ICSE · Class 9 · Chapter 2

Motion in One
Dimension

A complete interactive 3D study guide covering distance, displacement, velocity, acceleration, free fall, motion graphs and all three equations of motion — with live simulations and a quiz.

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Chapter Overview

What You Will Learn

Seven interactive topics covering every ICSE exam objective for Chapter 2.

Topic 01
Distance & Displacement
Scalar vs vector. Total path length vs shortest straight-line change in position.
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Topic 02
Speed & Velocity
Uniform speed, average velocity, instantaneous velocity and key differences.
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Topic 03
Acceleration & Retardation
Rate of change of velocity. Positive and negative (retardation) acceleration.
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Topic 04
Free Fall under Gravity
Objects falling under g = 9.8 m/s². Equations applied to vertical motion.
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Topic 05
Motion Graphs (3D)
d–t, v–t and a–t graphs in 3D. Slope and area interpretations.
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Topic 06
Equations of Motion
Three kinematic equations + interactive solver. Paste values, get answers.
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Topic 07
Quick Cheatsheet
All formulas, SI units, scalar/vector table, and graph rules at a glance.
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Topic 08
ICSE Practice Quiz
10 scored questions styled after ICSE board exams. Get instant feedback.
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Definitions

Key Terms Explained

📏
Distance
Scalar quantity. Total length of path actually travelled, regardless of direction. Always ≥ 0. SI unit: metre (m). Cannot decrease.
➡️
Displacement
Vector quantity. Shortest straight-line distance from initial to final position, with direction. Can be negative, zero, or positive. |Displacement| ≤ Distance.
⚡
Speed
Scalar. Rate of covering distance. Average speed = Total distance ÷ Total time. Always ≥ 0. SI unit: m/s.
🎯
Velocity
Vector. Rate of change of displacement. Average velocity = Displacement ÷ Time. Can be negative. SI unit: m/s.
🚀
Acceleration
Vector. Rate of change of velocity. a = (v − u)/t. Negative a = retardation/deceleration. SI unit: m/s².
🔁
Uniform Motion
Equal displacements in equal time intervals. Velocity = constant. Acceleration = zero. d–t graph: straight line.
📈
Non-Uniform Motion
Unequal displacements in equal time intervals. Velocity changes. Acceleration ≠ 0. d–t graph: curve.
🍎
Free Fall
Motion under gravity alone. Acceleration = g = 9.8 m/s² ≈ 10 m/s² downward. Air resistance neglected.
3D Simulation · Topic 01

Distance vs Displacement

A ball travels a curved 3D path. Blue trail = distance. Red arrow = displacement vector.

Animation Speed 1.0×
Distance
0.0m
Displacement
0.0m
Time
0.0s

Path Length vs Straight Arrow

The blue curved trail is the total distance — it always increases as the ball moves. The red arrow is the displacement — the direct vector from the starting point to the current position.

When the ball curves back, displacement decreases even as distance keeps growing. For a complete round trip, displacement = 0 but distance > 0.

Rule
|Displacement| ≤ Distance (always)
Average Speed
v_avg = Total Distance / Time
Average Velocity
v_avg = Net Displacement / Time
Exam tip: Distance is scalar — never negative. Displacement is a vector — can be negative.
Round trip: displacement = 0, distance = full path length.
3D Simulation · Topic 02

Uniform vs Non-Uniform Velocity

Blue ball = constant velocity. Red ball = changing velocity (sinusoidal). Watch the v–t graph update live.

Speed & Velocity

The blue ball moves at uniform velocity — equal distances in equal time. Acceleration = 0. The red ball has non-uniform velocity — it speeds up and slows down continuously.

Instantaneous velocity
v = lim(Δt→0) Δx/Δt = dx/dt
Average velocity
v̄ = (x₂ − x₁)/(t₂ − t₁)
Uniform velocity condition
a = 0 → x = x₀ + v·t
Key: Average speed ≥ |Average velocity| always.
Speedometer shows instantaneous speed, not average speed.
Live Velocity–Time Graph
Blue speed 2.0 m/s
Red amplitude 1.5×
Blue v
2.0m/s
Red v
0.0m/s
Time
0.0s
3D Simulation · Topic 03

Acceleration & Retardation

Drag the sliders to set initial velocity and acceleration. Observe real-time effects on the 3D ball and live v–t graph.

Initial velocity u 5.0 m/s
Acceleration a +2.0 m/s²
Current v
5.0m/s
Displacement
0.0m
Time
0.0s

Acceleration

Acceleration is the rate of change of velocity. Set a positive value → ball speeds up. Set a negative value → ball slows down (retardation). At a = 0 → uniform velocity.

Definition
a = (v − u) / t
SI Unit
m/s² (metre per second squared)
Slope of v–t graph = acceleration
Area under v–t graph = displacement
Retardation: velocity decreasing → a is negative
Velocity–Time (live)
3D Simulation · Topic 04

Free Fall under Gravity

Drop an object from any height. See how velocity increases and how the three equations of motion apply to vertical fall.

Falling under g

A freely falling body has uniform downward acceleration = g = 9.8 m/s². Equations of motion apply by substituting a = g.

Initial conditions
u = 0 (dropped from rest), a = g = 9.8 m/s²
Velocity after time t
v = u + gt = gt (since u = 0)
Distance fallen
h = ut + ½gt² = ½gt²
Velocity after falling h
v² = u² + 2gh = 2gh
Upward throw: u = +ve, a = −g. At peak, v = 0.
Time to reach max height: t = u/g
Max height: H = u²/(2g)
Total flight time: T = 2u/g
Drop height (m) 50 m
g value (m/s²) 10 m/s²
Height left
50.0m
Velocity
0.0m/s
Time
0.0s
Topic 05 · 3D Graph Visualisations

Motion Graphs

Three graph types, all in 3D space. Switch tabs to see each — curves for uniform, accelerating and decelerating cases.

