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Overview

(Unit Two→Linear Motion — Second Secondary, Term 1)

🌳 Concept tree

Chapter 2→Motion in a Straight Line

├── What you will learn

├── 1.1 Why study motion?

├── 1.2 Definition of motion

├── 1.3 Types of motion

├── 2.1 Why we need velocity

├── 2.2 Speed and velocity

├── 2.3 Converting units

├── 3.1 Uniform and non-uniform (variable) velocity

├── 3.2 Instantaneous and average velocity

├── 4.1 How to draw it

├── 4.2 Reading the displacement–time graph

├── 5.1 Definition

├── 5.2 A motion diagram: the car that gains 5 m/s every second

├── 5.3 Types of acceleration

├── 6.1 Reading the graph

├── 6.2 Analysing a multi-stage graph

├── 6.3 Extension — displacement from the v–t graph

├── 6.4 Motion (dot) diagrams

├── Short answers to typical discussion questions

├── Summary

Chapter 2: Motion in a Straight Line · 1 / 23

What you will learn

The meaning of motion, and its types (translational and periodic).
Speed and velocity; uniform and variable velocity; instantaneous and average velocity.
Drawing and reading the displacement–time graph.
Acceleration and its types (positive, zero, negative); drawing and reading the velocity–time graph.
Analysing motion diagrams and graphs of motion in a straight line.

Chapter 2: Motion in a Straight Line · 2 / 23

1.1 Why study motion?

Look around you some things are at rest and others are moving.
Ships, trains and aeroplanes can only run on a safe timetable if we can describe and analyse their motion.
Departure and arrival times follow from their speeds.
In this chapter we learn the concepts and quantities needed to describe motion; in the next chapter we study what causes motion (forces).

Chapter 2: Motion in a Straight Line · 3 / 23

1.2 Definition of motion

Motion is the change in the position of an object with time, relative to the position of another object (a reference).
Two points to notice:

Motion is relative. A passenger sitting in a train is at rest relative to the train but moving relative to the platform. Whenever you say an object moves, you must say relative to what.
If the position of an object does not change over a period of time, it is at rest.

When the object moves along a straight path, we call it motion in a straight line.
It is the simplest type of motion.
Example a train on a long straight stretch of railway track.
(Think if you look at a film strip showing the positions of a mouse at equal time intervals, the mouse is moving if its position relative to the background is different in the successive pictures.)

Chapter 2: Motion in a Straight Line · 4 / 23

1.3 Types of motion

Motion can be classified into two main types:

Comparison aspect	Main idea	Example	Keywords
Translational motion	motion that has a starting point and an end point. Examples: motion in a straight line projectile motion (a thrown ball) the motion of a car train ship or aeroplane	—	startingpointExamplesstraight
Periodic motion	motion that repeats itself at equal intervals of time. Examples: circular motion (a point on a wheel) vibrational (oscillating) motion	a swinging pendulum or a vibrating tuning-fork prong	repeatsitselfequalintervals

Practice classification

Motion	Type
Pendulum of a clock	periodic (vibrational)
A projectile	translational
A train going from Cairo to Aswan	translational
Prong of a sounding tuning fork	periodic (vibrational)
The Earth’s revolution around the Sun	periodic (circular)

Our study in this unit concentrates on force and motion and their life applications; it prepares the way for the mathematical proof of laws in mathematics and applied mathematics.

Chapter 2: Motion in a Straight Line · 5 / 23

2.1 Why we need velocity

We often say "fast" or "slow", but these words are not scientific.
To describe motion exactly we need a quantity — velocity (and speed).
Velocity

is the displacement covered by the object in one second, or the rate of change of displacement with time.
Its units m/s (SI) or km/h.

velocity v = displacement ÷ time = Δd / Δt

Here Δd (the change in position) is the displacement and Δt is the time taken.

Chapter 2: Motion in a Straight Line · 6 / 23

2.2 Speed and velocity

In a car, the speedometer in front of the driver shows, for example, "80 km/h".
It tells us how fast the car moves but nothing about its direction.
This quantity is called speed.
To describe the motion completely we need the direction too, for example "80 km/h toward the east" — this is velocity.

Point	Speed	Velocity
Definition	distance covered per unit time	displacement per unit time
Type of quantity	scalar (magnitude only)	vector (magnitude and direction)
Sign	always positive	positive in one direction negative in the opposite direction

So the speedometer measures speed, not velocity, because it does not give the direction of travel.
Note in the problems of this book, when we say "velocity" we use it as the quantity that gives the complete description of the motion; for straight-line motion the direction is just a plus or minus sign.

