Chapter 2: Motion in One Dimension

ICSE Class 9 Physics — Complete Notes for School Exams | Reference: S. Chand Concise Physics by R.K. Bansal

📖 To the Student: Motion is the foundation of Physics. From a rolling ball to a falling raindrop — everything that moves follows the principles in this chapter. Understanding motion in a straight line (one dimension) will prepare you for all of kinematics. Focus on the definitions, the three equations of motion, and the interpretation of graphs. These are guaranteed exam topics!

1. Rest and Motion

Core Definitions A body is said to be in motion if it changes its position with respect to its surroundings (a reference point) as time passes.

A body is said to be at rest if it does not change its position with respect to its surroundings as time passes.

Key Concept: Motion and Rest are Relative Rest and motion are relative concepts — a body may be at rest with respect to one observer and in motion with respect to another.

Example: A passenger sitting in a moving train is:

At rest relative to another passenger sitting beside them (same reference).
In motion relative to a person standing on the platform.

∴ There is no absolute rest or absolute motion.

Types of Motion

Type	Description	Example
Translatory / Linear	Body moves along a path (straight or curved) without rotating	Car on road, falling stone
Rectilinear	Special case — motion along a straight line only	Freely falling ball, train on track
Circular	Body moves along a circular path	Earth around the Sun, fan blades
Oscillatory / Vibratory	Body moves back and forth about a fixed mean position	Pendulum, guitar string
Random	Irregular, unpredictable movement	Dust particles in air, Brownian motion

Uniform vs. Non-Uniform Motion

Uniform vs Non-Uniform Uniform Motion: Equal distances are covered in equal intervals of time (however small). Speed is constant.

Non-Uniform Motion: Unequal distances are covered in equal intervals of time. Speed keeps changing.

Scalar vs Vector quantities and types of motion

Fig. 1 — Scalar quantities have only magnitude; vector quantities have both magnitude and direction. Motion can be uniform, non-uniform, or rest.

2. Scalar and Vector Quantities

Scalar Quantities Quantities described by magnitude (size) alone, with no direction.
Examples: Distance, Speed, Time, Mass, Temperature, Energy, Work, Power.

Vector Quantities Quantities that have both magnitude AND direction. Represented by arrows.
Examples: Displacement, Velocity, Acceleration, Force, Momentum, Weight.

Scalar	Corresponding Vector
Distance ($d$)	Displacement ($s$)
Speed ($v$)	Velocity ($\vec{v}$)
Mass ($m$)	Weight ($W = mg$, downward)
Energy (J)	Force ($\vec{F}$)
—	Acceleration ($\vec{a}$)

3. Distance and Displacement

Difference between distance and displacement

Fig. 2 — Distance is the total path length (scalar); displacement is the shortest straight-line path from start to finish (vector).

Distance Distance is the total length of path actually travelled by a body, regardless of direction.

Symbol: $d$ or $s$ | SI Unit: metre (m)
Type: Scalar — always positive or zero

Displacement Displacement is the shortest straight-line distance from the initial position to the final position, along with its direction.

Symbol: $s$ or $\Delta x$ | SI Unit: metre (m)
Type: Vector — can be positive, negative, or zero

Key Differences: Distance vs. Displacement

Point	Distance	Displacement
Nature	Scalar	Vector
Defined as	Total path length	Shortest path from start to end + direction
Value	Always ≥ 0	Can be +, −, or 0
Comparison	Distance ≥ Displacement	Displacement ≤ Distance
When equal?	When body moves in a straight line without turning back
When displacement = 0?	When body returns to its starting point (e.g., full circular trip)

Example 1 ICSE Q: A boy walks 4 m East, then 3 m North. Find his (a) distance, (b) displacement.

(a) Distance = total path = 4 + 3 = 7 m
(b) Displacement = shortest path = $\sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = $ 5 m (in the North-East direction)
Note: Displacement (5 m) < Distance (7 m) ✓

🧠 Memory Trick Think of it like GPS vs. an odometer:
• Odometer (car dashboard) → counts every twist and turn = Distance
• GPS → shows direct point-to-point = Displacement

A body completing one full lap of a track: Distance = circumference, Displacement = 0.

