Physics Chapter 1 — Motion | Class 9 CBSE Science 2026-27

Section 1.1 – 1.2 · The Basics

Rest, Motion & the Reference Point

A body is said to be in motion when its position changes continuously with time, with respect to a stationary object chosen as a reference point. From atoms to galaxies, motion is everywhere — it is a universal phenomenon.

🚌 Why rest and motion are "relative"

Whether a body is at rest or moving depends entirely on the reference point you choose. A passenger sitting in a moving bus is at rest relative to other passengers, but in motion relative to a person standing on the road. Neither observer is "wrong" — they have simply chosen different reference points.

This is why we always say motion is relative. To describe any motion we must first fix what we are measuring it against.

A motorcycle changes its position (A → B → C) with respect to the fixed tree — so it is in motion along a straight line.

🧭

In this chapter we study only rectilinear motion — motion of objects along a straight line.

Section 1.3 · Physical Quantities

Scalar & Vector Quantities

Every physical quantity used to describe motion is either a scalar or a vector, depending on whether direction is part of its definition.

📏 Scalar Quantity

Fully described by magnitude (size) alone — no direction. Always positive or zero.

Examples: distance, speed, mass, time, volume, density, energy, temperature.

➡️ Vector Quantity

Needs both magnitude and direction. May be positive, negative or zero. Written with an arrow above the symbol.

Examples: displacement, velocity, acceleration, force, momentum — e.g. v⃗, a⃗, F⃗.

The same "50 km/h" becomes a velocity once we add a direction.

⚠️

Key rule: vectors pointing in different directions cannot simply be added like ordinary numbers — they follow special (geometric) rules. Scalars add by plain arithmetic.

Section 1.4 · The First Big Idea

Distance & Displacement

These two words feel similar in everyday speech, but in physics they are sharply different — and the difference is tested in almost every exam.

Distance

The length of the actual path covered by a moving body, regardless of direction. A scalar. Always positive; never zero for a body that has moved.

Displacement

The shortest straight-line distance from the initial to the final position, with direction. A vector. Can be positive, negative or zero.

Worked picture: 3 m N → 4 m E → 6 m S

Net displacement OR = √(east² + south²) = √(4² + 3²) = 5 m. Distance = 13 m.

Given

Path: 3 m North, then 4 m East, then 6 m South.

Distance

3 + 4 + 6 = 13 m (just add the path lengths)

Net legs

North 3 − South 6 = 3 m South; East = 4 m

Formula

Displacement = √(x² + y²) (Pythagoras on the net legs)

Substitute

√(4² + 3²) = √(16 + 9) = √25

Answer

Distance = 13 m · Displacement = 5 m

Point of difference	Distance	Displacement
Nature	Scalar (magnitude only)	Vector (magnitude + direction)
Measures	Actual path length	Shortest gap, initial → final
Sign	Always positive	Can be +, − or 0
For a moving body	Never zero	Can be zero (round trip)
Golden rule: Distance ≥ \|Displacement\|, always (equal only for straight-line motion without a turn)

💡

Exam favourite: in a full round of a circular track or any return-to-start trip, displacement = 0 while distance is large.

Section 1.5 · Classifying Motion

Uniform & Non-Uniform Motion

We classify motion by whether equal distances are covered in equal intervals of time — however small those intervals are.

Uniform Motion

Equal distances in equal time intervals. Example: a car covering 50 km every hour, or 5 km every 6 minutes. Its distance–time graph is a straight line.

Non-Uniform Motion

Unequal distances in equal time intervals. Example: a freely falling stone (covering more distance each second), or a train pulling out of a station. Its distance–time graph is a curve.

Uniform motion: equal 50 km steps in each equal 1-hour interval.

⚡

When speed increases with time → accelerated motion. When speed decreases → retarded motion.

Section 1.6 · How Fast?

Speed

Speed tells us how fast a body moves — the distance it covers per unit time. It is a scalar.

Speed

v = s / t

s = distance, t = time. SI unit: m/s (m s⁻¹). Other units: km/h, cm/s.
Speed is a scalar — zero or positive, never negative.
Conversion: 1 km/h = 1000 m ÷ 3600 s = 5/18 m/s

The four types of speed

1. Uniform

const

Equal distances in equal time intervals.