Chapter 2: Motion in a Straight Line · 7 / 23

2.3 Converting units

1 km/h = 1000 m / 3600 s = 5/18 m/s.
So multiply km/h by 5/18 to get m/s; multiply m/s by 18/5 (= 3.6) to get km/h.
Examples:

Comparison aspect	Main idea	Keywords
•	36 km/h = 36 × 5/18 = 10 m/s
•	25 m/s = 25 × 3.6 = 90 km/h
Example 1.	A cyclist covers 450 m in 90 s along a straight road. Find the velocity in m/s and in km/h	cyclistcoversalongstraight
v = Δd/Δt = 450/90 = 5 m/s; in km/h	5 × 3.6 = 18 km/h
Example 2.	A train moves at a velocity of 72 km/h. How far does it go in 25 min if the velocity is uniform?	trainmovesvelocitydoes
•	72 km/h = 20 m/s t = 25 × 60 = 1500 s Δd = vΔt = 20 × 1500 = 30 000 m = 30 km	1500
(Or in km/h	72 × 25/60 = 30 km.)
Example 3.	An athlete runs 6000 m in 40 min along a straight track. The average velocity is 6000/(40 × 60) = 2.5 m/s = 9 km/h. If he keeps this average velocity he covers 2.5 × 3600 = 9000 m in one hour	athleteruns6000along

Chapter 2: Motion in a Straight Line · 8 / 23

3.1 Uniform and non-uniform (variable) velocity

Imagine a runner photographed at equal time intervals (for example every second).

Comparison aspect	Main idea	Keywords
Uniform velocity	the object covers equal displacements in equal intervals of time. Its velocity has a constant magnitude and a constant direction (a straight line). In the picture the positions are equally spaced	objectcoversequaldisplacements
Non-uniform (variable) velocity	the object covers unequal displacements in equal intervals of time. The velocity changes in magnitude or in direction or both. The positions in the picture are unequally spaced	objectcoversunequaldisplacements

Chapter 2: Motion in a Straight Line · 9 / 23

3.2 Instantaneous and average velocity

In real journeys a car does not keep a fixed velocity it speeds up, slows down, or stops at lights.
To describe such motion we use two quantities.
Instantaneous velocity (v)

— the velocity of the body at a certain moment.
It can be read from the speedometer at that moment (as a speed).
On a displacement–time graph it is the slope of the tangent to the curve at the point corresponding to that moment.
Average velocity (v̄) the total displacement from the starting point to the end point divided by the total time:
v̄ = (final position − initial position) / total time = Δd / Δt
On a displacement–time graph it is the slope of the straight line (chord) joining the starting point and the end point.
Example 1 — average velocity of a two-part trip. A bus travels 120 km east in 1.5 h and then 60 km east in 0.5 h. Total displacement = 180 km; total time = 2.0 h.
v̄ = 180/2 = 90 km/h east. (This is neither 80 km/h nor 120 km/h: do not average the two part-velocities, 80 and 120, unless the times are equal.)

✍️ Worked examples

Example 2 — return trip. A student walks 600 m east in 5 min and then 600 m back west in 5 min.
Displacement = 0 → average velocity = 0.
Distance = 1200 m, so average speed = 1200/600 s = 2 m/s.
This shows that speed and velocity are different quantities.
Example 3 — how long? A car moves with an average velocity of 15 m/s. How long does it take to cover 4.5 km? t = 4500/15 = 300 s = 5 min.

Chapter 2: Motion in a Straight Line · 10 / 23

4.1 How to draw it

Watch a moving object and record its displacement at different times. Then:

Mark the time on the horizontal axis and the displacement on the vertical axis (with a suitable scale).
For each time, draw a vertical line from the time axis and a horizontal line from the displacement axis; their crossing point is one data point.
Repeat for all the readings.
Draw the best straight line (or smooth curve) passing near the points.
Find the velocity as the slope of the line.

