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Physics Revision Notes
Leaving Cert Higher Level Physics
Chapter 1: Linear Motion & Vectors
These notes teach Linear Motion & Vectors clearly in simple English and then push into the deeper Higher Level reasoning. The aim is to build understanding first and exam confidence second.
Subtopics Covered
- Chapter 1: Linear Motion & Vectors
- Objectives — What you need to know
- Definitions — Exam-ready terminology
- Concepts — The core theory
- Visual Understanding — Real diagrams
- Formulae — Equations, units, and when to use them
What This Pack Includes
- Structured physics notes formatted for ExamsLogic website reading
- Exam-focused diagrams, equations, and worked examples
- Practical notes and mark scheme style guidance from the source files
- Independent study guidance based on the official curriculum
- Print-friendly layout for future PDF export when needed
This publication is an independent educational resource developed by ExamsLogic and compiled by experienced educators. It is based on publicly available official curricula, including Cambridge, Pearson Edexcel, IB, and the Irish Leaving Certificate. This product is not endorsed by, affiliated with, or sponsored by any examination board or governing authority. All registered trademarks remain the property of their respective owners.
Chapter 1: Linear Motion & Vectors
1. Objectives — What you need to know
Simple explanation: Treat these objectives as your finish line. If you can do each one without help, the chapter is in strong shape for the exam.
- Distinguish between vector and scalar quantities.
- Define displacement, velocity, and acceleration using exam-ready wording.
- Use and derive the three equations of motion for constant acceleration.
- Solve Higher Level problems involving vertical motion and gravity.
- Describe an experiment to measure constant acceleration.
- Interpret velocity-time graphs using gradient and area.
2. Definitions — Exam-ready terminology
Simple explanation: Learn these definitions in short, exact sentences. In physics, one missing word can turn a full-mark definition into a half-mark answer.
| Term | ILC HL Definition | Exam warning |
|---|---|---|
| Scalar | A physical quantity that has magnitude only. | Do not mention direction. |
| Vector | A physical quantity that has both magnitude and direction. | Direction must be stated or implied. |
| Displacement | Distance in a specified direction. | Not always equal to total distance travelled. |
| Velocity | The rate of change of displacement with respect to time. | Velocity is vector; speed is scalar. |
| Acceleration | The rate of change of velocity with respect to time. | Negative acceleration means velocity is decreasing in the chosen positive direction. |
3. Concepts — The core theory
Simple explanation: This chapter is about how objects move, how we describe that motion, and how forces change what happens next.
Linear motion is the study of objects moving along a straight line. At Higher Level, the key skill is not just memorising equations — it is choosing the correct direction, identifying hidden data, and linking graphs to motion.
Acceleration due to gravity is normally taken as g = 9.8 m s-2 downwards. If upward is chosen as the positive direction, gravity becomes a = -9.8 m s-2. That sign choice is where many marks vanish like socks in a washing machine.
4. Visual Understanding — Real diagrams
Simple explanation: Read the diagram like a story. Ask what is changing, what stays constant, and which label explains the main physics idea.
A. Scalar vs Vector
Exam point: “20 m” is distance. “20 m east” is displacement.
B. Velocity-time graph
C. Object thrown upwards
D. Practical setup: measuring acceleration with light gates
5. Formulae — Equations, units, and when to use them
Simple explanation: Do not treat the formula box like decoration. First decide what each symbol means in the question, then check that your units make sense.
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ut + ½at²
v² = u² + 2as
| Symbol | Meaning | Unit | Use carefully when... |
|---|---|---|---|
| u | Initial velocity | m/s | The object starts from rest, so u = 0. |
| v | Final velocity | m/s | The object comes to rest, so v = 0. |
| a | Acceleration | m/s² | Gravity may be positive or negative depending on chosen direction. |
| s | Displacement | m | It is not always the same as distance. |
| t | Time | s | Must be in seconds. |
Interactive Simulators
Teacher voice: These tools are here to slow the topic down. Change one value at a time and watch which part of the answer changes with it.
