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Rajasthan (RBSE)Class 9 Science★ 7–9 marks · RBSE Class 9 Science (new NCF — Curiosity)

Describing Motion Around Us

The new NCF Class 9 Science chapter on motion. Reference points and position; distance vs displacement; speed, velocity and acceleration; uniform and non-uniform motion; distance–time and velocity–time graphs; the three equations of motion; and uniform circular motion — with fully solved board-pattern numericals.

⏱ 38 min readHard difficulty✓ Free · NCERT-aligned

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The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Define reference point, position, rest and motion, and explain why motion is relative
2Distinguish distance (scalar) from displacement (vector) and compute each for a path
3Define and calculate speed, velocity and acceleration with correct SI units
4Classify motion as uniform or non-uniform and link it to acceleration
5Interpret and draw distance–time and velocity–time graphs (slope = speed/acceleration; area = distance)
6Apply the three equations of motion, including free fall (a = g)

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Why this chapter matters

Motion is the gateway to all of mechanics. The three equations of motion and the graph skills you build here are reused in 'How Forces Affect Motion', in Class 11 kinematics, and directly in JEE/NEET. It is also the most numerical-heavy chapter of Class 9 Science, so it is a reliable source of board marks if you drill the problems.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Speed, distance and time (Class 6–7 Curiosity)

Basic speed = distance/time and units.

Describing Motion Around Us (RBSE Class 9 · Science)

A train, a falling ball, a child on a swing — nature is full of motion. To study it, scientists strip away the mess and look at the simplest forms first: motion in a straight line. This chapter teaches you to describe motion not just in words, but in numbers, equations and graphs.

RBSE note (2026-27). Class 9 uses the new NCF (Curiosity) Science textbook. This chapter, Describing Motion Around Us, is the new book's treatment of kinematics (it builds on speed/distance from the Grade 6–7 Curiosity books). BSER (Ajmer) sets the exam.

1. Reference point, rest and motion

To describe where an object is, we choose a fixed reference point (origin) and state the distance and direction from it — that is the object's position.

An object is in motion if its position changes with time relative to the reference point.
It is at rest if its position does not change.

Motion is relative: a passenger sitting in a moving train is at rest relative to the train but in motion relative to the platform.

2. Distance and displacement

Quantity	Meaning	Type
Distance	total path length covered	scalar (magnitude only)
Displacement	shortest (straight-line) distance from start to finish, with direction	vector (magnitude + direction)

Distance is always ≥ |displacement|. If you walk 3 m east then 4 m west, distance = 7 m but displacement = 1 m west. For a complete round trip, displacement = 0 while distance ≠ 0.

3. Speed, velocity and acceleration

Speed = distance per unit time (scalar): $speed = \frac{distance}{time}$

Velocity = displacement per unit time (vector — speed in a stated direction): $v = \frac{displacement}{time}$

Unit of both: m/s. Average speed = total distance / total time.

Acceleration = rate of change of velocity: $a = \frac{v - u}{t}$

where u = initial velocity, v = final velocity, t = time. Unit: m/s². Acceleration is positive when speeding up, negative (retardation/deceleration) when slowing down, and zero for uniform velocity.

4. Uniform and non-uniform motion

Uniform motion: equal distances in equal intervals of time → constant velocity, zero acceleration.
Non-uniform motion: unequal distances in equal intervals → velocity changes → there is acceleration.

5. Graphs of motion

Distance–time graph

A straight line through changing distance = uniform speed; its slope = speed.
A line parallel to the time axis (flat) = the object is at rest.
A curve (getting steeper) = non-uniform, accelerated motion.

Velocity–time graph

A line parallel to the time axis = uniform velocity (zero acceleration).
A straight slanting line = uniform acceleration; its slope = acceleration.
The area under a velocity–time graph = distance travelled.

6. The three equations of motion (uniform acceleration)

For motion with constant acceleration a, initial velocity u, final velocity v, time t and displacement s:

$v = u + a t$ $s = u t + \frac{1}{2} a t^{2}$ $v^{2} = u^{2} + 2 a s$

These are the workhorses for every numerical in this chapter. For a freely falling body, a = g ≈ 9.8 m/s² (downward); for a body thrown up, a = −g until it stops momentarily at the top.

7. Uniform circular motion

When an object moves in a circle at constant speed, it is in uniform circular motion. Although the speed is constant, the direction of motion changes continuously — so the velocity changes, which means the motion is accelerated. The speed of an object going once around a circle of radius r in time T is:

$v = \frac{2 π r}{T}$

Examples: the moon around the earth, a stone whirled on a string, the tip of a clock's second hand.

8. Worked example

A car starts from rest and accelerates uniformly at 2 m/s² for 5 s. Find (a) its final velocity and (b) the distance covered.