Chapter 2: Motion in a Straight Line · 11 / 23

4.2 Reading the displacement–time graph

slope = Δ(displacement) / Δ(time) = velocity

Shape of the graph	Meaning
Straight line sloping upward	uniform velocity (positive)
Steeper straight line	greater velocity
Horizontal line	displacement does not change → the object is at rest (velocity = 0)
Straight line sloping downward	uniform velocity in the negative direction (moving back)
Curve	variable velocity the slope changes from point to point
Slope of a tangent at a point	instantaneous velocity at that moment
Slope of the chord joining two points	average velocity between those two moments

Example 1.→A straight line on a displacement–time graph passes through (t = 2 s, d = 10 m) and (t = 8 s, d = 40 m). Velocity = slope = (40 − 10)/(8 − 2) = 30/6 = 5 m/s.

Comparison aspect	Main idea	Keywords
Example 2.	A curved d–t graph: at t = 1 s the object is at 1 m and at t = 5 s it is at 25 m. The average velocity between these moments is (25 − 1)/(5 − 1) = 6 m/s. The instantaneous velocity at t = 3 s needs a tangent drawn at t = 3 s if the tangent rises 12 m in 2 s the instantaneous velocity is 6 m/s	curvedgraphobjectaverage
Example 3 (graph in words).	A walker moves from the origin for 10 s rising to 20 m then the graph is horizontal for 5 s then it goes back down to 0 at t = 25 s. Velocities: first stage 20/10 = +2 m/s second stage 0 (rest) third stage (0 − 20)/10 = −2 m/s (returning). The total displacement at the end is zero though the distance covered is 40 m	walkermovesoriginrising

Chapter 2: Motion in a Straight Line · 12 / 23

5.1 Definition

We have seen that velocity can change in magnitude, direction or both.
Motion in which velocity changes with time is called accelerated motion.
The physical quantity that expresses the change of velocity with time is acceleration (a).
Examples of change of velocity:

at the beginning of a journey the velocity increases;
on a curve the direction of the velocity changes;
at the end the velocity decreases until the vehicle stops.

Chapter 2: Motion in a Straight Line · 13 / 23

5.2 A motion diagram: the car that gains 5 m/s every second

A car starting from rest has the following speedometer readings:

time (s)	0	1	2	3	4
velocity (m/s)	0	5	10	15	20

The velocity increases by 5 m/s every second — a constant rate. This rate is the acceleration:

a = Δv / Δt = (v₂ − v₁) / (t₂ − t₁)

Check with the table→between t = 1 s and t = 2 s, a = (10 − 5)/(2 − 1) = 5 m/s².

Acceleration

is the change in the velocity of the body per unit time, i.e.
the time rate of change of velocity.
Unit m/s² (SI) or km/h².
It is a vector quantity.
Its dimensional formula is LT⁻².
Meaning of "5 m/s²" every second the velocity changes by 5 m/s.

Chapter 2: Motion in a Straight Line · 14 / 23

5.3 Types of acceleration

Take the direction of the velocity as the positive direction.
A small ball rolling on a smooth surface with different slopes shows all three cases:

Positive acceleration(velocity increases): e.g. the ball rolling down an inclined plane.

Zero acceleration(velocity constant): e.g. the ball rolling on a smooth horizontal surface. The change in velocity is zero.

Negative acceleration(velocity decreases — often called deceleration or retardation): e.g. the ball rolling up an inclined plane.

General rule about signs→(very important):

If velocity and acceleration have the same direction (same sign), the object speeds up — even if both are negative.
If they have opposite directions (opposite signs), the object slows down.
If acceleration = 0, the velocity is constant. It is not necessarily zero (e.g. a car at a steady 20 m/s has zero acceleration).
Life application.

Comparison aspect	Main idea	Keywords
•	Inside every car there are three controls that change the velocity: the accelerator (gas) pedal increases the magnitude the brake pedal decreases it and the steering wheel changes its direction	Insideeverythreecontrols
•	Each of them produces an acceleration	themproducesacceleration
Example 1.	A car increases its velocity uniformly from 10 m/s to 30 m/s in 8 s. a = (30 − 10)/8 = +2.5 m/s²	increasesvelocityuniformly
Example 2 — braking.	A car moving at 25 m/s brakes uniformly and its velocity falls to 5 m/s in 4 s. a = (5 − 25)/4 = −5 m/s² (deceleration of 5 m/s²)	movingbrakesuniformlyvelocity
Example 3 — time to reach a velocity.	A motorcycle starts from rest with a = 3 m/s². Time to reach 90 km/h (= 25 m/s)? From a = Δv/Δt: t = 25/3 = 8.3 s	motorcyclestartsrest

✍️ Worked examples→Take upward as positive: a ball leaves the hand at +20 m/s and, because it slows down, has a negative acceleration (−10 m/s²); after 1 s the velocity is +10 m/s, after 2 s it is 0, and after 3 s it is −10 m/s (moving down, now speeding up because velocity and acceleration are both negative).