2. Non-uniform

varies

Unequal distances in equal time intervals.

3. Average

s/t

Total distance ÷ total time for the whole journey.

4. Instantaneous

now

Speed at a single instant — read off the speedometer.

Average Speed

average speed = total distance / total time

⚠️ It is NOT the ordinary average of the speeds (unless equal times are spent at each speed).

🚗 Speedometer vs Odometer

The speedometer shows the vehicle's instantaneous speed (usually km/h). The odometer records the total distance the vehicle has travelled. Two different instruments, two different jobs.

Section 1.7 · How Fast & Which Way?

Velocity

Velocity is speed in a stated direction — the rate of change of displacement. It is a vector.

Velocity & Average Velocity

velocity = displacement / time

SI unit: m/s. A vector (can be +, − or 0).
When velocity changes uniformly: average velocity = (u + v) / 2

Uniform velocity

const

Both magnitude and direction stay constant → straight-line path.

Variable velocity

changes

Magnitude or direction (or both) changes with time.

Average velocity

(u+v)/2

For uniformly changing velocity only.

Property	Speed	Velocity
Type	Scalar	Vector
Based on	Distance	Displacement
Direction	Not specified	Specified
Sign	Zero or positive	+, − or 0
When equal?	Average speed = average velocity only for straight-line motion without turning

Section 1.8 · How Velocity Changes

Acceleration

Acceleration is the rate of change of velocity with time — how quickly a body speeds up or slows down.

Acceleration

a = (v − u) / t

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

➕ Positive / Uniform Acceleration

Velocity increases by equal amounts in equal time intervals. The classic example is free fall, where a = g ≈ 9.8 m/s² (downward). Acceleration acts in the direction of motion.

➖ Negative Acceleration (Retardation)

Velocity decreases with time — also called deceleration. Example: a vehicle braking. Acceleration acts opposite to the direction of motion, so we write it with a minus sign.

Uniform vs Non-uniform acceleration

Uniform acceleration: velocity changes by equal amounts in equal times (e.g. free fall). Non-uniform acceleration: velocity changes by unequal amounts in equal times.

Given

A body slows from 40 m/s to rest in 5 s.

Formula

a = (v − u) / t

Substitute

a = (0 − 40) / 5

Calculate

a = −40 / 5 = −8 m/s² — the minus sign means retardation.

Answer

Acceleration = −8 m/s² (retardation of 8 m/s²)

Section 1.9 · The Toolkit

The Kinematic Equations of Motion

For a body moving in a straight line with uniform acceleration, three equations connect the five quantities u, v, a, t and s. Master these and you can solve almost any numerical in this chapter.

The Three Equations of Motion

1. v = u + at (velocity–time)

2. s = ut + ½at² (position–time)

3. v² = u² + 2as (position–velocity)

Bonus (JEE-useful): distance in the nth second Sₙ = u + ½a(2n − 1)

Which equation do I pick? — the "missing variable" trick

Equation	Contains	Variable it leaves OUT	Use when…
v = u + at	u, v, a, t	s (distance)	distance not needed/given
s = ut + ½at²	u, a, t, s	v (final velocity)	final velocity not needed/given
v² = u² + 2as	u, v, a, s	t (time)	time not needed/given

🧭 The 4 setup rules before every numerical

Same units everywhere. Convert km/h → m/s (× 5/18) before substituting.

"Starts from rest" → u = 0.

"Comes to rest / stops" → v = 0.

"Uniform velocity" → a = 0; and for slowing down, write a as negative.

Where do these equations come from? (graph derivation)

Take a velocity–time graph: a straight line rising from initial velocity u (at t = 0) to final velocity v at time t. The slope of this line is the acceleration a.

① v = u + at: slope a = (v − u)/t ⟹ rearrange ⟹ v = u + at.

② s = ut + ½at²: distance = area under the line = rectangle (u × t) + triangle (½ × t × (v − u)). Since v − u = at, area = ut + ½at².

③ v² = u² + 2as: distance = area of trapezium = ½(u + v)t. From ①, t = (v − u)/a. Substitute ⟹ s = (v² − u²)/2a ⟹ v² = u² + 2as.

Section 1.10 · The Deep Dive

Graphical Representation of Motion

Graphs let us "see" motion. The two key graphs are the position–time (distance–time) graph and the velocity–time (speed–time) graph. Two ideas unlock everything: SLOPE and AREA.

