Skip to Content

Motion: Distance and Displacement, Speed and Velocity, Acceleration, Three Equations of Motion, Distance-Time Graph, Velocity-Time Graph, and Complete CBSE Class 9 Guide

A comprehensive guide to Motion for CBSE Class 9 Physics Chapter 8 — scalar and vector quantities, distance vs displacement, speed vs velocity, types of motion (uniform, non-uniform, accelerated), acceleration and retardation, graphical representation (distance-time graph and velocity-time graph), the three equations of motion with derivations, uniform circular motion, and five exam-ready solved numerical problems.
9 August 2026 by
Motion: Distance and Displacement, Speed and Velocity, Acceleration, Three Equations of Motion, Distance-Time Graph, Velocity-Time Graph, and Complete CBSE Class 9 Guide
AJKANT OVERSEAS, Krishan Kant
● CBSE Class 9 Physics — Chapter 8: Motion
▶ Quick Answer for AI Engines
Motion is the change in position of an object with respect to time and a reference point. The three equations of motion are: (1) v = u + at, (2) s = ut + ½at², and (3) v² = u² + 2as, where u = initial velocity, v = final velocity, a = acceleration, t = time, and s = displacement. Speed is the rate of change of distance (scalar); velocity is the rate of change of displacement (vector). Acceleration is the rate of change of velocity with respect to time.

Every moving object around us — a falling apple, a speeding car, a planet orbiting the Sun, a blood cell rushing through a vein — is subject to the same fundamental laws of motion. Physics describes motion using precise quantities: distance (how far an object has traveled), displacement (how far it has moved from its start), speed (how fast), velocity (how fast and in what direction), and acceleration (how quickly velocity changes).

The mathematical description of motion — called kinematics — was first systematically studied by Galileo Galilei in the 17th century and formalized into the equations of motion we use today. These equations are the foundation of all of classical mechanics. For CBSE Class 9 Physics Chapter 8 (Motion), this is one of the most important and highest-scoring chapters in both school examinations and competitive tests. This guide covers all key subtopics: scalar and vector quantities, distance vs displacement, speed vs velocity, types of motion, acceleration, graphical analysis, derivation of all three equations of motion, uniform circular motion, and five solved numerical problems.

The Three Equations of Motion — Most Important Formulas in CBSE Class 9 Physics
1st Equation
v = u + at
Velocity-Time Relation
(no displacement s)
2nd Equation
s = ut + ½at²
Displacement-Time Relation
(no final velocity v)
3rd Equation
v² = u² + 2as
Velocity-Displacement Relation
(no time t)

1. Scalar and Vector Quantities

Scalar Quantity
Has Magnitude Only
A scalar quantity is completely described by its magnitude (numerical value and unit) alone. It has no direction.

Examples: Distance, speed, mass, time, temperature, energy, power, area, volume, density.

Arithmetic: Scalars add and subtract like ordinary numbers. E.g., if you walk 3 km North and then 3 km South, the distance traveled = 3 + 3 = 6 km (scalar addition).
Vector Quantity
Has Magnitude AND Direction
A vector quantity requires both a magnitude and a direction for complete description.

Examples: Displacement, velocity, acceleration, force, momentum, weight, electric field, magnetic field.

Arithmetic: Vectors add by the vector addition law (triangle law or parallelogram law), not simple arithmetic. E.g., 3 km North + 3 km South = 0 displacement (vectors cancel).

2. Distance vs Displacement

Distance
Total Path Length
Scalar quantity (no direction)
Definition: Distance is the total length of the path traveled by a moving object, regardless of direction.

• Always positive (or zero). Never negative.
• Can never decrease as the object moves.
• Depends on the actual path taken between two points.

Example: If a person walks 4 km East, then 3 km North, the distance = 4 + 3 = 7 km.
Displacement
Shortest Straight-Line Distance
Vector quantity (has direction)
Definition: Displacement is the shortest straight-line distance from the initial position to the final position of an object, measured in a specific direction.

• Can be positive, negative, or zero.
• Can be zero even if distance is not (e.g., circular trip returning to start).
• Displacement ≤ Distance (always).