4. Speed and Velocity

Fig. 3 — Speed is scalar (how fast), velocity is vector (how fast + which direction), acceleration measures the rate of change of velocity.

Speed

Speed — Definition & Formula Speed is defined as the distance travelled per unit time. $$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \implies v = \frac{d}{t}$$

SI Unit: metre per second (m/s) | Type: Scalar
Conversion: 1 km/h = $\dfrac{5}{18}$ m/s 1 m/s = $\dfrac{18}{5}$ km/h = 3.6 km/h

Type	Definition	Formula
Uniform Speed	Equal distances in equal time intervals	Speed = constant
Non-Uniform (Variable) Speed	Unequal distances in equal time intervals	Speed varies
Average Speed	Total distance ÷ Total time	$v_{avg} = \dfrac{\text{Total Distance}}{\text{Total Time}}$
Instantaneous Speed	Speed at a particular instant	Reading on speedometer at that moment

Velocity

Velocity — Definition & Formula Velocity is defined as the displacement per unit time. $$\text{Velocity} = \frac{\text{Displacement}}{\text{Time}} \implies v = \frac{s}{t}$$

SI Unit: m/s (with direction) | Type: Vector
Average Velocity $= \dfrac{\text{Total Displacement}}{\text{Total Time}}$
For uniform acceleration: $v_{avg} = \dfrac{u + v}{2}$

Speed vs. Velocity — Critical Differences

Comparison Table

Speed	Velocity
Scalar	Vector
= Distance / Time	= Displacement / Time
Always ≥ 0	Can be +, −, or 0
Never negative	Can be negative (means opposite direction)
Speed ≥ \|Velocity\|	\|Velocity\| ≤ Speed

Key Insight: A car going around a circular track at 60 km/h has constant speed but continuously changing velocity (because direction changes).

Example 2 Numerical ICSE Q: A car travels 150 km in 3 hours. Find its average speed in (a) km/h, (b) m/s.

(a) Average speed $= \dfrac{150}{3} = $ 50 km/h
(b) In m/s: $50 \times \dfrac{5}{18} = \dfrac{250}{18} \approx $ 13.9 m/s

Example 3 — Tricky Average Speed Important Q: A man covers the first half of his journey at 40 km/h and the second half at 60 km/h. Find his average speed for the whole journey.

Warning: Average speed ≠ (40+60)/2 = 50 km/h (this is wrong!)
Let total distance = $2d$ km.
$t_1 = \dfrac{d}{40}$ h $t_2 = \dfrac{d}{60}$ h Total time $= \dfrac{d}{40}+\dfrac{d}{60} = \dfrac{3d+2d}{120} = \dfrac{5d}{120} = \dfrac{d}{24}$
Average speed $= \dfrac{2d}{d/24} = 2d \times \dfrac{24}{d} = $ 48 km/h

5. Acceleration and Retardation

Acceleration — Definition & Formula Acceleration is the rate of change of velocity, i.e., change in velocity per unit time. $$a = \frac{v - u}{t} = \frac{\text{Final velocity} - \text{Initial velocity}}{\text{Time taken}}$$

SI Unit: m/s² | Type: Vector
Positive $a$: velocity increasing (speeding up)
Negative $a$ (Retardation / Deceleration): velocity decreasing (slowing down)
Zero $a$: velocity is constant (uniform motion)

Uniform vs Non-Uniform Acceleration Uniform Acceleration: Velocity changes by equal amounts in equal time intervals. Acceleration is constant.
Example: Freely falling body — accelerates at $g = 9.8$ m/s² constantly.

Non-Uniform Acceleration: Velocity changes by unequal amounts in equal time intervals.
Example: A car in heavy city traffic.

Example 4 Numerical Q: A car starts from rest and reaches 72 km/h in 10 seconds. Find its acceleration.

$u = 0$ m/s, $v = 72 \times \dfrac{5}{18} = 20$ m/s, $t = 10$ s
$a = \dfrac{v-u}{t} = \dfrac{20-0}{10} = $ 2 m/s²

Example 5 — Retardation Numerical Q: A train moving at 90 km/h comes to rest in 30 seconds. Find its retardation.