🔑 The two master keys

Slope of a distance–time graph = speed.
Slope of a velocity–time graph = acceleration.
Area under a velocity–time graph = distance / displacement.

A · Distance–Time (Position–Time) Graphs — three cases

Case 1 — Body at rest

The position does not change with time, so the graph is a horizontal line parallel to the time axis.

Case 2 — Uniform velocity (slope gives velocity)

Equal distances in equal times ⟹ a straight slanting line. The slope = velocity. Steeper line = faster body.

Calculation: velocity = displacement ÷ time = (x₂ − x₁)/(t₂ − t₁) = tan θ = slope.

Case 3 — Non-uniform motion (accelerated / retarded)

If velocity changes, the graph is a curve. An increasing slope (curve bending upward) means the body is accelerating; a decreasing slope means it is retarding. The slope of the tangent at any point gives the instantaneous velocity there.

Conclusions from a distance–time graph

1) Line parallel to time axis → body is at rest. 2) Straight slanting line → uniform velocity (slope = velocity). 3) Curve → variable velocity (accelerated or retarded); slope of tangent = velocity at that instant.

B · Velocity–Time (Speed–Time) Graphs — three cases

Case (i) — Constant velocity (no acceleration)

Velocity is unchanging ⟹ a horizontal line parallel to the time axis (a = 0). The area of the rectangle below it = distance travelled.

Calculation of distance

distance = velocity × time = area of rectangle

Case (ii) — Uniform acceleration (straight slanting line)

Velocity increases by equal amounts in equal times ⟹ a straight line inclined to the time axis. From this single graph we can read off three things:

1. Acceleration = slope = (v − u)/t = tan θ.

2. If the body starts from rest (line from origin): distance = area of the triangle = ½ × base × height = ½ × t × v.

3. If initial velocity is not zero (line starts above origin): distance = area of the trapezium = ½ (u + v) × t. This is exactly how equation 2 is derived.

From a uniform-acceleration v–t graph

a = slope · distance = ½(u + v)t = area of trapezium

Case (ii-b) — Uniform retardation (line sloping down)

If velocity decreases uniformly to zero, the line slopes downward. The slope (and hence acceleration) is negative → the body undergoes uniform retardation. Distance is still the area under the line.

Case (iii) — Non-uniform acceleration (curve)

If velocity changes irregularly, the v–t graph is a curve. The acceleration at any instant = slope of the tangent at that point; a steeper tangent means greater acceleration. The area under the curve still gives the distance.

Calculation summary — slope & area at a glance

Want to find…	From distance–time graph	From velocity–time graph
Velocity / speed	slope of the line	read the y-value directly
Acceleration	—	slope of the line
Distance / displacement	read the y-value	area under the line

Conclusions from a velocity–time graph

1) Line parallel to time axis → uniform velocity, a = 0; area gives distance. 2) Straight slanting line → uniform acceleration (slope = a; positive slope = speeding up, negative = retardation); area gives distance. 3) Curve → non-uniform (variable) acceleration; slope of tangent = a at that instant; area gives distance.

Worked graph problems

Graph Example 1

A cyclist rides at a uniform 8 m/s for 8 s, then stops paddling and the cycle comes to rest in the next 10 s. Find (a) the retardation, (b) distance at uniform velocity, (c) distance while slowing, (d) average velocity. [5 Marks]

(a) Formula

retardation = slope of BC = (8 − 0)/(18 − 8) = 8/10 = 0.8 m/s²

(b) Formula

distance at uniform v = area of rectangle = 8 × 8 = 64 m

distance while slowing = area of triangle = ½ × 8 × 10 = 40 m

(d) Formula

average velocity = total distance/total time = (64 + 40)/18 = 104/18 ≈ 5.77 m/s

Answer

0.8 m/s² · 64 m · 40 m · 5.77 m/s

Graph Example 2

A car's v–t graph rises from rest to 40 km/h in 2 h (A→B), stays at 40 km/h for 2 h (B→C), then falls to 0 in 2 h (C→D). Find the acceleration in the first and last 2 hours, and the total distance. [5 Marks]

First 2 h

a = (40 − 0)/2 = 20 km/h²

Last 2 h

a = (0 − 40)/2 = −20 km/h² (retardation)