Chapter 2: Motion in a Straight Line · 15 / 23

6.1 Reading the graph

Plot time on the horizontal axis and velocity on the vertical axis.
For the car in the table of Lesson 5 the graph is a straight line through the origin, which means that the velocity increases at a uniform rate, i.e.
uniform acceleration .
The slope of the velocity–time graph = acceleration.

Shape of the v–t graph	Meaning
Horizontal line (above the axis)	constant velocity → a = 0 (uniform velocity)
Horizontal line on the time axis (v = 0)	at rest
Straight line sloping upward	uniform positive acceleration
Straight line sloping downward	uniform negative acceleration (deceleration)
Steeper line	larger magnitude of acceleration
Curved line	non-uniform acceleration

Chapter 2: Motion in a Straight Line · 16 / 23

6.2 Analysing a multi-stage graph

A car moves in a straight line, and its velocity is recorded for 30 s.
Suppose the graph is: A(0 s, 0) → B(10 s, 25 m/s) → C(20 s, 25 m/s) → D(30 s, 0).
Analyse each stage:

Stage	AB	BC	CD
Initial velocity vᵢ	0	25 m/s	25 m/s
Final velocity v_f	25 m/s	25 m/s	0
Change Δv	+25 m/s	0	−25 m/s
Duration	10 s	10 s	10 s
Acceleration a = Δv/t	+2.5 m/s²	0	−2.5 m/s²
Description	speeding up uniformly	constant velocity	slowing down uniformly to rest (braking)

Chapter 2: Motion in a Straight Line · 17 / 23

6.3 Extension — displacement from the v–t graph

(Not stated in the book text, but needed for graph questions like "find the total distance covered".)
The area between the v–t graph and the time axis equals the displacement (distance for motion in one direction).
For the graph above:

Comparison aspect	Main idea	Keywords
AB	½ × 10 × 25 = 125 m
BC	10 × 25 = 250 m
CD	½ × 10 × 25 = 125 m
•	Total distance = 500 m average velocity = 500/30 = 16.7 m/s	Totaldistanceaveragevelocity
•	For constant acceleration the following results (all follow from the straight-line graph) are useful	constantaccelerationfollowingresults

v = u + at (this is just the definition of acceleration
a = (v − u)/t),

s = ½(u + v)t (area of a trapezoid),

s = ut + ½at², v² = u² + 2as.
Example. A car starts from rest and accelerates at 2 m/s² for 10 s. Then v = 0 + 2 × 10 = 20 m/s and the distance s = ½ × 2 × 10² = 100 m (check: area of the triangle ½ × 10 × 20 = 100 m).

Chapter 2: Motion in a Straight Line · 18 / 23

6.4 Motion (dot) diagrams

A motion diagram shows the positions of an object at equal time intervals as dots.

Equal spacing between dots → uniform velocity.
Spacing that increases → speeding up (positive acceleration).
Spacing that decreases → slowing down (negative acceleration).
Each of these corresponds to a graph: equal spacing ↔ horizontal v–t line (straight d–t line); increasing spacing ↔ rising v–t line (upward-curving d–t line); decreasing spacing ↔ falling v–t line (flattening d–t curve).
Example 1. From a v–t graph, the line passes through (2 s, 6 m/s) and (6 s, 18 m/s). a = (18 − 6)/(6 − 2) = 3 m/s².
Example 2. A train slows from 30 m/s to 10 m/s in 20 s. a = (10 − 30)/20 = −1 m/s². Its v–t line slopes downward with slope −1.

Example 3 (both quantities negative).→A cart moves to the left (negative) with velocity −4 m/s and has acceleration −2 m/s² (also to the left). After 3 s: v = −4 + (−2)(3) = −10 m/s. The speed has increased from 4 to 10 m/s, because velocity and acceleration have the same sign.