Example: 4 km East + 3 km North → Displacement = √(4²+3²) = √25 = 5 km (North-East).
PropertyDistanceDisplacement
TypeScalar (magnitude only)Vector (magnitude + direction)
DefinitionTotal path length traveledShortest straight-line path from start to end
Symbolds (or x, or Δx)
SignAlways positive or zeroCan be positive, negative, or zero
Can be zero?Only if object does not moveYes — if object returns to starting point
RelationDistance ≥ |Displacement||Displacement| ≤ Distance
Unitmetre (m)metre (m)

3. Speed vs Velocity

Speed
Rate of Change of Distance
Speed = Distance / Time
Definition: Speed is the distance traveled by an object per unit time. It is a scalar quantity — it has magnitude only, no direction.
Unit: m/s (SI) or km/h

Average Speed: Total distance / Total time
Uniform Speed: Equal distances in equal time intervals

Example: A car that travels 100 km in 2 hours has an average speed of 50 km/h, regardless of direction.
Velocity
Rate of Change of Displacement
Velocity = Displacement / Time
Definition: Velocity is the displacement of an object per unit time. It is a vector quantity — it has both magnitude and direction.
Unit: m/s (SI) — with direction specified

Average Velocity: Total displacement / Total time
Uniform Velocity: Equal displacements in equal time intervals in the same direction

Example: A car going 50 km/h East has velocity = 50 km/h (East). The same car going 50 km/h West has velocity = -50 km/h (if East is positive).

4. Types of Motion

▶▶
Uniform Motion
An object is in uniform motion when it travels equal distances in equal time intervals in the same direction. The speed is constant and direction is constant → velocity is constant → acceleration = 0. Example: A train moving at exactly 80 km/h on a straight track.
▶≃
Non-Uniform Motion
An object is in non-uniform motion when it covers unequal distances in equal time intervals. Speed and/or direction changes over time. Example: A car in city traffic — accelerating, decelerating, stopping. A falling stone (speed increases every second).
▶▶▶
Uniformly Accelerated Motion
A special case of non-uniform motion where the acceleration is constant (velocity changes by equal amounts in equal time intervals). Example: A ball rolling down an inclined plane, a stone falling under gravity (a = g = 9.8 m/s²). The three equations of motion apply to this type.
Uniform Circular Motion
An object moves in a circle at constant speed. Though the speed is constant, the velocity continuously changes direction → the object is accelerating (centripetal acceleration). Example: Earth orbiting the Sun, a satellite in circular orbit, a stone on a string.
▼▼
Retardation (Deceleration)
When the velocity of an object decreases over time, the acceleration is negative (opposite to the direction of motion). This is called retardation or deceleration. Example: Braking car, ball thrown upward (deceleration due to gravity while rising).
📈
Periodic / Oscillatory Motion
An object repeats its motion after a fixed time interval (period T). Example: A simple pendulum swinging back and forth, a vibrating guitar string, Earth rotating on its axis once every 24 hours.

5. Acceleration and Retardation

⚡ Acceleration
a = (v - u) / t    Units: m/s²
Definition: Acceleration is the rate of change of velocity with respect to time. It is a vector quantity (has both magnitude and direction).

a = (v − u) / t
where: u = initial velocity (m/s), v = final velocity (m/s), t = time taken (s), a = acceleration (m/s²)

Positive acceleration: Velocity is increasing (object speeding up) in the direction of motion.
Negative acceleration (retardation / deceleration): Velocity is decreasing (object slowing down). Acceleration is in the direction opposite to velocity.
Zero acceleration: Velocity is constant (uniform motion) OR object is at rest.

Uniform acceleration: Equal changes in velocity in equal time intervals. The velocity-time graph is a straight line. The three equations of motion are derived for uniform acceleration.

Non-uniform acceleration: Velocity changes by different amounts in equal time intervals. Velocity-time graph is a curve (not a straight line).

SI unit of acceleration: metre per second squared (m/s²). Other units: cm/s², km/h².

Standard acceleration due to gravity: g = 9.8 m/s² ≈ 10 m/s² (downward, toward Earth’s centre).