Motion Tool 1
Speed Calculator
Use this to connect the basic formula to real numbers.
Speed
6.00 m/s
Motion Tool 2
Constant Acceleration Displacement
This one helps with SUVAT by showing how the variables work together.
Displacement using s = ut + ½at2
36.00 m
6. Worked Examples — Step-by-step walkthroughs
Simple explanation: Move through the example line by line. The goal is not only to get the answer, but to see why that method was the sensible choice.
Exam StandardExample 1: A stone is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the maximum height reached.
Step 1: Extract data.
u = 20 m/s
v = 0 m/s (at maximum height, it stops momentarily)
a = -9.8 m/s² (gravity acts downwards)
s = ?
Step 2: Choose equation. We need s and do not have t, so use v² = u² + 2as.
Step 3: Solve.
0 = (20)² + 2(-9.8)(s)
0 = 400 - 19.6s
19.6s = 400
s = 20.4 m
Exam note: The answer is positive because height is measured upwards from the launch point.
Step 1: Extract data.
u = 20 m/s
v = 0 m/s (at maximum height, it stops momentarily)
a = -9.8 m/s² (gravity acts downwards)
s = ?
Step 2: Choose equation. We need s and do not have t, so use v² = u² + 2as.
Step 3: Solve.
0 = (20)² + 2(-9.8)(s)
0 = 400 - 19.6s
19.6s = 400
s = 20.4 m
Exam note: The answer is positive because height is measured upwards from the launch point.
H1 ChallengeExample 2: A car accelerates from 12 m/s to 28 m/s over a distance of 160 m. Find its acceleration.
Data: u = 12 m/s, v = 28 m/s, s = 160 m, a = ?
Equation: v² = u² + 2as
28² = 12² + 2(a)(160)
784 = 144 + 320a
640 = 320a
a = 2.0 m/s²
Data: u = 12 m/s, v = 28 m/s, s = 160 m, a = ?
Equation: v² = u² + 2as
28² = 12² + 2(a)(160)
784 = 144 + 320a
640 = 320a
a = 2.0 m/s²
7. Examiner Tips — How to maximise marks
Simple explanation: These are the small habits that protect marks under pressure, especially when the maths is easy but the wording is sneaky.
The SUVAT List: Always write down u, v, a, s, t before solving. It helps you choose the equation and may earn method marks.
Graph Questions: For velocity-time graphs, write these two lines before calculating:
Gradient = acceleration
Area under graph = displacement
Derivations: Start from known equations and show each substitution clearly. Do not jump straight to the final formula.
Graph Questions: For velocity-time graphs, write these two lines before calculating:
Gradient = acceleration
Area under graph = displacement
Derivations: Start from known equations and show each substitution clearly. Do not jump straight to the final formula.
8. Common Mistakes — Where students drop marks
Simple explanation: This section matters because students often understand the science but still lose marks through a sign error, unit slip, or definition mix-up.
1. Units: Forgetting to convert km/h to m/s. Divide by 3.6.
2. Missing the square: Writing s = ut + ½at instead of s = ut + ½at².
3. Sign error: Using +9.8 m/s² for an upward journey when upward has been chosen as positive.
4. Graph confusion: Saying the area under a velocity-time graph gives acceleration. It gives displacement.
2. Missing the square: Writing s = ut + ½at instead of s = ut + ½at².
3. Sign error: Using +9.8 m/s² for an upward journey when upward has been chosen as positive.
4. Graph confusion: Saying the area under a velocity-time graph gives acceleration. It gives displacement.
9. Examiner Traps — Hidden catches
Simple explanation: Pause on these trigger words. Examiners like to hide the method inside one or two ordinary-looking phrases.
“Starts from rest” → u = 0
“Comes to a halt / stop” → v = 0
“Dropped from a height” → u = 0 and a = 9.8 m/s² downward
“Thrown upwards” → acceleration is still downward
“Uniform acceleration” → SUVAT equations are allowed
“Comes to a halt / stop” → v = 0
“Dropped from a height” → u = 0 and a = 9.8 m/s² downward
“Thrown upwards” → acceleration is still downward
“Uniform acceleration” → SUVAT equations are allowed
10. Practical Skills — Lab experiments
Simple explanation: In a practical answer, imagine the examiner cannot see your setup. You need to state the apparatus, the measurement, the precaution, and the reason clearly.
