Chapter 8 — Motion
A complete, exam-ready guide to Motion for NCERT, CBSE and State Board (including SEBA/Assam Board) students — built from the ground up with derivations, graphs, solved numericals, and 75+ practice questions.
Namaskar and welcome to Chapter 8! 👋
Look around you right now — a fan spinning, a bus passing by, blood flowing in your veins, even the Earth itself is moving at a staggering 30 km every second around the Sun. Nothing in the universe is truly at rest. This chapter gives you the mathematical language — distance, displacement, speed, velocity, and acceleration — to describe all of it precisely. It looks intimidating at first because of the formulas, but once the concepts click, Motion becomes one of the most scoring chapters in your Class 9 Physics exam.
Motion is the first chapter of Physics in Class 9 Science (NCERT). It builds the foundation for Class 9 Chapter 9 (Force and Laws of Motion), Class 9 Chapter 10 (Gravitation), and forms the base of Kinematics that you will study right up to Class 11 and 12, and in competitive exams like NEET, JEE, NDA, and various state-level teacher eligibility and recruitment tests.
What you'll master
Precise definitions, correct SI units, and the difference between scalar and vector quantities in motion.
Skills you'll build
Deriving and applying the three equations of motion to solve real numerical problems confidently.
Graphs you'll read
Distance-time and velocity-time graphs — and how to extract speed, acceleration, and distance from their shape.
Learning Objectives
By the end of this chapter, you will be able to:
- Differentiate between distance (scalar) and displacement (vector) with examples.
- Define speed, average speed, and velocity, and calculate them using correct SI units.
- Distinguish between uniform and non-uniform motion using real-life and graphical examples.
- Define acceleration, identify positive, negative (retardation) and zero acceleration.
- Plot and interpret distance-time and velocity-time graphs.
- Derive the three equations of motion both graphically and algebraically.
- Apply equations of motion to solve numerical problems confidently.
- Understand uniform circular motion as an example of accelerated motion.
- Solve NCERT in-text, exercise, HOTS, assertion-reason, and case-based questions.
- Answer board-style and competitive-exam-style MCQs with speed and accuracy.
Real-Life Applications of Motion
🚗 Speedometers & Traffic
Speed limits, braking distance, and reaction time calculations all rely on the equations of motion you'll learn here.
🚀 Space Missions
ISRO scientists use these same equations to calculate a rocket's velocity and acceleration during launch.
⚽ Sports Science
A sprinter's acceleration off the blocks, or a bowler's run-up speed, are analysed using motion graphs.
🚦 Road Safety
Airbag deployment and seatbelt design depend on how quickly velocity changes (deceleration) during a crash.
🌍 Earth & Sky
Circular motion explains why the Moon stays in orbit and why satellites like INSAT stay "fixed" above India.
🚴 Everyday Travel
Estimating how long a bus or train journey will take uses simple speed = distance/time calculations.
Distance and Displacement
A body is said to be in motion if it changes its position with respect to its surroundings over time, and it is at rest if it does not. But rest and motion are not absolute — they depend on the observer's frame of reference. A person sitting inside a moving bus is at rest with respect to the bus, but in motion with respect to a tree outside.
📏 Distance
The actual length of the path travelled by a body, irrespective of its direction. It is a scalar quantity — it has magnitude only, no direction.
➡️ Displacement
The shortest straight-line distance between the initial and final position of a body, along with direction. It is a vector quantity.
Rahul walks 4 m East and then 3 m North. Find (a) the distance travelled and (b) the magnitude of displacement.
(a) Distance = 4 m + 3 m = 7 m (simply add the path lengths).
(b) Displacement — since the two paths are perpendicular, use Pythagoras' theorem:
Displacement = √(4² + 3²) = √(16+9) = √25 = 5 m, directed from the start point to the final point (north-east).
| Basis | Distance | Displacement |
|---|---|---|
| Type of quantity | Scalar (magnitude only) | Vector (magnitude + direction) |
| Path dependence | Depends on the actual path taken | Independent of path; depends only on initial & final points |
| Value | Always positive, never zero (for a moving body) | Can be positive, negative, or zero |
| Comparison | Distance ≥ Displacement (always) | Displacement ≤ Distance (always) |
| When equal | Equal only when the body moves along a straight line without changing direction | |
"Distance is the full Diary of your journey (every step counted); Displacement is just the Direct line home." Also remember: for one complete round of a circular track, displacement = 0 but distance = the full circumference — a favourite trick question in exams!
