km/h → 50 km/h).

11 · Board Exam Practice

Previous Year CBSE & State Board Questions

Using s = ½gt²: t = √(2s/g). t₁ = √(2×150/10) = √30 ≈ 5.48 s. t₂ = √(2×100/10) = √20 ≈ 4.47 s. Ratio t₁:t₂ = √30:√20 = √3:√2 ≈ 1.22:1.

A freely falling body undergoes uniformly accelerated (non-uniform velocity, uniform acceleration) motion, since g is constant. Its velocity-time graph is a straight line with a positive, constant slope (equal to g), passing through the origin if dropped from rest.

Total distance = 8+6+4 = 18 km. Total time = 3 h. Average speed = 18/3 = 6 km/h.

See the full step-by-step derivation using the area under a velocity-time graph (trapezium method) provided in the Equations of Motion section above.

Uniform motion: a train moving on a straight stretch of track at a fixed speed between two stations (e.g., Guwahati to Chaparmukh on an open stretch) covers equal distances in equal times. Non-uniform motion: a ferry crossing the Brahmaputra changes its speed continuously depending on current and traffic, covering unequal distances in equal time intervals.

12 · Exam Alert

Common Mistakes Students Make

⚠️ Mistake 1

Treating distance and displacement as always equal. Fix: They are equal only for straight-line motion in one direction without any turning back.

⚠️ Mistake 2

Forgetting to convert km/h to m/s before applying equations of motion. Fix: Multiply by 5/18 to convert km/h → m/s (and by 18/5 for the reverse).

⚠️ Mistake 3

Assuming acceleration is always positive. Fix: Deceleration (retardation) is simply negative acceleration — carry the correct sign into the equations.

⚠️ Mistake 4

Confusing the slope of a distance-time graph with the slope of a velocity-time graph. Fix: Distance-time slope = speed; velocity-time slope = acceleration. Never mix the two.

⚠️ Mistake 5

Believing uniform circular motion has zero acceleration because speed is constant. Fix: Remember velocity is a vector — direction change alone causes acceleration.

⚠️ Mistake 6

Using average velocity formula v_avg = (u+v)/2 even when acceleration is NOT uniform. Fix: This shortcut formula is valid strictly for uniform acceleration only.

13 · Last-Minute Revision

Quick Revision Notes & Key Formulas

Quantity	Formula	SI Unit	Nature
Speed	Distance / Time	m/s	Scalar
Velocity	Displacement / Time	m/s	Vector
Average Velocity	(u + v) / 2	m/s	Vector
Acceleration	(v − u) / t	m/s²	Vector
1st Equation	v = u + at	—	—
2nd Equation	s = ut + ½at²	—	—
3rd Equation	v² = u² + 2as	—	—
Speed in Circular Motion	v = 2πr / t	m/s	Scalar

🗺 Mind Map: Motion at a Glance

MOTION

→

Distance / Displacement

→

Speed / Velocity

→

Acceleration

→

Equations of Motion

→

Graphs

💡 One-Line Recap for Every Topic

1 Distance = path length; Displacement = shortest path with direction.
2 Speed = scalar rate; Velocity = vector rate.
3 Acceleration = rate of change of velocity.
4 Uniform motion → straight-line distance-time graph.
5 Slope of v-t graph = acceleration; area under v-t graph = displacement.
6 Three equations of motion apply only under uniform acceleration.
7 Uniform circular motion is always accelerated motion.

14 · Doubts Cleared

Frequently Asked Questions

Is Chapter 8 Motion easy for Class 9 CBSE board exams?

Yes, with regular practice of numericals and graphs, Motion is one of the most scoring chapters in Class 9 Physics because it follows clear, consistent formulas.

How many marks does Motion carry in the CBSE Class 9 exam?

It typically forms a significant portion of the Physics unit, with questions ranging from 1-mark MCQs to 5-mark long-answer numericals and graph-based questions almost every year.

What is the difference between speed and velocity in simple words?

Speed only tells you "how fast," while velocity tells you "how fast and in which direction." Speed is a scalar; velocity is a vector.

Can displacement be greater than distance?

No, never. Displacement is always less than or equal to distance, since it represents the shortest possible path.

Do the equations of motion work for non-uniform acceleration?

No. The three standard equations of motion are valid only when acceleration is uniform (constant). For non-uniform acceleration, calculus-based methods are required (studied in higher classes).

What is the best way to remember the three equations of motion?

Link them to the variable each one omits: the first equation has no 's', the second has no 'v', and the third has no 't'. Practising 8–10 numericals daily also builds strong recall.

Is this chapter useful for competitive exams like NTSE or Olympiads?

Absolutely. Motion forms the foundation for kinematics, which appears in NTSE, Science Olympiads, and later in Class 11 Physics as well.

🚀 Final Word

Keep Moving Forward

Motion is not just a chapter to memorise — it is the language scientists use to describe everything from a falling apple to a rocket leaving Earth's atmosphere. Every formula you've learned here trains you to observe the world with a scientist's eye: to ask not just "what is moving?" but "how fast, in what direction, and how is that changing?" Keep questioning, keep calculating, and keep that curiosity in motion — the next chapter, Force and Laws of Motion, builds directly on everything you've mastered here.

— Team Jnaanangkur, The Learning Hub

Jnaanangkur – The Learning Hub

Class 9 Science · Chapter 8 · Motion — NCERT / CBSE / SEBA (Assam Board) aligned study material.

↑