6. Distance-Time Graph and Velocity-Time Graph

📈 Distance-Time Graph (d-t graph)

d (m) | / | / | ________/ (Non-uniform speed: curve) | / | / (Uniform speed: straight line) | / |___________/ ____________ t (s) 0 Slope of d-t graph = Distance / Time = SPEED Horizontal line (slope=0) = Object at rest Straight line (constant slope) = Uniform speed Curve (increasing slope) = Increasing speed (acceleration)
Key interpretations of the distance-time graph:
Slope = Speed: A steeper slope means greater speed.
Horizontal line (slope = 0): Object is at rest (distance not changing with time).
Straight line with positive slope: Uniform speed (equal distance in equal time).
Upward curve (increasing slope): Speed is increasing (non-uniform, accelerated motion).
Downward curve (decreasing slope): Speed is decreasing (non-uniform, decelerated motion).
Note: A d-t graph can never have a negative slope (distance never decreases).

📊 Velocity-Time Graph (v-t graph) — Most Important for CBSE

v (m/s) | B_________C (Constant velocity: horizontal) | / | / \ (Deceleration: negative slope) | / | / (Uniform acceleration: positive slope) | / |___/A_____________________t (s) 0 Slope of v-t graph = (v-u)/t = ACCELERATION Area under v-t graph = Displacement (distance) OA (slope up, straight): uniformly accelerated motion BC (horizontal): uniform velocity (a=0) C onwards (slope down): uniformly decelerated (retardation)
Key interpretations of the velocity-time graph:
Slope = Acceleration: Steep positive slope = large acceleration; negative slope = deceleration.
Horizontal line (slope = 0): Uniform velocity (acceleration = 0).
Straight line with positive slope: Uniform acceleration.
Straight line with negative slope: Uniform retardation (deceleration).
Area under the v-t graph = Displacement: For a rectangle (uniform v): s = v×t. For a triangle (uniform a from rest): s = ½×v×t. This is how the 2nd equation of motion is derived!

7. Three Equations of Motion — Derivation and Use

The three equations of motion apply to an object moving with uniform acceleration (constant acceleration a) in a straight line. Variables: u = initial velocity, v = final velocity, a = acceleration (constant), t = time, s = displacement.

1st Equation of Motion: v = u + at (Velocity-Time Relation)
Derivation (from definition of acceleration):
Acceleration = Change in velocity / Time taken
a = (v − u) / t
Multiplying both sides by t:
at = v − u
Rearranging:
v = u + at

What it tells us: Given initial velocity u, acceleration a, and time t, find the final velocity v.
When to use: When displacement s is not required.
Special cases: u=0 (starts from rest): v = at. a=0 (uniform motion): v = u.
v = u + at
2nd Equation of Motion: s = ut + ½at² (Displacement-Time Relation)
Derivation (graphical method — area under v-t graph):
The displacement s = area under the v-t graph (trapezium with vertices at (0,u), (t,v), (t,0), (0,0)).
Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (u + v) × t
Substituting v = u + at:
s = ½ × (u + u + at) × t = ½ × (2u + at) × t
s = ut + ½at²

What it tells us: Given u, a, and t, find the displacement s.
When to use: When final velocity v is not required.
Special cases: u=0 (starts from rest): s = ½at². a=0 (uniform motion): s = ut.
s = ut + ½at²
3rd Equation of Motion: v² = u² + 2as (Velocity-Displacement Relation)
Derivation (eliminating t from 1st and 2nd equations):
From 1st equation: t = (v − u)/a
Substituting into s = ½(u+v)t:
s = ½(u+v) × (v−u)/a = (v²−u²)/(2a)
Rearranging:
2as = v² − u²
v² = u² + 2as

What it tells us: Given u, a, and s, find final velocity v (or vice versa).
When to use: When time t is not given or required.
Special cases: u=0: v² = 2as → v = √(2as). a=0: v = u (uniform motion, v doesn’t change with distance).
v² = u² + 2as

8. Uniform Circular Motion

↻ Uniform Circular Motion
v = 2πr/T    (v = speed, r = radius, T = period)
Definition: Uniform circular motion is the motion of an object moving in a circular path at constant speed.

