Describing Motion Around Us – Class 9 Science Chapter 4 Notes & NCERT PDF

Class 9ScienceChapter 4NCERT book: Exploration - Science Class 9

Notes and a simple summary of Chapter 4 of the NCERT Class 9 Science book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 4: Describing Motion Around Us

Think It Over

How much distance should we keep from the truck ahead so that we do not crash if it brakes suddenly? Does this distance depend on how fast we are moving?

By the end of this chapter you can answer this with numbers, graphs and equations.

Introduction

Everything in nature moves — butterflies, tides, dust in a sunbeam. To study such complex motion, scientists first look at idealised, simple forms: motion in a straight line and uniform circular motion. Earlier you met distance, time and speed; now you add displacement, average velocity and average acceleration, and learn to describe motion in words, numbers, equations and graphs.

Key Concepts

1. Position, rest and motion

To describe position we need a fixed reference point (origin O), plus the distance and direction from it.

If the position changes with time, the object is in motion; if not, it is at rest.

For straight-line motion, direction is shown by + and – signs (right of O is usually +, left is –).

An instant is a single clock reading; a time interval is the gap between two readings.

2. Distance travelled vs displacement

Displacement is the net change in position of an object between two instants. It has a magnitude and a direction.

Book example: an athlete runs from O to A (100 m) and back to B (40 m from O). Total distance = 100 m + 60 m = 160 m, but displacement = 40 m in the + direction.

Total distance needs only a number (a scalar); displacement needs number + direction (a vector). SI unit of both: metre (m).

Magnitude of displacement ≤ total distance. They are equal only when the object moves in one direction without turning back.

A ball thrown up that returns to the thrower has zero displacement but non-zero distance.

3. Average speed and average velocity

Average speed = total distance ÷ time interval (no direction).

Average velocity v_av = displacement ÷ time interval (has direction). Unit of both: m s⁻¹ (also km h⁻¹).

Uniform motion: equal distances in equal time intervals (constant speed). Otherwise it is non-uniform.

Book example (Sarang): 25 m pool, swims there and back (50 m) in 50 s → average speed 1 m s⁻¹, average velocity 0 m s⁻¹ (back where he started).

India's contribution: the idea speed = distance ÷ time appears in the Aryabhatiya (5th century CE). A Ganitakaumudi (14th century CE) problem: two postmen 210 yojanas apart walk towards each other at 9 and 5 yojanas/day → they meet after 210 ÷ 14 = 15 days.

The speedometer shows (nearly) the magnitude of velocity at an instant.

4. Average acceleration

Average acceleration a = (v − u) ÷ (t₂ − t₁) — change in velocity divided by the time interval. SI unit m s⁻².

Speeding up → acceleration in the direction of velocity; slowing down → acceleration opposite to velocity (negative sign).

Book example (bus): 36 km h⁻¹ (10 m s⁻¹) → 54 km h⁻¹ (15 m s⁻¹) in 10 s: a = 0.5 m s⁻². Braking from 15 m s⁻¹ to 0 in 5 s: a = –3 m s⁻².

A very fast object can have zero acceleration if its velocity is constant.

Free fall: velocity rises by 9.8 m s⁻¹ every second (9.8, 19.6, 29.4, 39.2 m s⁻¹…), so acceleration is constant: g = 9.8 m s⁻².

5. Graphs of motion

Position–time graph: a straight line → constant velocity; a curve → changing velocity (accelerated motion). The slope gives velocity.

Velocity–time graph: a line parallel to the time axis → constant velocity, zero acceleration; a straight sloping line → constant acceleration (rising = speeding up, falling = slowing down). The slope gives acceleration and the area under the graph gives displacement.

Book example: velocity rising from 5 to 10 m s⁻¹ between 10 s and 20 s → displacement = rectangle (5 × 10 = 50 m) + triangle (½ × 10 × 5 = 25 m) = 75 m.

6. Kinematic equations (constant acceleration only)

v = u + at

s = ut + ½at²

v² = u² + 2as

u = initial velocity, v = final velocity, a = acceleration, t = time interval, s = displacement.

Derived from the v–t graph: slope gives v = u + at; area (rectangle + triangle) gives s = ut + ½at²; eliminating t gives v² = u² + 2as.

Book example (braking, a = –4 m s⁻²): at 54 km h⁻¹ the car travels 28.1 m before stopping; at 108 km h⁻¹ it travels 112.5 m — double the speed, four times the braking distance. This is why safe distance must grow with speed. India is developing V2V (vehicle-to-vehicle) communication to warn drivers of collisions.

7. Motion in a plane — uniform circular motion

Motion in two dimensions: overtaking, a kicked ball, a satellite.

In one revolution of radius R, distance = 2πR, displacement = 0. Average speed v = 2πR ÷ T.

Uniform circular motion: constant speed on a circular path. The direction of velocity changes continuously (along the tangent), so the motion is accelerated even though speed is constant.

Activity 4.5: a marble running inside a tape ring flies off in a straight line when the ring is lifted — it keeps moving in the direction it had at that instant.

Worked Numericals

Braking distance (book Example 4.8): u = 15 m s⁻¹, v = 0, a = –4 m s⁻². From v² = u² + 2as → 0 = 225 – 8s → s = 225/8 = 28.1 m. For u = 30 m s⁻¹: s = 900/8 = 112.5 m.