Distance

area = ½(2)(40) + (2)(40) + ½(2)(40) = 40 + 80 + 40

Answer

+20 km/h², −20 km/h², total distance = 160 km

Graph Example 3

A car has uniform acceleration. Speed (m/s): 5, 10, 15, 20, 25, 30 at times (s): 0, 10, 20, 30, 40, 50. Find (a) acceleration and (b) distance in 50 s. [4 Marks]

(a) slope

a = (30 − 5)/(50 − 0) = 25/50 = 0.5 m/s²

(b) area

distance = ½(u + v)t = ½(5 + 30)(50) = ½ × 35 × 50

Answer

a = 0.5 m/s² · distance = 875 m

Section 1.11 · Going in Circles

Uniform Circular Motion & Closed Paths

When a body moves around a fixed point along a circular path at constant speed, it is in uniform circular motion. Even though the speed is constant, the direction changes at every instant — so the velocity changes, which means the motion is accelerated.

🔄 Why "constant speed" still means "accelerated"

Velocity is a vector (speed + direction). On a circle, the direction of motion is always along the tangent, so it keeps turning. A change in direction is a change in velocity, and any change in velocity is acceleration. This acceleration points toward the centre and is called centripetal acceleration; the force producing it is the centripetal force.

Velocity v is always tangent to the circle; the centripetal acceleration a points toward the centre O.

Speed in uniform circular motion

v = 2πr / t

In one full revolution the body covers the circumference 2πr in time t (the time period). r = radius.

Why a closed path becomes a circle

Watch an athlete run a closed track and count how often they change direction: on a square they turn at 4 corners, on a hexagon at 6, on an octagon at 8. As the number of sides increases, the turns get more frequent and the sides get shorter. When the number of sides becomes very large (sides → 0 length), the shape becomes a circle, and the direction changes continuously. That is why circular motion is the limit of motion along a closed polygon — and why direction (hence velocity) is always changing.

More sides → more frequent turns → in the limit, a circle with continuously changing direction.

Calculations on a circular / closed path

Circular Example 1

A cyclist goes once around a circular track of diameter 105 m in 5 minutes. Find the speed. (π = 22/7) [3 Marks]

Given

diameter = 105 m → r = 52.5 m; t = 5 min = 300 s

Formula

v = 2πr / t (distance in one round = circumference)

Substitute

v = 2 × (22/7) × 52.5 / 300

Calculate

= 330 / 300 = 1.1 m/s

Answer

Speed = 1.1 m/s

Circular Example 2

A circular track has circumference 314 m, with AB as a diameter (north–south). A cyclist rides from A to B along the curve at a constant 15.7 m/s. Find (a) distance, (b) displacement, (c) average velocity. [3 Marks]

(a) Distance

half the circumference = 314 / 2 = 157 m

(b) Displacement

straight A→B = diameter = circumference/π = 314 / 3.14 = 100 m (south)

t = distance/speed = 157 / 15.7 = 10 s

= displacement/time = 100 / 10 = 10 m/s

Answer

157 m · 100 m (south) · 10 m/s

Closed-Path Example 3

A body covers a distance l along a semicircular path. Find the magnitude of displacement and the ratio distance : displacement. [3 Marks]

Given

semicircle, distance = arc = πr = l

Radius

from πr = l → r = l/π

Displacement

straight across = diameter = 2r = 2l/π

Ratio

distance : displacement = πr : 2r = π : 2

Answer

Displacement = 2l/π · ratio = π : 2 ≈ 1.57 : 1

Circular Example 4

A cyclist on a circular track of radius 50 m completes one revolution in 4 minutes. Find (a) average speed and (b) average velocity. (π = 3.14) [3 Marks]

Given

r = 50 m; t = 4 min = 240 s

(a) Avg speed

= 2πr/t = 2 × 3.14 × 50 / 240 = 314/240 ≈ 1.31 m/s

(b) Avg velocity

one full revolution → displacement = 0 → 0 / 240 = 0 m/s

Answer

Average speed ≈ 1.31 m/s · Average velocity = 0

🌍 Real examples of uniform circular motion

The Moon revolving around the Earth (force = Earth's gravity).
The Earth revolving around the Sun (force = Sun's gravity).
An artificial satellite orbiting the Earth.
A stone whirled on a string; tips of a clock's second, minute and hour hands.