Is it accelerated? YES — even though the speed is constant, the direction of velocity changes continuously (velocity is always tangent to the circle). Since velocity (a vector) is changing, the object IS accelerating. This acceleration is directed toward the centre of the circle and is called centripetal acceleration (a⍾ = v²/r).

Key quantities:
Speed (v) = constant = 2πr/T (circumference / period)
Period (T) = time for one complete revolution
Frequency (f) = 1/T (revolutions per second, Hz)
Angular velocity (ω) = 2π/T = 2πf (radians per second)
Centripetal force (F⍾) = mv²/r = mω²r (directed inward, toward centre)

Important note: In uniform circular motion, speed is constant but velocity is NOT constant (direction keeps changing). So the kinetic energy is constant, but momentum is not constant.

Examples of uniform circular motion:
• Earth orbiting the Sun (approximately circular orbit).
• Moon orbiting Earth.
• A satellite in circular orbit.
• A stone tied to a string and whirled in a horizontal circle.
• The tip of a clock hand (minute or second hand).
• A car going around a circular roundabout at constant speed.

9. Solved Numerical Problems

Q1. A car starts from rest and attains a velocity of 72 km/h in 10 seconds. Find the acceleration of the car.
Given: u = 0 (starts from rest)  |  v = 72 km/h = 72 × (1000/3600) = 20 m/s  |  t = 10 s
Using v = u + at:
20 = 0 + a × 10
a = 20/10
a = 2 m/s² (the car accelerates at 2 m/s²)
Q2. A train is moving at 54 km/h. Brakes are applied and the train comes to rest in 5 seconds. Find the retardation and the distance traveled during braking.
Given: u = 54 km/h = 15 m/s  |  v = 0 (comes to rest)  |  t = 5 s
Retardation (a): v = u + at → 0 = 15 + a×5 → a = -15/5 = -3 m/s² (retardation = 3 m/s²)
Distance: s = ut + ½at² = 15×5 + ½×(-3)×25 = 75 − 37.5
Retardation = 3 m/s²  |  Distance = 37.5 m
Q3. A stone is thrown vertically upward with a velocity of 19.6 m/s. Find the maximum height reached and the time taken to reach that height. (g = 9.8 m/s²)
Given: u = 19.6 m/s (upward)  |  v = 0 (at maximum height, velocity = 0)  |  a = −9.8 m/s² (deceleration due to gravity)
Time: v = u + at → 0 = 19.6 + (−9.8)×t → t = 19.6/9.8 = 2 s
Max height: v² = u² + 2as → 0 = (19.6)² + 2×(−9.8)×s
→ s = (19.6)² / (2×9.8) = 384.16/19.6
Maximum height = 19.6 m  |  Time to reach top = 2 s
Q4. An object covers a distance of 100 m in 5 s. The initial velocity was 10 m/s. Find the acceleration and the final velocity.
Given: u = 10 m/s  |  s = 100 m  |  t = 5 s
Using s = ut + ½at²:
100 = 10×5 + ½×a×25
100 = 50 + 12.5a
12.5a = 50 → a = 4 m/s²
Final velocity: v = u + at = 10 + 4×5 = 10 + 20
a = 4 m/s²  |  v = 30 m/s
Q5. A cyclist goes around a circular track of radius 70 m in 44 seconds. Calculate the speed of the cyclist and the centripetal acceleration. (π = 22/7)
Given: r = 70 m  |  T = 44 s  |  π = 22/7
Speed: v = 2πr/T = 2 × (22/7) × 70 / 44 = (2 × 22 × 70) / (7 × 44) = 3080/308 = 10 m/s
Centripetal acceleration: a⍾ = v²/r = (10)²/70 = 100/70
Speed v = 10 m/s  |  Centripetal acceleration = 100/70 ≈ 1.43 m/s²

10. Frequently Asked Questions (FAQ)

What are the three equations of motion? State them and their variables.