Chapter 2: Motion in a Straight Line · 19 / 23

Short answers to typical discussion questions

Comparison aspect	Main idea	Keywords
•	A ball rolls on the floor slows down and stops — do the velocity and the acceleration have the same sign? No. The velocity is in the direction of motion (say +) the acceleration is opposite (−) because the velocity decreases (friction slows the ball)	ballrollsfloorslows
If an object’s acceleration is zero, is its velocity zero?	No. Zero acceleration means constant velocity. Example: a car travelling steadily at 60 km/h on a straight level road has a = 0 but v ≠ 0	meansconstantExampletravelling
Why does the average velocity differ from the instantaneous velocity?	The average is the overall displacement ÷ total time the instantaneous is the value at one moment	overalldisplacementtotaltime

Chapter 2: Motion in a Straight Line · 20 / 23

Summary

Motion = change of position with time relative to another object; types: translational and periodic.
Velocity v = Δd/Δt (vector, m/s); speed is the scalar magnitude. 1 km/h = 5/18 m/s.
Uniform velocity equal displacements in equal times; variable velocity: unequal.
Instantaneous velocity = slope of tangent on d–t; average velocity = total displacement/total time = slope of chord.
Acceleration a = Δv/Δt (m/s², LT⁻²); positive, zero or negative; same signs → speeding up; opposite signs → slowing down.
d–t graph slope = velocity; v–t graph slope = acceleration (area = displacement).
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Test yourself on this lesson

The lesson’s question bank is split across the points above; each point has its own questions underneath.
To work through the whole bank on one page:

Chapter 2: Motion in a Straight Line · 21 / 23

Key Scientific Tables & Anatomical Diagrams Extracted from Official Curricula

💡 Core Diagrams & Reference Data→All data, reference comparison tables, and anatomical figures have been accurately digitized from official curriculum textbooks.

🔬 figure_diagram_1

Scientific details and precise observational breakdown:

(4). x-2y-1=0 oo X=2y+l

Substituting in the second equation :

« (2y+1)P?-y(2y+1)=0

o4y?+4y4+1-2y?-y=0

2 2y?4+3y+1=0

nys-dorys-1

From (1)→«. X=OorxX=-1

+. The $.S.= {(0 – 4) 515-1}

(5). x+2y=4

Substituting in the second equation :

Chapter 2: Motion in a Straight Line · 22 / 23

Part 23

📖

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Distance vs Displacement	Distance is the total length of the path traveled (scalar always positive). Displacement is the shortest directed straight line from the initial position to the final position (vector)	totallengthpathtraveled
Velocity vs Speed	Velocity is the rate of change of displacement with respect to time (vector: v = Δd/Δt). Speed is the distance covered per unit time (scalar)	ratechangedisplacementrespect
Uniform Acceleration (a)	The constant rate of change of velocity in equal time intervals: a = (v_f – v_i) / t, expressed in m/s². Positive indicates speeding up in the positive direction; negative indicates deceleration.	constantratechangevelocity
Free Fall Under Gravity	The motion of an object subjected solely to gravitational attraction neglecting air resistance. All objects experience identical downward acceleration (g ≈ 9.8 m/s²)	motionobjectsubjectedsolely

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspect	Main idea	Keywords
st Kinematic Equation	vf = vi + a t
nd Kinematic Equation	d = vi t + (1/2) a t²
rd Kinematic Equation	vf² = vi² + 2 a d
Free fall from rest (vi = 0)	vf = g t \| d = (1/2) g t² \| vf² = 2 g d
Maximum projectile height	hmax = vi² / (2g) \| Time to peak: t = vi / g	hmaxTimepeak

🔍 In-Depth Explanations, Deductions & Core Concepts:

Why do a feather and a coin fall with identical acceleration in a vacuum? Because gravitational acceleration (g) is independent of the mass, shape, or density of the falling object when air resistance is eliminated.
Can an object have zero velocity but non-zero acceleration? Yes, at the peak of a vertical projectile throw, instantaneous velocity is zero (v_f = 0), but downward acceleration remains g = 9.8 m/s².

Equation	Mathematical Formula	Key Physical Relation	Missing Variable
First Equation	v_f = v_i + a t	Velocity-Time	Displacement (d)
Second Equation	d = v_i t + 1/2 a t²	Displacement-Time	Final Velocity (v_f)
Third Equation	v_f² = v_i² + 2 a d	Velocity-Displacement	Time (t)

💡 High-Yield Exam Traps & Key Takeaways:

Comparison aspect	Main idea	Keywords
If an object starts from rest	vi = 0
If an object comes to a halt / brakes applied	vf = 0 and acceleration a is negative	accelerationnegative
•	Time of ascent equals time of descent back to the original launch level	Timeascentequalsdescent

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