Problem-Solving Masterclass

Every Formula, Applied Step by Step

The goal here is not the answer — it's the method. Each problem follows the same routine: Given → Formula → Substitute → Calculate → Answer. Train your eye to spot which formula a problem is asking for.

M1 · uses v = u + at

A car starts from rest and accelerates uniformly at 2 m/s² for 8 s. Find its final velocity.

Given

u = 0 (from rest), a = 2 m/s², t = 8 s; find v

Formula

distance not needed → use v = u + at

Substitute

v = 0 + 2 × 8

Calculate

v = 16 m/s

Answer

Final velocity = 16 m/s

M2 · uses s = ut + ½at²

A motorbike at 18 km/h accelerates at 5 m/s² for 5 s. Find the distance covered.

Given

u = 18 km/h, a = 5 m/s², t = 5 s; find s

Convert

u = 18 × 5/18 = 5 m/s

Formula

final velocity not needed → use s = ut + ½at²

Substitute

s = 5×5 + ½×5×5²

Calculate

s = 25 + 62.5 = 87.5 m

Answer

Distance = 87.5 m

M3 · uses v² = u² + 2as

A bus moving at 20 m/s brakes and stops after 50 m. Find the retardation.

Given

u = 20 m/s, v = 0 (stops), s = 50 m; find a

Formula

time not given → use v² = u² + 2as

Substitute

0² = 20² + 2·a·50

Calculate

0 = 400 + 100a → a = −4 m/s²

Answer

Retardation = 4 m/s²

M4 · two equations together

A scooter brakes to rest in 1.5 s with a retardation of 5 m/s². Find the initial speed in m/s and km/h.

Given

v = 0, t = 1.5 s, a = −5 m/s²; find u

Formula

v = u + at

Substitute

0 = u + (−5)(1.5)

Calculate

u = 7.5 m/s = 7.5 × 18/5 = 27 km/h

Answer

Initial speed = 7.5 m/s = 27 km/h

M5 · average speed (the trap)

A car covers 20 km at 60 km/h and 40 km at 80 km/h. Find the average speed.

Formula

average speed = total distance / total time (NOT average of speeds)

Times

t₁ = 20/60 = 1/3 h; t₂ = 40/80 = 1/2 h; total = 5/6 h

Substitute

v = 60 ÷ (5/6) = 60 × 6/5

Answer

Average speed = 72 km/h (the "simple average" 70 is wrong!)

M6 · free fall

A ball is dropped from rest. Find its velocity and the distance fallen after 3 s (g = 9.8 m/s²).

Given

u = 0, a = g = 9.8 m/s², t = 3 s

Velocity

v = u + at = 0 + 9.8×3 = 29.4 m/s

Distance

s = ut + ½at² = 0 + ½×9.8×9 = 44.1 m

Answer

v = 29.4 m/s · distance = 44.1 m

M7 · vertical throw (v = 0 at top)

A stone is thrown up at 19.6 m/s. How high does it rise? (g = 9.8 m/s²)

Given

u = 19.6 m/s, v = 0 (at the top), a = −9.8 m/s² (up positive)

Formula

time not asked → use v² = u² + 2as

Substitute

0 = (19.6)² + 2(−9.8)s = 384.16 − 19.6s

Answer

s = 384.16/19.6 = 19.6 m

M8 · train crossing a bridge

A 100 m long train at 60 km/h crosses a 1 km bridge. Find the time.

Distance

train length + bridge = 100 + 1000 = 1100 m

Convert

v = 60 × 5/18 = 50/3 m/s

Formula

t = distance / v = 1100 ÷ (50/3) = 1100 × 3/50

Answer

t = 66 s = 1 min 6 s

M9 · nth-second formula

A body starts from rest with a = 2 m/s². Find the distance covered in the 5th second.

Formula

Sₙ = u + ½a(2n − 1)

Substitute

u = 0, a = 2, n = 5 → S₅ = 0 + ½×2×(2×5 − 1)

Calculate

= 1 × 9 = 9 m

Answer

Distance in 5th second = 9 m

M10 · JEE-style (relative motion)

Two bodies move toward each other and the gap closes at 8 m/s. Moving in the same direction at the same individual speeds, the gap opens at 4 m/s. Find each speed.