$u = 90 \times \dfrac{5}{18} = 25$ m/s, $v = 0$, $t = 30$ s
$a = \dfrac{0-25}{30} = -0.83$ m/s²
Retardation = 0.83 m/s² (retardation is always expressed as a positive value)

6. Equations of Motion (Uniform Acceleration)

Three equations of motion with velocity-time graph derivation

Fig. 4 — All three equations of motion are derived from the v-t graph. The slope gives acceleration; the shaded area gives displacement.

Variables Used $u$ = initial velocity (m/s) | $v$ = final velocity (m/s) | $a$ = acceleration (m/s²) | $s$ = displacement (m) | $t$ = time (s)

First Equation: $v = u + at$

Derivation By definition of acceleration: $$a = \frac{v-u}{t} \implies at = v - u$$ $$\boxed{v = u + at}$$ Use: To find final velocity or time when acceleration is uniform.

Second Equation: $s = ut + \dfrac{1}{2}at^2$

Derivation For uniform acceleration, average velocity $= \dfrac{u+v}{2}$
Displacement = average velocity × time: $$s = \frac{u+v}{2} \times t$$ Substitute $v = u + at$: $$s = \frac{u + u + at}{2} \times t = \frac{2u + at}{2} \times t$$ $$\boxed{s = ut + \frac{1}{2}at^2}$$ Use: To find displacement when time is known.

Third Equation: $v^2 = u^2 + 2as$

Derivation From 1st equation: $t = \dfrac{v-u}{a}$. Substitute into $s = \dfrac{u+v}{2} \times t$: $$s = \frac{u+v}{2} \times \frac{v-u}{a} = \frac{v^2 - u^2}{2a} \implies 2as = v^2 - u^2$$ $$\boxed{v^2 = u^2 + 2as}$$ Use: To find velocity or displacement when time is not given or required.

THREE EQUATIONS OF MOTION — Quick Reference

$v = u + at$      (use when time is known, find $v$)
$s = ut + \dfrac{1}{2}at^2$      (use when time is known, find $s$)
$v^2 = u^2 + 2as$      (use when time is NOT given)

🧠 Choosing the Right Equation

Time given, need velocity → $v = u + at$
Time given, need displacement → $s = ut + \frac{1}{2}at^2$
No time given/needed → $v^2 = u^2 + 2as$
Body starts from rest: $u = 0$ → equations become $v = at$, $s = \frac{1}{2}at^2$, $v^2 = 2as$
Body comes to rest: $v = 0$ → substitute and solve

Example 6 — 1st & 2nd Equations Numerical ICSE Q: A bus starts from rest with uniform acceleration of 2 m/s². Find: (a) velocity after 5 s, (b) distance covered in 5 s.

Given: $u = 0$, $a = 2$ m/s², $t = 5$ s

(a) $v = u + at = 0 + 2 \times 5 = $ 10 m/s
(b) $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 25 = $ 25 m

Example 7 — 3rd Equation Numerical Important Q: A car travelling at 20 m/s applies brakes and decelerates at 4 m/s². Find the distance it covers before stopping.

$u = 20$ m/s, $v = 0$ (stops), $a = -4$ m/s²
Using $v^2 = u^2 + 2as$:
$0 = 400 + 2(-4)s \implies 8s = 400 \implies s = $ 50 m

Example 8 — Finding Time Numerical Q: A stone is thrown upward at 19.6 m/s. How long to reach maximum height? ($g = 9.8$ m/s²)

At max height: $v = 0$, $a = -g = -9.8$ m/s²
$v = u + at \implies 0 = 19.6 - 9.8t \implies t = $ 2 s

7. Distance-Time (d-t) Graphs

Fig. 5 — Five types of d-t graphs. The slope at any point gives instantaneous speed. A steeper slope means greater speed; horizontal line means rest.