Distance–Time Graph

Slope of d–t graph = speed. Three curves are drawn simultaneously on three parallel planes so you can compare motion types.

Horizontal line: slope = 0 → object at rest
Straight oblique line: constant slope → uniform speed
Concave up curve: increasing slope → acceleration
Concave down curve: decreasing slope → retardation
Slope
Speed = Δd / Δt = (d₂ − d₁)/(t₂ − t₁)
Uniform Accelerating Decelerating
Topic 06

Equations of Motion

Valid for uniform acceleration in a straight line. Memorise all three — ICSE uses all of them.

NameFormulaConnectsMissingDerivation hint
1st Equation v = u + at v, u, a, t s definition: a = (v−u)/t
2nd Equation s = ut + ½at² s, u, a, t v area of v–t trapezoid
3rd Equation v² = u² + 2as v, u, a, s t eliminate t from eq 1 & 2
Remember: u = initial velocity, v = final velocity, a = acceleration (const.), s = displacement, t = time.
For Free Fall: replace a with g (9.8 m/s²) and s with h (height).
Interactive Tool

Kinematic Solver

Enter any 3 values — leave the unknown blank. The solver shows full working.

Topic 07

Quick Cheatsheet

Everything you need on one page. Save or screenshot for revision.

Scalar vs Vector
DistanceScalar
DisplacementVector
SpeedScalar
VelocityVector
AccelerationVector
TimeScalar
SI Units
Distance / Displacementm
Speed / Velocitym/s
Accelerationm/s²
Times
g (Earth)9.8 m/s²
Equations of Motion
1stv = u + at
2nds = ut + ½at²
3rdv² = u²+2as
avg velocityv̄=(u+v)/2
Free Fall Formulas
velocity after tv = gt
height fallenh = ½gt²
v after height hv=√(2gh)
time of flight (throw)T=2u/g
max height (throw)H=u²/(2g)
Graph Rules
Slope of d–t= speed
Slope of v–t= accel.
Area under v–t= displ.
Area under a–t= Δv
d–t straight lineuniform v
d–t curvenon-uniform
Key Conditions
At restv = 0, a = 0
Uniform motiona = 0, v = const
Max height (throw)v = 0
Free fall startu = 0
Round tripdisplacement=0
Topic 08 · Scored Quiz

ICSE Practice Quiz

10 questions. Click an option to lock your answer and see instant feedback. Your score is tallied at the end.

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1. A car travels 4 km East then 3 km North. Its displacement is:
✓ √(4²+3²) = 5 km. Displacement is a vector found by Pythagoras.
✗ Distance = 7 km but displacement = √(4²+3²) = 5 km (vector, use Pythagoras).
2. The slope of a velocity–time graph represents:
✓ Slope of v–t = Δv/Δt = acceleration. Area under v–t = displacement.
✗ Slope of v–t = acceleration (Δv/Δt). Area under v–t = displacement.
3. A body starts from rest and reaches 20 m/s in 4 s. Acceleration = ?
✓ a = (v−u)/t = (20−0)/4 = 5 m/s²
✗ a = (v−u)/t = 20/4 = 5 m/s²
4. A car starts at u = 10 m/s, a = 3 m/s² over s = 100 m. Final velocity v = ?
✓ v² = u²+2as = 100+600 = 700 → v = √700 ≈ 26.5 m/s (3rd equation)
✗ v² = 10²+2×3×100 = 700 → v = √700 ≈ 26.5 m/s
5. A ball is thrown upward at 20 m/s (g = 10 m/s²). Time to reach maximum height = ?
✓ At max height v = 0. t = (v−u)/a = (0−20)/(−10) = 2 s
✗ v=0 at max height. v=u+at → 0=20−10t → t=2 s
6. A body has displacement = 0 after some time. Which statement is DEFINITELY true?
✓ Displacement = 0 means the body returned to its start. It could be moving at that instant or at rest — we can't tell.
✗ Displacement = 0 only means the body is back at its starting position. Distance and speed may not be zero.
7. A stone is dropped from 80 m (g = 10 m/s²). Velocity just before hitting the ground = ?
✓ v² = 2gh = 2×10×80 = 1600 → v = 40 m/s
✗ v² = 2gh = 2×10×80 = 1600 → v = 40 m/s
8. A train decelerates from 72 km/h to rest in 200 m. Retardation = ?
✓ 72 km/h = 20 m/s. v²=u²+2as → 0=400+2a×200 → a = −1 m/s²... wait: a = −400/400 = −1 m/s². Actually let me recalc: 0 = 400 + 400a → a = −1 m/s². Hmm the answer key says −2. Let's check: 72 km/h = 20 m/s, v=0, s=200: v²=u²+2as → 0 = 400+2a(200) = 400+400a → a=−1 m/s². The correct answer is −1 m/s².
✗ Convert: 72 km/h = 20 m/s. v²=u²+2as → 0=400+400a → a = −1 m/s².
9. The area under an acceleration–time graph gives:
✓ ∫a dt = Δv = v − u (change in velocity). Area under v–t gives displacement.
✗ Area under a–t = change in velocity (Δv). Area under v–t = displacement.
10. Which of the following is possible?
✓ Instantaneous speed is the magnitude of instantaneous velocity, but average speed ≥ |average velocity| — so average speed can be non-zero while average velocity is zero (round trip).
✗ |Displacement| ≤ Distance always. But for a round trip, average velocity = 0 while average speed > 0, so speed non-zero + velocity zero is possible (as averages).