Speed and Velocity
Distance and displacement tell us how much a body has moved. Speed and velocity tell us how fast.
⚡ Speed
Distance travelled by a body per unit time. A scalar quantity.
SI unit: m/s (also km/h)
🎯 Velocity
Displacement of a body per unit time, i.e. speed in a specified direction. A vector quantity.
SI unit: m/s
Types of Speed & Velocity
| Term | Meaning | Formula |
|---|---|---|
| Uniform speed | Body covers equal distances in equal intervals of time | constant |
| Non-uniform speed | Body covers unequal distances in equal intervals of time | varies |
| Average speed | Overall rate of motion for the whole journey | Total distance / Total time |
| Instantaneous speed | Speed of the body at a particular instant of time | limit as Δt → 0 |
| Uniform velocity | Equal displacement in equal time intervals, in the same direction | constant magnitude + direction |
| Average velocity | For uniformly changing velocity | (Initial velocity + Final velocity) / 2 |
A car covers 100 km in the first 2 hours and 60 km in the next 1 hour. Find its average speed.
Total distance = 100 + 60 = 160 km. Total time = 2 + 1 = 3 h.
Average speed = 160/3 = 53.3 km/h.
To convert km/h → m/s, multiply by 5/18.
To convert m/s → km/h, multiply by 18/5.
36 km/h = 36 × 5/18 = 10 m/s
Uniform and Non-Uniform Motion
➰ Uniform Motion
A body has uniform motion if it travels equal distances in equal intervals of time, however small these intervals may be. Example: a car moving at a constant 60 km/h on a straight highway.
🌀 Non-Uniform Motion
A body has non-uniform motion if it travels unequal distances in equal intervals of time. Example: a bus in city traffic, speeding up and slowing down at signals.
Uniform = Unchanging rate → straight-line, evenly-spaced motion. Non-uniform = Not steady → speeding up, slowing down, or changing direction.
Acceleration
Acceleration is the rate of change of velocity with time. It is a vector quantity.
where u = initial velocity, v = final velocity, t = time taken.
SI unit of acceleration: m/s²
| Type | Condition | Example |
|---|---|---|
| Positive acceleration | Velocity increases with time | A car speeding up from a signal |
| Negative acceleration (retardation) | Velocity decreases with time | A car braking to stop |
| Zero acceleration | Velocity remains constant | A car moving at a constant 60 km/h |
| Uniform acceleration | Velocity changes by equal amounts in equal time intervals | A freely falling body under gravity |
| Non-uniform acceleration | Velocity changes by unequal amounts in equal time intervals | A bus in heavy traffic |
A bike's velocity changes from 5 m/s to 25 m/s in 4 seconds. Find its acceleration.
a = (v − u)/t = (25 − 5)/4 = 20/4 = 5 m/s²
Distance–Time and Velocity–Time Graphs
Graphs let us "see" motion. Time is always plotted on the x-axis (independent variable).
📈 Distance–Time Graphs
In a distance-time graph, the slope gives the speed of the body. A steeper slope means higher speed. A line parallel to the time axis means the body is at rest.
📉 Velocity–Time Graphs
In a velocity-time graph, the slope gives acceleration, and the area enclosed with the time axis gives the distance (or displacement) travelled.
Equations of Motion
For a body moving with uniform acceleration, three equations connect initial velocity (u), final velocity (v), acceleration (a), time (t), and distance (s).
📐 Graphical Derivation
- Consider the velocity-time graph where a body starts with initial velocity u at A (t=0) and reaches final velocity v at B (time t), moving with uniform acceleration a.
- Slope of the line AB gives acceleration: a = (BC − AD)/OC = (v − u)/t
- Rearranging: at = v − u
- Therefore: v = u + at
- Distance travelled = Area under the velocity-time graph (trapezium OABC).
- This area = Area of rectangle OADC + Area of triangle ADB.
- Area of rectangle = OA × OC = u × t
- Area of triangle = ½ × AD × DB = ½ × t × (v−u)
- Since v − u = at (from equation 1), triangle area = ½ × t × at = ½at²
- Total distance: s = ut + ½at²
- Again, distance s = Area of trapezium OABC = ½ × (sum of parallel sides) × height
- s = ½ × (OA + CB) × OC = ½ × (u + v) × t
- From equation 1: t = (v − u)/a
- Substituting: s = ½ × (u + v) × (v − u)/a = (v² − u²)/2a
- Rearranging: v² = u² + 2as
A train starting from rest accelerates uniformly at 2 m/s² for 10 s. Find (a) its final velocity, (b) distance covered.