Key Rules for d-t Graphs — MEMORIZE

Slope of d-t graph = Speed $\left(\text{slope} = \dfrac{\Delta d}{\Delta t} = v\right)$
Steeper slope → Greater speed
Horizontal line → Body at rest (speed = 0)
Straight diagonal line → Uniform speed
Curve bending upward (concave up) → Accelerated motion (speed increasing)
Curve bending downward (concave down) → Decelerated motion (speed decreasing)
Distance can never decrease — line never goes downward in a d-t graph

Shape of d-t Graph	Type of Motion	Speed
Horizontal straight line	Body at rest	Speed = 0
Straight diagonal line (positive slope)	Uniform motion	Constant speed
Upward curve (parabola opening up)	Uniformly accelerated	Increasing uniformly
Downward curve (flattening)	Uniformly decelerated	Decreasing uniformly
Irregular curve	Non-uniform motion	Varies irregularly

Example 9 — Reading a d-t Graph ICSE Q: A d-t graph shows a body covering 60 m in 12 seconds (straight line). Find its speed.

Speed = slope $= \dfrac{\Delta d}{\Delta t} = \dfrac{60}{12} = $ 5 m/s (uniform — straight line)

8. Velocity-Time (v-t) Graphs

Fig. 6 — Six types of v-t graphs. Slope = acceleration; area under the graph = displacement. These are the most tested graphs in ICSE exams!

Key Rules for v-t Graphs — MOST IMPORTANT

Slope of v-t graph = Acceleration $\left(\text{slope} = \dfrac{\Delta v}{\Delta t} = a\right)$
Area under v-t graph = Displacement
Horizontal line → Uniform velocity (acceleration = 0)
Straight line going up → Uniform acceleration (positive slope)
Straight line going down → Uniform deceleration (negative slope)
Curved line → Non-uniform acceleration
Line touching x-axis (v = 0) → Body momentarily at rest

Calculating Displacement from v-t Graph (Area Rule)

Area = Displacement Triangle area (body starts from rest or ends at rest): $A = \dfrac{1}{2} \times \text{base} \times \text{height}$

Rectangle area (uniform velocity): $A = \text{length} \times \text{breadth} = v \times t$

Trapezoid area (uniform acceleration with initial velocity): $A = \dfrac{1}{2}(u+v) \times t$

Feature	d-t Graph	v-t Graph
Y-axis	Distance (m)	Velocity (m/s)
Slope gives	Speed	Acceleration
Area gives	—	Displacement
Horizontal line	Rest (speed = 0)	Uniform velocity (a = 0)
Straight diagonal	Uniform speed	Uniform acceleration
Curve bending upward	Acceleration	Non-uniform acceleration (a increasing)

Example 10 — v-t Graph Full Analysis Numerical Important Q: A v-t graph shows: velocity increases from 0 to 30 m/s in 6 s, then stays at 30 m/s for 4 s, then decreases to 0 in 3 s. Find: (a) acceleration in each phase, (b) total displacement.

(a) Phase 1 (0–6 s): $a_1 = \dfrac{30-0}{6} = $ 5 m/s²
Phase 2 (6–10 s): $a_2 = 0$ m/s² (constant velocity)
Phase 3 (10–13 s): $a_3 = \dfrac{0-30}{3} = $ −10 m/s² (retardation = 10 m/s²)

(b) Total Displacement = Total area under v-t graph:
Phase 1 (triangle): $\frac{1}{2} \times 6 \times 30 = 90$ m
Phase 2 (rectangle): $4 \times 30 = 120$ m
Phase 3 (triangle): $\frac{1}{2} \times 3 \times 30 = 45$ m
Total = 90 + 120 + 45 = 255 m

9. Free Fall and Acceleration Due to Gravity

Free fall and acceleration due to gravity diagram

Fig. 7 — A freely falling body accelerates downward at g = 9.8 m/s². The increasing gap between positions shows the body speeds up as it falls.

Free Fall When a body falls under gravity alone (no air resistance), it is called free fall.

Key Facts:

All freely falling bodies, regardless of mass, fall with the same acceleration (proved by Galileo).
This acceleration is the acceleration due to gravity, symbol $g$.
$g = 9.8$ m/s² ≈ 10 m/s² (directed downward) at Earth's surface.
$g$ is slightly greater at the poles than at the equator.
$g$ decreases as altitude increases.