Given

towards each other: u + v = 8; same direction: u − v = 4

Method

add the two equations: 2u = 12 → u = 6

Then

v = 8 − 6 = 2 m/s

Answer

Speeds = 6 m/s and 2 m/s

Textbook Exercise · Fully Solved

Exercise — Every Type Covered

This section works through the chapter's exercise the way it is set in the book — first the quick conceptual answers (MCQ keys, fill-ups, match, VSA), then the numericals solved in full with the formula shown every time and, where useful, an alternate / shortcut method.

A · Multiple-Choice — answer key with reasoning

Question	Answer	Why
Greatest height of a body thrown up with velocity u	u²/2g	v²=u²−2gh, v=0 at top → h=u²/2g
Magnitude of displacement is always…	≤ distance travelled	shortest path can't exceed actual path
Boy: 2 rounds of circular track r=7 m	88 m distance, 0 displacement	2×2πr = 2×2×(22/7)×7 = 88 m; back to start → 0
Speedometer measures…	instantaneous speed	odometer measures distance
Retardation is expressed in…	m s⁻²	same unit as acceleration
Slope of velocity–time graph gives…	acceleration	(area gives distance)
v–t graph: horizontal straight line →	constant velocity	no change in velocity
Displacement ∝ (time)² → motion has…	uniform acceleration	s = ut + ½at² with u=0 → s ∝ t²
Bus 20 m/s, a = 4 m/s², after 2 s speed =	28 m/s	v = u+at = 20+4×2
Motorcycle 20 m/s → rest in 5 s, deceleration =	−4 m/s²	a = (0−20)/5

B · Assertion–Reason & conceptual one-liners

Displacement can be zero while distance is not — true; when the body returns to its start, initial and final positions coincide.
Uniform circular motion is accelerated — true; the direction of velocity changes continuously even at constant speed.
Speed can never be negative — true; it is a scalar (magnitude only), so it is zero or positive.
A body can have zero velocity and non-zero acceleration — true; e.g. at the top of a vertical throw.
Distance and displacement are equal only when the car moves on a straight road without turning.

C · Quick fill-ups & match (key terms)

Distance travelled by a moving body cannot be…	zero
Instantaneous speed of a vehicle is shown by the…	speedometer
v–t graph for non-uniform acceleration is a…	curved line
Velocity = slope of the…	position–time graph
Direction keeps changing along a…	circular path
Speed → scalar; Velocity → vector; "always positive" → distance/speed; "can be ±/0" → displacement/velocity	match keys

D · Numericals — full solution, formula + alternate method

Q1 · 5 Marks

A body starting from rest accelerates uniformly at 5 m/s². Find the distance covered in 20 s — by two methods.

Given

u = 0, a = 5 m/s², t = 20 s

Method 1 (equation)

s = ut + ½at² = 0 + ½×5×20² = ½×5×400 = 1000 m

Method 2 (graph)

v at 20 s = at = 100 m/s; distance = area of triangle = ½ × 20 × 100 = 1000 m ✓

Answer

Distance = 1000 m (both methods agree)

Q2 · 5 Marks

A body at 36 km/h accelerates at 4 m/s² for 15 s. Find the final velocity and distance — verify distance by an alternate method.

Convert

u = 36 × 5/18 = 10 m/s

Velocity

v = u + at = 10 + 4×15 = 70 m/s

Method 1 (s = ut+½at²)

= 10×15 + ½×4×225 = 150 + 450 = 600 m

Method 2 (v²=u²+2as)

70² = 10² + 8s → 4900 − 100 = 8s → s = 600 m ✓

Answer

v = 70 m/s · distance = 600 m

Q3 · 3 Marks

A cyclist at 15 m/s brakes to avoid a wall 18 m away. Find the minimum deceleration.

Given

u = 15 m/s, v = 0, s = 18 m

Formula

time not given → v² = u² + 2as

Substitute

0 = 225 + 2a(18) → 36a = −225

Answer

a = −6.25 m/s² → deceleration of 6.25 m/s²

Q4 · 5 Marks

On a 120 km track a train does the first 30 km at 30 km/h. What speed for the next 90 km gives an average of 60 km/h?