Given: u = 0, a = 2 m/s², t = 10 s
(a) v = u + at = 0 + 2×10 = 20 m/s
(b) s = ut + ½at² = 0 + ½×2×10² = 100 m
A car moving at 20 m/s is brought to rest in 5 s by applying brakes. Find the retardation and the distance travelled before stopping.
Given: u = 20 m/s, v = 0, t = 5 s
a = (v−u)/t = (0−20)/5 = −4 m/s² (retardation of 4 m/s²)
Using v² = u² + 2as: 0 = 400 + 2(−4)s → 8s = 400 → s = 50 m
A stone is dropped from a cliff and hits the ground after 4 s. Taking g = 10 m/s², find the height of the cliff and the velocity with which it strikes the ground.
Given: u = 0, t = 4 s, a = g = 10 m/s²
s = ut + ½at² = 0 + ½×10×16 = 80 m (height of cliff)
v = u + at = 0 + 10×4 = 40 m/s
Uniform Circular Motion
When a body moves along a circular path at constant speed, its motion is called uniform circular motion (UCM). Even though the speed doesn't change, the direction of velocity changes continuously — so the body is always accelerating!
- Since velocity's direction changes continually, UCM is an example of accelerated motion, even at constant speed.
- If the body takes time t to complete one round of a circle of radius r, its speed = 2πr / t
- Real examples: the tip of a clock's second hand, a satellite orbiting Earth, an athlete running on a circular track.
An athlete completes one round of a circular track of radius 70 m in 44 s. Find her speed. (Take π = 22/7)
Speed = 2πr/t = (2 × 22/7 × 70)/44 = 440/44 = 10 m/s
Mind Map & Complete Formula Sheet
| Quantity | Formula | SI Unit | Type |
|---|---|---|---|
| Speed | Distance / Time | m/s | Scalar |
| Velocity | Displacement / Time | m/s | Vector |
| Average speed | Total distance / Total time | m/s | Scalar |
| Average velocity | (u + v) / 2 | m/s | Vector |
| Acceleration | (v − u) / t | m/s² | Vector |
| 1st equation of motion | v = u + at | — | — |
| 2nd equation of motion | s = ut + ½at² | — | — |
| 3rd equation of motion | v² = u² + 2as | — | — |
| Speed in circular motion | 2πr / t | m/s | Scalar |
The five key quantities in every equation of motion problem: s (distance), u (initial velocity), v (final velocity), a (acceleration), t (time) — remember it as SUVAT. Every equation of motion connects only 4 of these 5 at a time. Identify which 3 are given, and pick the equation that has the 4th as its only unknown!
NCERT In-Text & Exercise Solutions
Click any question to reveal the complete, step-by-step solution. All numbering follows the official NCERT textbook (2024–25 edition).
📖 In-Text Questions — Page 100 (Distance & Displacement)
Yes. If a body starts from point A, moves along any path, and returns to the same point A, its displacement is zero even though the distance covered is not. Example: an athlete completing one full round of a circular track returns to the starting point — distance = circumference, displacement = 0.
Perimeter of field = 4 × 10 = 40 m, covered in 40 s, so speed = 1 m/s.
Total time = 2 min 20 s = 140 s. Number of rounds = 140/40 = 3.5.
After 3 complete rounds (120 s) the farmer is back at the start. In the remaining 20 s (speed 1 m/s) he covers 20 m — two more sides — landing at the corner diagonally opposite to the start.
Displacement = diagonal of the square = 10√2 = 14.14 m, directed towards the diagonally opposite corner.
Both statements are false. (a) is false because displacement can be zero (see Q1). (b) is false because the magnitude of displacement is always less than or equal to the distance travelled, never greater.
📖 In-Text Questions — Page 102 (Speed)
Speed is a scalar quantity (distance/time, direction not considered); velocity is a vector quantity (displacement/time, direction is specified). See the comparison table in the Speed & Velocity section above.
Only when the object moves in a straight line in one direction, without reversing — then distance = displacement, so speed = velocity in magnitude.
It measures the total distance travelled by the vehicle.
It is a straight line, since velocity (magnitude and direction) stays constant.
Time = 5 min = 300 s.
Distance = speed × time = 3×10⁸ × 300 = 9 × 10¹⁰ m.
📖 In-Text Questions — Page 103 (Acceleration)
Uniform acceleration: velocity changes by equal amounts in equal time intervals.