Given: u = 0, a = 2 m/s², t = 5 s.

(a) v = u + at = 0 + 2 × 5 = 10 m/s.

(b) s = ut + ½at² = 0 + ½ × 2 × 5² = ½ × 2 × 25 = 25 m.

9. Quick recap

Motion is described relative to a reference point; it is relative.
Distance (scalar, path length) vs displacement (vector, shortest with direction).
Speed uses distance; velocity uses displacement; acceleration = (v − u)/t.
Uniform motion → zero acceleration; non-uniform → there is acceleration.
Distance–time slope = speed; velocity–time slope = acceleration; area under v–t graph = distance.
Three equations: v = u + at; s = ut + ½at²; v² = u² + 2as. Free fall: a = g ≈ 9.8 m/s².
Uniform circular motion is accelerated because direction (hence velocity) keeps changing.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Speed

speed = distance / time

Scalar; unit m/s.

Velocity

v = displacement / time

Vector; speed in a stated direction.

Acceleration

a = (v − u) / t

Unit m/s²; negative = retardation.

First equation

v = u + at

Velocity after time t under uniform acceleration.

Second equation

s = ut + ½at²

Displacement in time t.

Third equation

v² = u² + 2as

Links velocity and displacement (no t).

Free fall

a = g ≈ 9.8 m/s²

Downward; use −g for upward throw.

Circular speed

v = 2πr / T

Once around a circle of radius r in time T.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Treating distance and displacement as the same

✓ Distance is total path (scalar); displacement is the shortest straight line with direction (vector). For a round trip displacement = 0 but distance ≠ 0.

WATCH OUT

✗ Forgetting that uniform circular motion is accelerated

✓ Speed is constant but DIRECTION changes, so velocity changes — hence there IS acceleration (centripetal).

WATCH OUT

✗ Using the wrong equation of motion

✓ Pick by what's missing: no s → v=u+at; no v → s=ut+½at²; no t → v²=u²+2as.

WATCH OUT

✗ Wrong sign for deceleration or upward throw

✓ Slowing down → a is negative. Body thrown up → a = −g (9.8 m/s² downward) until it stops at the top.

WATCH OUT

✗ Reading area under a distance–time graph as distance

✓ Distance = area under the VELOCITY–time graph. On a distance–time graph the SLOPE gives speed.

WATCH OUT

✗ Not converting km/h to m/s

✓ Multiply km/h by 5/18 to get m/s before using equations (e.g. 36 km/h = 10 m/s).

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

In-text (position & motion)

Reference point, distance vs displacement

Questions

In-text (speed, velocity, acceleration)

Definitions, units, simple numericals

Questions

In-text (graphs)

Interpreting distance–time and velocity–time graphs

Questions

End-of-chapter Ex

End-of-chapter exercise

Mostly numericals using the three equations of motion

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Distance/displacement

A boy walks 4 m east, then 3 m west. Find the distance and the displacement.

▶ Show solution

Distance = 4 + 3 = 7 m.

Displacement = 4 − 3 = 1 m towards east.

✦ Answer: distance 7 m; displacement 1 m east.

Q2EASY· Speed

Convert 36 km/h into m/s.

▶ Show solution

Step 1 — multiply by 5/18.

Step 2 — 36 × 5/18 = 10 m/s.

✦ Answer: 10 m/s.

Q3EASY· Acceleration

A car's velocity rises from 5 m/s to 20 m/s in 3 s. Find its acceleration.

▶ Show solution

Step 1 — a = (v − u)/t = (20 − 5)/3.

Step 2 — a = 15/3 = 5 m/s².

✦ Answer: 5 m/s².

Q4MEDIUM· Equations of motion

A body starts from rest and accelerates at 4 m/s² for 6 s. Find the distance covered.

▶ Show solution

Step 1 — s = ut + ½at², u = 0.

Step 2 — s = 0 + ½ × 4 × 6² = ½ × 4 × 36.

Step 3 — s = 72 m.

✦ Answer: 72 m.

Q5MEDIUM· Free fall

A stone is dropped from rest. What is its velocity after 2 s? (g = 9.8 m/s²)

▶ Show solution

Step 1 — v = u + at, u = 0, a = g = 9.8.

Step 2 — v = 0 + 9.8 × 2 = 19.6 m/s.

✦ Answer: 19.6 m/s (downward).

Q6MEDIUM· Graphs

On a velocity–time graph, a body moves at constant 10 m/s for 5 s. What does the area under the graph represent and what is its value?

▶ Show solution

Step 1 — Area under a v–t graph = distance travelled.

Step 2 — Area = 10 m/s × 5 s = 50 m.

✦ Answer: distance = 50 m.