Total time

= total distance/avg speed = 120/60 = 2 h

Time for 30 km

= 30/30 = 1 h → time left = 2 − 1 = 1 h

Required speed

= remaining distance/time = 90/1

Answer

90 km/h

Q5 · 3 Marks

The second's hand of a clock is 10.5 cm long. Find the speed of its tip. (π = 22/7)

Given

r = 10.5 cm = 0.105 m; T = 60 s (one round per minute)

Formula

v = 2πr/T

Substitute

v = 2 × 22/7 × 0.105 / 60 = 0.66/60

Answer

v ≈ 0.011 m/s

Q6 · 3 Marks

A boy runs along the sides of a square field of side 100 m. Find the maximum magnitude of his displacement.

Reasoning

max displacement = straight line across the diagonal of the square

Formula

diagonal = √(100² + 100²) = 100√2

Answer

Maximum displacement = 100√2 ≈ 141.4 m

Q7 · 3 Marks (case-based)

A car's speed (m/s) at times 0, 2, 4, 6, 8, 10 s is 4, 8, 12, 16, 20, 24. Find (a) the acceleration and (b) the distance in 10 s.

(a) slope

a = (24 − 4)/(10 − 0) = 2 m/s² (uniform acceleration)

(b) area

distance = ½(u + v)t = ½(4 + 24)(10) = 140 m

Answer

a = 2 m/s² · distance = 140 m

Quick Reference

Formula & Fact Sheet

Everything in one place for a 5-minute pre-exam revision.

Concept	Formula / Rule	Note
Speed	v = s / t	scalar, m/s
Average speed	total distance / total time	not the simple average
Velocity	displacement / time	vector, m/s
Average velocity	(u + v) / 2	uniform acceleration
Acceleration	a = (v − u) / t	vector, m/s²
1st equation	v = u + at	no s
2nd equation	s = ut + ½at²	no v
3rd equation	v² = u² + 2as	no t
nth-second distance	Sₙ = u + ½a(2n−1)	distance in nth second
Circular speed	v = 2πr / t	r = radius, t = period
Max height (thrown up)	h = u² / 2g	v = 0 at top
km/h → m/s	× 5/18	m/s → km/h: × 18/5
d–t graph slope	= speed	—
v–t graph slope	= acceleration	—
v–t graph area	= distance	triangle / trapezium
Distance vs displacement	distance ≥ \|displacement\|	equal only if no turn

Smart Study

12 Exam & JEE-Foundation Tips

Habits that turn a slow solver into a fast, accurate one.

Convert units first

Change every speed to m/s (× 5/18) before substituting. Mixed units cause most wrong answers.

Always write the formula

State it before substituting — CBSE awards a mark for the correct formula on its own.

"Stops" → v=0; "from rest" → u=0

These phrases hand you a value and hint at which equation to use.

No time given → use v²=u²+2as

If a problem gives u, v, a or s but not t, the third equation is the shortcut.

Distance ≥ displacement

Equal only for straight-line motion without turning. Any return trip → smaller (often zero) displacement.

Average speed ≠ average of speeds

Use total distance ÷ total time. The mean of two speeds is usually wrong.

Slope vs area

d–t slope = speed; v–t slope = acceleration; v–t area = distance. Never mix these.

Circular motion is accelerated

Constant speed, changing direction = acceleration. A classic 1-mark conceptual question.

Units on every answer

m/s for velocity, m/s² for acceleration. A bare number loses a mark.

Retardation = negative a

When slowing down, write a with a minus sign, then state retardation as its magnitude.

Solve graphs by geometry

Triangles, rectangles and trapeziums under v–t lines give distance fast — often faster than equations.

Cross-check by a 2nd method

JEE habit: confirm a numerical with an alternate route (equation vs graph). Two methods, one answer = confidence.

CBSE Pattern Practice

Practice Question Bank

MCQ · Assertion–Reason · VSA (1M) · SA (2M / 3M) · LA (5M) · Application (4M) — covering every question type in the chapter exercise, each with a step-by-step solution. Use the filters to drill one type at a time.

⚡ Beyond the Textbook

JEE Foundation Challenge Set

Original questions written to stretch you one level above the board exam — same Class 9 toolkit (the three equations, graphs, relative motion), but sharper reasoning. Try each before opening the solution.

These are bonus problems — tap a card's button to reveal the worked solution.