The three equations of motion for uniform acceleration are:

1st Equation: v = u + at (Velocity-Time relation)
2nd Equation: s = ut + ½at² (Displacement-Time relation)
3rd Equation: v² = u² + 2as (Velocity-Displacement relation)

Variables:
• u = initial velocity of the object (m/s)
• v = final velocity of the object (m/s)
• a = acceleration of the object (m/s²) — must be constant (uniform acceleration)
• t = time elapsed (s)
• s = displacement of the object during time t (m)

Which equation to use: Choose the equation that contains the three known quantities and the one unknown you need to find. If t is not given or needed, use the 3rd equation. If v is not given or needed, use the 2nd equation.

What is the difference between distance and displacement?

Distance is the total length of the path traveled by an object, regardless of direction. It is a scalar quantity (magnitude only). Distance is always positive or zero and can never decrease as an object moves.

Displacement is the shortest straight-line distance from the initial position to the final position of an object, in a specific direction. It is a vector quantity (magnitude + direction). Displacement can be positive, negative, or zero.

Key distinction: An object can have zero displacement while still having traveled a large distance. For example, if you walk 400 m around a circular track and return to your starting point, your distance = 400 m but your displacement = 0 m (same start and end point).

Relation: |Displacement| ≤ Distance. They are equal only when the motion is in a straight line in one direction without any reversal.

What is the difference between speed and velocity?

Speed is the distance traveled by an object per unit time. It is a scalar quantity (magnitude only, no direction). Speed is always positive. Formula: Speed = Distance/Time. Unit: m/s.

Velocity is the displacement of an object per unit time. It is a vector quantity (magnitude + direction). Velocity can be positive, negative, or zero. Formula: Velocity = Displacement/Time. Unit: m/s (with direction).

Example: A car driving in a circle at a constant 60 km/h has constant speed = 60 km/h, but its velocity is continuously changing (direction changes at every point).

Average speed vs average velocity: Average speed = Total distance / Total time. Average velocity = Total displacement / Total time. For a round trip (returning to start): average speed > 0 but average velocity = 0 (displacement = 0).

What does the slope of a velocity-time graph represent?

The slope (gradient) of a velocity-time graph represents the acceleration of the object.
Slope = Rise/Run = Change in velocity / Change in time = (v − u) / t = acceleration (a)

Interpreting v-t graph slopes:
Positive slope (upward line): Positive acceleration (velocity increasing) — object speeding up in the positive direction.
Zero slope (horizontal line): Zero acceleration (constant velocity) — uniform motion.
Negative slope (downward line): Negative acceleration or retardation (velocity decreasing) — object slowing down.

Additionally: The area under the velocity-time graph equals the displacement of the object during that time interval. This is because area = velocity × time = displacement. For uniform velocity (rectangle): s = v × t. For uniformly increasing velocity (triangle): s = ½ × v × t. This is how the second equation of motion (s = ut + ½at²) is derived graphically.

Is uniform circular motion an accelerated motion? Why?

Yes, uniform circular motion is accelerated motion, even though the speed is constant.

Reason: Acceleration is defined as the rate of change of velocity (not speed). Velocity is a vector quantity (it has both magnitude and direction). In uniform circular motion, the speed (magnitude of velocity) remains constant, but the direction of velocity continuously changes at every point along the circular path (velocity is always tangent to the circle). Since the direction of the velocity vector is changing, the velocity is changing — therefore, there IS acceleration.

This acceleration is called centripetal acceleration (a⍾ = v²/r), and it is directed toward the centre of the circle at every instant. The corresponding force keeping the object in circular motion is the centripetal force (F = mv²/r), also directed toward the centre.

Examples: A stone on a string whirled in a circle, the Moon orbiting Earth, a car turning around a curved road — all are accelerated even if speed is constant, because direction is continuously changing.

Source Motion & Mechanics Lab Equipment from Ambala

AJKANT Overseas manufactures and supplies ticker tape timers, inclined planes (with pulley and trolley), stopwatches, metre scales, vernier calipers, force meters (spring balances), mass sets, and complete CBSE Class 9 mechanics and motion practical lab kits. Factory-direct from Ambala, India. Trusted by schools, colleges, and government institutions across India and 25+ countries.

Request Mechanics Lab Equipment Quote →