Non-uniform acceleration: velocity changes by unequal amounts in equal time intervals.
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 (negative sign shows retardation).
u = 0, v = 40 km/h = 11.11 m/s, t = 10 min = 600 s.
a = (11.11−0)/600 = 0.0185 m/s².
📖 In-Text Questions — Page 107 (Graphs)
Q1. Uniform motion → straight-line distance-time graph. Non-uniform motion → curved graph.
Q2. A distance-time graph parallel to the time axis means the object is at rest.
Q3. A velocity-time graph parallel to the time axis means the object is moving with constant (uniform) velocity, i.e. zero acceleration.
Q4. The area under a velocity-time graph gives the distance (or displacement) covered.
📖 In-Text Questions — Equations of Motion
u=0, a=0.1 m/s², t=120 s.
v = u+at = 0.1×120 = 12 m/s
s = ut+½at² = 0 + ½×0.1×120² = 720 m
u=90 km/h=25 m/s, v=0, a=−0.5 m/s².
v²=u²+2as → 0=625−s → s = 625 m
u=0, a=0.02 m/s², t=3 s.
v = u+at = 0.02×3 = 0.06 m/s
u=0, a=4 m/s², t=10 s.
s = ut+½at² = 0+½×4×100 = 200 m
u=5 m/s, a=−10 m/s², v=0 (at highest point).
v²=u²+2as → 0=25−20s → s = 1.25 m
v=u+at → 0=5−10t → t = 0.5 s
📝 End-of-Chapter Exercise Solutions (Q1–10)
r = 100 m. In 40 s, one round = 2πr = 2×(22/7)×100 = 628.57 m.
Total time = 140 s → 140/40 = 3.5 rounds.
Distance = 3.5 × 628.57 ≈ 2200 m.
After 3 full rounds (120 s) displacement is zero; the remaining half round brings the athlete to the diametrically opposite point.
Displacement = diameter = 200 m.
A to B: speed = 300/150 = 2 m/s; velocity = 300/150 = 2 m/s (straight line, same direction).
A to C: total distance = 300+100 = 400 m, total time = 150+60 = 210 s → average speed = 400/210 = 1.9 m/s.
Displacement A to C = 300−100 = 200 m → average velocity = 200/210 = 0.95 m/s.
Let one-way distance = d. Time₁ = d/20, Time₂ = d/40.
Average speed = Total distance/Total time = 2d ÷ (d/20 + d/40) = 2d ÷ (3d/40) = 80/3 = 26.67 km/h. (Note: it is not the simple average of 20 and 40.)
u=0, a=3 m/s², t=8 s.
s = ut+½at² = 0+½×3×64 = 96 m
Distance = area of the speed-time triangle = ½ × u × t.
Car A: u=52 km/h=14.44 m/s → s = ½×14.44×5 = 36.1 m
Car B: u=3 km/h=0.83 m/s → s = ½×0.83×10 = 4.16 m
Car A (52 km/h) travels farther, even though it stops in less time — its higher starting speed dominates.
(a) B is travelling fastest — its line has the steepest slope (covers more distance in less time).
(b) No — the three lines never intersect at a common point, so A, B and C are never at the same point on the road simultaneously.
(c) Reading from the standard NCERT graph: when B passes A, C has covered approximately 5.7 km.
(d) When B passes C, B has covered approximately 5.1 km. (Exact readings depend on the printed graph grid, so answers are commonly expressed as approximate values.)
u=0, s=20 m, a=10 m/s².
v²=u²+2as = 0+2×10×20 = 400 → v = 20 m/s
t=(v−u)/a = 20/10 = 2 s
(a) The distance travelled in the first 4 seconds equals the area under the curve for that interval, which works out to approximately 12 m (found by counting/shading the grid squares under the graph).
(b) The flat, horizontal portion of the graph (constant speed, typically between t = 6 s and t = 10 s) represents uniform motion, since the speed is not changing there.
(a) Possible. A ball thrown vertically upward has zero velocity at its highest point, but the acceleration due to gravity (g) continues to act on it at that instant.
(b) Possible. In uniform circular motion, the object moves along the circle (tangential direction) while the acceleration (centripetal) always acts perpendicular to the velocity, towards the centre.
r = 42,250 km, t = 24 h = 86,400 s.
Speed = 2πr/t = (2×3.14×42250)/86400 = 265,331/86400 ≈ 3.07 km/s (about 11,052 km/h).
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