Q7HARD· Third equation

A car moving at 20 m/s brakes and stops in 50 m. Find its (uniform) retardation.

▶ Show solution

Step 1 — v² = u² + 2as, with v = 0, u = 20, s = 50.

Step 2 — 0 = 20² + 2a(50) → 0 = 400 + 100a.

Step 3 — a = −400/100 = −4 m/s².

✦ Answer: retardation = 4 m/s².

Q8HARD· Circular motion

A stone tied to a string is whirled in a circle of radius 0.5 m, completing one round in 1 s. Find its speed, and state why the motion is accelerated.

▶ Show solution

Step 1 — v = 2πr/T = 2 × 3.14 × 0.5 / 1.

Step 2 — v = 3.14 m/s.

Step 3 — Although the speed is constant, the direction (hence velocity) changes continuously, so the motion is accelerated.

✦ Answer: 3.14 m/s; accelerated because direction/velocity changes.

Q9HARD· Vertical throw

A ball is thrown straight up at 19.6 m/s. Find (a) the time to reach the top and (b) the maximum height. (g = 9.8 m/s²)

▶ Show solution

Step 1 — At the top v = 0; a = −9.8 m/s², u = 19.6.

Step 2 — v = u + at → 0 = 19.6 − 9.8t → t = 2 s.

Step 3 — v² = u² + 2as → 0 = 19.6² − 2(9.8)s → s = 384.16/19.6 = 19.6 m.

✦ Answer: (a) 2 s, (b) 19.6 m.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Position is distance + direction from a reference point; motion is relative.
•Distance = path length (scalar); displacement = shortest straight line + direction (vector); distance ≥ |displacement|.
•Speed = distance/time; velocity = displacement/time; both in m/s.
•Acceleration a = (v − u)/t (m/s²); negative = retardation; uniform velocity → a = 0.
•Distance–time slope = speed; velocity–time slope = acceleration; area under v–t = distance.
•Equations: v = u + at; s = ut + ½at²; v² = u² + 2as. Choose by the missing quantity.
•Free fall a = g ≈ 9.8 m/s²; upward throw uses a = −g.
•Uniform circular motion: constant speed but changing direction → accelerated; v = 2πr/T.

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 7–9 marks

Question type	Marks each	Typical count	What it tests
MCQ / Assertion–Reason	1	1–2	Scalar/vector, units, graph reading
Short answer / numerical	2	2	Acceleration, single-equation numericals, graph area
Short answer	3	1	Third equation; circular motion reasoning
Long / numerical	4–5	1	Multi-step problem (vertical throw, graphs)

Prep strategy

Memorise the three equations and choose by the missing variable
Always convert km/h → m/s (×5/18) and write units at every step
Practise sketching and reading both graph types; remember area under v–t = distance
Drill 15+ numericals mixing free fall, braking and vertical throw

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Safe following distance

Braking distance grows with speed (v² = u² + 2as) — exactly why you keep more gap behind a fast truck.

Speedometer vs odometer

The speedometer shows instantaneous speed; the odometer adds up distance — the two quantities of this chapter.

Lift / elevator

You feel heavier as it accelerates up and lighter as it decelerates — changing velocity in action.

Satellite and moon orbits

Uniform circular motion: constant speed, continuously changing direction, hence centripetal acceleration.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

List u, v, a, t, s for every numerical, then pick the equation missing the unwanted variable.
Convert all data to SI units (km/h → m/s) before substituting.
Mind the signs: deceleration and upward motion use negative acceleration.
For graph questions, state slope = speed/acceleration and area (v–t) = distance explicitly.
Always write the final answer with its unit and direction where relevant.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Relative velocity in one dimension (two trains, river–boat problems).
Centripetal acceleration a = v²/r and the force that causes it.
Non-uniform acceleration and the idea of instantaneous velocity (slope of the tangent).

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 9 Annual (BSER Ajmer)Very high — numericals + a graph almost every year

NTSE / NMMSHigh — kinematics MCQs and graph reading

JEE FoundationVery high — direct base for Class 11 kinematics

Science Olympiad (NSO)High — equations of motion and graphs

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes. Class 9 (2026-27) uses the new NCF NCERT 'Curiosity' Science book — the same as CBSE. 'Describing Motion Around Us' is its kinematics chapter. BSER Ajmer sets the RBSE paper.

Never. Displacement is the shortest path, so |displacement| ≤ distance. They are equal only for motion in a straight line without reversing direction.

Yes — in uniform circular motion the speed is constant but the direction (and hence velocity) changes continuously, so the body is accelerating.

Identify the quantity NOT given/asked: if displacement is absent use v = u + at; if final velocity is absent use s = ut + ½at²; if time is absent use v² = u² + 2as.

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