Equations for Free Fall

Body dropped from rest ($u = 0$, $a = g$ downward):
$v = gt$ $h = \dfrac{1}{2}gt^2$ $v^2 = 2gh$

Body thrown vertically upward ($a = -g$, upward positive):
Max height: $H = \dfrac{u^2}{2g}$ Time to reach top: $t = \dfrac{u}{g}$ Total flight time: $T = \dfrac{2u}{g}$

Example 11 — Free Fall Numerical ICSE Q: A stone is dropped from a cliff 80 m high. Find: (a) time to reach the ground, (b) velocity on impact. ($g = 10$ m/s²)

$u = 0$, $h = 80$ m, $a = g = 10$ m/s²

(a) $h = \frac{1}{2}gt^2 \implies 80 = 5t^2 \implies t^2 = 16 \implies t = $ 4 s
(b) $v = gt = 10 \times 4 = $ 40 m/s
Verify: $v^2 = 2gh = 2(10)(80) = 1600 \implies v = 40$ m/s ✓

Example 12 — Upward Throw Numerical Important Q: A ball is thrown vertically upward at 29.4 m/s. Find: (a) maximum height, (b) total time of flight. ($g = 9.8$ m/s²)

$u = 29.4$ m/s, $v = 0$ at max height, $a = -9.8$ m/s²

(a) $H = \dfrac{u^2}{2g} = \dfrac{29.4^2}{2 \times 9.8} = \dfrac{864.36}{19.6} = $ 44.1 m
(b) Time to top: $t = \dfrac{u}{g} = \dfrac{29.4}{9.8} = 3$ s Total flight time $= 2t = $ 6 s

10. Formula Quick Reference

Quantity	Formula	SI Unit	Type
Speed	$v = d/t$	m/s	Scalar
Average Speed	Total distance / Total time	m/s	Scalar
Velocity	$v = s/t$	m/s	Vector
Acceleration	$a = (v-u)/t$	m/s²	Vector
1st Equation	$v = u + at$	—	—
2nd Equation	$s = ut + \frac{1}{2}at^2$	—	—
3rd Equation	$v^2 = u^2 + 2as$	—	—
Free fall height	$h = \frac{1}{2}gt^2$ (if $u=0$)	m	—
Max height (thrown up)	$H = u^2/(2g)$	m	—
Total flight time	$T = 2u/g$	s	—
Unit conversion	1 km/h = 5/18 m/s \| 1 m/s = 18/5 km/h	—	—

11. Practice Problems — All Types

Type A: Conceptual Questions

Q1 — Conceptual ICSE Q: Can a body have (a) zero velocity but non-zero acceleration? (b) constant speed but changing velocity?

(a) Yes. A ball thrown vertically upward has $v = 0$ at its highest point, yet still has downward acceleration $g = 9.8$ m/s² (gravity does not switch off at the peak).

(b) Yes. A body moving in a circular path with constant speed — the magnitude of velocity (speed) stays constant, but the direction changes continuously, so velocity changes.

Q2 — Distance vs Displacement ICSE Q: A person walks 3 km East, then 4 km North. Find the distance and displacement.

Distance = 3 + 4 = 7 km
Displacement = $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = $ 5 km in the North-East direction (at 53.1° from East)

Type B: Speed and Velocity Problems

Q3 Numerical Q: A cyclist rides 300 m in 60 s, rests for 30 s, then rides back 300 m in 90 s. Find: (a) average speed, (b) average velocity.

Total distance $= 600$ m, Total time $= 180$ s, Net displacement $= 0$
(a) Average speed $= 600/180 = $ 3.33 m/s
(b) Average velocity $= 0/180 = $ 0 m/s (returns to start)

Q4 — Train Problem Numerical Q: A train 200 m long passes a pole in 10 s. How long will it take to pass a bridge 800 m long?