Car from rest (exercise): u = 0, v = 24 m s⁻¹, t = 6 s → a = 24/6 = 4 m s⁻²; s = ½ × 4 × 6² = 72 m.

Motorbike stopping (exercise): u = 28 m s⁻¹, s = 98 m, v = 0 → 0 = 784 + 196a → a = –4 m s⁻²; t = (0 – 28)/(–4) = 7 s.

Road trip (Pause and Ponder): 200 km north in 3 h, then 200 km south in 2 h → average speed = 400 km ÷ 5 h = 80 km h⁻¹; displacement 0 → average velocity 0.

Stairs (exercise): 4 floors up (12 m) then 2 floors down (6 m), each floor 3 m → distance 18 m, displacement 6 m upward.

Circular track (own example): R = 35 m, one lap in 22 s → v = 2 × (22/7) × 35 ÷ 22 = 10 m s⁻¹.

Activities in the Chapter

Activity 4.1: ball thrown up from O to B and back — fill a table of distance and displacement; conclude that |displacement| ≤ distance.

Activity 4.2: look up 0–100 km h⁻¹ times of cars and calculate their average acceleration.

Activity 4.3: plot a position–time graph on graph paper for a vehicle covering 20 m every second (a straight line).

Activity 4.4: draw a triangle on the graph to find the slope (change in position ÷ change in time) = velocity.

Activity 4.5: marble in a tape ring shows velocity is along the tangent in circular motion.

Key Terms

Reference point (संदर्भ बिंदु) — fixed point from which position is measured

Displacement (विस्थापन) — net change in position, with direction

Scalar (अदिश) — needs only magnitude (distance, speed)

Vector (सदिश) — needs magnitude and direction (displacement, velocity, acceleration)

Average speed (औसत चाल) — total distance ÷ time

Average velocity (औसत वेग) — displacement ÷ time

Acceleration (त्वरण) — rate of change of velocity

Uniform motion (एकसमान गति) — equal distances in equal times

Slope (ढाल) — steepness of a graph line

Kinematic equations (गति के समीकरण) — relate s, t, u, v and a

Acceleration due to gravity, g (गुरुत्वीय त्वरण) — 9.8 m s⁻²

Uniform circular motion (एकसमान वृत्तीय गति) — constant speed on a circle

Tangent (स्पर्श रेखा) — direction of velocity on a circle

Real-life Connections

Keeping a safe distance on roads; braking distance rises with the square of speed and also depends on wet roads, worn tyres, fog, night driving and reaction time.

A speedometer reads instantaneous speed; odometers and fuel use depend on total distance, not displacement.

Merry-go-rounds, satellites and vehicles turning on circular roads are all accelerated motion.

Common Misconceptions

"Displacement and distance are the same." — Only when motion is in one direction without turning back.

"Zero velocity means zero acceleration." — At the top of a vertical throw the ball is momentarily at rest but still accelerating (g).

"Constant speed means no acceleration." — Not in circular motion, where direction keeps changing.

"A fast object must be accelerating." — Acceleration depends on how quickly velocity changes, not on how fast it is.

Using kinematic equations when acceleration is not constant — they are valid only for constant acceleration.

Forgetting to convert km h⁻¹ to m s⁻¹ (multiply by 1000/3600, e.g. 54 km h⁻¹ = 15 m s⁻¹).

Revise, Reflect, Refine — Hints

Draw a straight line with O marked before solving any distance/displacement question.

Always write u, v, a, s, t with signs before choosing an equation.

For graph questions: slope of position–time = velocity; slope of velocity–time = acceleration; area under velocity–time = displacement.

Two parallel straight position–time lines mean equal velocities (same slope).

💡 Key Learning Points

✓Distinguish distance (scalar) from displacement (vector) and average speed from average velocity.
✓Calculate average acceleration and interpret its sign (speeding up vs slowing down).
✓Read position–time and velocity–time graphs: slope gives velocity/acceleration, area gives displacement.
✓Apply v = u + at, s = ut + ½at² and v² = u² + 2as to motion with constant acceleration, including braking distance.
✓Explain why uniform circular motion is accelerated even though the speed is constant.

👨‍🏫 Teaching Tips

→Mark a number line on the classroom floor with chalk and have a student walk the athlete example (O→A→B) while the class records distance and displacement.
→Use the Sarang swimming-pool example to spark debate: how can average velocity be zero when he swam hard for 50 s?
→Have groups do Activity 4.3 on graph paper, then swap graphs and find slopes of each other's lines.
→Show the braking-distance numbers (28.1 m vs 112.5 m) and link them to road-safety posters from 'The Journey Beyond'.
→Drop a ball beside a metre scale while recording slow-motion video on a phone to show velocity increasing each second (g).
→Do the marble-in-a-tape-ring activity (4.5) as a class demonstration and ask students to predict the path before lifting the ring.
→Insist on writing u, v, a, s, t with signs before solving any numerical; check units (km h⁻¹ → m s⁻¹) first.

📋 Assessment Questions

Can the student find distance and displacement for a to-and-fro motion on a straight line?
Can the student convert km h⁻¹ to m s⁻¹ and calculate acceleration with the correct sign?
Can the student identify uniform/non-uniform motion from position–time and velocity–time graphs and compute slope and area?
Can the student choose the right kinematic equation and solve for braking distance or stopping time?
Can the student explain why a body in uniform circular motion is accelerating?

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