Speed $= \dfrac{200}{10} = 20$ m/s
Distance to pass bridge $= 200 + 800 = 1000$ m
Time $= \dfrac{1000}{20} = $ 50 s

Type C: Equations of Motion

Q5 — Distance in nth Second Numerical Important Q: A car starts from rest with acceleration 3 m/s². Find the distance covered in the 5th second.
Formula: $s_n = u + \dfrac{a}{2}(2n-1)$

$u = 0$, $a = 3$ m/s², $n = 5$:
$s_5 = 0 + \dfrac{3}{2}(2\times5-1) = \dfrac{3}{2} \times 9 = $ 13.5 m

Q6 — Two cars ICSE Q: Two cars start from rest. Car A has acceleration 4 m/s²; Car B has 2 m/s². After how many seconds will A be 100 m ahead of B?

After time $t$: $s_A = \frac{1}{2}(4)t^2 = 2t^2$ $s_B = \frac{1}{2}(2)t^2 = t^2$
$s_A - s_B = t^2 = 100 \implies t = $ 10 s

Type D: Graph Problems

Q7 — v-t Graph Reading ICSE Q: A v-t graph: velocity rises from 0 to 20 m/s in 5 s, stays at 20 m/s for 3 s, then falls to 0 in 2 s. Find (a) accelerations, (b) total distance.

(a) Phase 1: $a = 20/5 = $ 4 m/s² | Phase 2: $a = $ 0 | Phase 3: $a = -20/2 = $ −10 m/s²
(b) Area: $\frac{1}{2}(5)(20) + (3)(20) + \frac{1}{2}(2)(20) = 50 + 60 + 20 = $ 130 m

Type E: Free Fall

Q8 — Two balls Numerical Important Q: A ball is dropped and another is thrown downward at 10 m/s from 80 m height. Which hits first? ($g = 10$ m/s²)

Ball 1 (dropped): $80 = 5t_1^2 \implies t_1 = 4$ s
Ball 2 (thrown): $80 = 10t_2 + 5t_2^2 \implies t_2^2 + 2t_2 - 16 = 0 \implies t_2 \approx 3.12$ s
$t_2 < t_1$ → Ball 2 hits first

Q9 — Classic ICSE ICSE Important Q: A body thrown up takes 4 s to return to the starting point. Find: (a) initial velocity, (b) maximum height. ($g = 10$ m/s²)

Time to reach top $= 4/2 = 2$ s. At top, $v = 0$:
(a) $0 = u - 10(2) \implies u = $ 20 m/s
(b) $H = u^2/(2g) = 400/20 = $ 20 m

Type F: Mixed Problems

Q10 — Two towns meeting ICSE Q: Towns A and B are 600 km apart. A car leaves A at 60 km/h and a bus leaves B at 90 km/h simultaneously, heading toward each other. When and where do they meet?

Relative closing speed $= 60 + 90 = 150$ km/h
Time to meet $= \dfrac{600}{150} = $ 4 hours
Distance from A $= 60 \times 4 = $ 240 km from A (360 km from B)

12. Common Mistakes & Exam Tips

⚠️ Common Mistakes to Avoid 1. Confusing speed (scalar) with velocity (vector) — speed has no direction.
2. Confusing distance and displacement — a body returning to start: distance > 0, displacement = 0.
3. Not converting km/h → m/s before using equations (multiply by 5/18).
4. Using $a = +g$ for a body thrown upward — should be $a = -g$ (gravity opposes upward motion).
5. For "comes to rest," setting $u = 0$ instead of $v = 0$.
6. Thinking average speed $= \dfrac{v_1+v_2}{2}$ always — only valid when time at each speed is equal, not distance.
7. Confusing slope rules: slope of d-t = speed; slope of v-t = acceleration.
8. In free fall, thinking heavier objects fall faster — ALL objects fall at the same $g$.

🎯 Exam Strategy for ICSE

Definitions (1–2 marks): Write exact definition + formula + SI unit.
Numericals: Always write: Given → Find → Formula → Substitution → Answer with unit.
Graph questions: Identify slope (acceleration) and area (displacement) correctly — appear every year.
Derivations (4–5 marks): Know all 3 equations of motion — step-by-step derivation.
Convert units first: Always convert to SI before substituting into formulas.
Sign convention: Choose one direction as positive. If upward = +, then downward displacement is −.
$g$ value: Use $g = 9.8$ m/s² unless question explicitly says to use 10 m/s².
High-value topics: Equations of motion, v-t graph, d-t graph, free fall — guarantee 70% of marks.