Experiment: To measure constant acceleration using light gates.
Apparatus: trolley, sloped track, two light gates, timer/data logger, card of known length.
Method: Release the trolley from rest. Measure initial velocity at gate 1, final velocity at gate 2, and the time between gates.
Calculation: a = (v - u) / t
Precautions: Use a smooth track, measure the card length accurately, release without pushing, repeat and average.
Improvement: Use a data logger to reduce reaction-time error.
Apparatus: trolley, sloped track, two light gates, timer/data logger, card of known length.
Method: Release the trolley from rest. Measure initial velocity at gate 1, final velocity at gate 2, and the time between gates.
Calculation: a = (v - u) / t
Precautions: Use a smooth track, measure the card length accurately, release without pushing, repeat and average.
Improvement: Use a data logger to reduce reaction-time error.
11. Exam Questions — Structured practice with marks
Simple explanation: Treat these as rehearsal, not decoration. Before solving, say out loud what topic the question is really testing.
FoundationQ1. [6 marks] A car travelling at 15 m/s accelerates uniformly at 2 m/s² for 6 seconds. Calculate:
(a) its final velocity. [3]
(b) the distance travelled during this time. [3]
Exam StandardQ2. [12 marks] A train accelerates uniformly from rest to 15 m/s in 20 s. It then travels at 15 m/s for 40 s, before decelerating uniformly to rest in 10 s. Draw a velocity-time graph and use it to calculate the total distance travelled.
H1 ChallengeQ3. [6 marks] Explain why a ball thrown vertically upwards can have zero velocity but non-zero acceleration at its highest point.
12. MCQs — With “why the others are wrong”
Simple explanation: Use MCQs to test understanding, not guessing speed. Try to explain why each wrong option is wrong before you check the answer.
1. Which of the following is a scalar quantity?
A. Displacement B. Velocity C. Mass D. Force
| A | Wrong. Displacement includes direction, so it is a vector. |
| B | Wrong. Velocity includes direction, so it is a vector. |
| C | Correct. Mass has magnitude only. |
| D | Wrong. Force has magnitude and direction. |
2. The area under a velocity-time graph represents:
A. Acceleration B. Distance/displacement C. Time D. Momentum
| A | Wrong. Acceleration is the gradient of a velocity-time graph. |
| B | Correct. Area = velocity × time = displacement. |
| C | Wrong. Time is shown on the x-axis. |
| D | Wrong. Momentum requires mass × velocity. |
13. Past Paper Style Question
Simple explanation: This is where theory turns into exam judgement. Notice which facts are given, which facts are implied, and which formula or principle unlocks the question.
Note: This is written in the style of ILC Higher Level mechanics questions. When using real past-paper questions, add the exact year, section, and question number after verifying the source.
Question [12 marks]: A train accelerates uniformly from rest to 15 m/s in 20 seconds, travels at constant speed for 40 seconds, then decelerates uniformly to rest in 10 seconds. Draw a velocity-time graph and calculate the total distance travelled.
14. Critical Thinking — H1 Application
Simple explanation: This is the stretch section. The maths or reasoning is a bit sharper here, but the same core ideas still do the heavy lifting.
Derivation Challenge: Using v = u + at and s = average velocity × time, derive s = ut + ½at².
Concept question: Can an object have zero velocity and still be accelerating?
Answer: Yes. At the highest point of vertical motion, velocity is momentarily zero, but gravity is still acting downwards, so acceleration is not zero.
Answer: Yes. At the highest point of vertical motion, velocity is momentarily zero, but gravity is still acting downwards, so acceleration is not zero.
15. Last-Minute Revision Box
Simple explanation: Use this as your final scan before an exam: key facts, key equations, and the traps you are most likely to forget.
If the exam is tomorrow, remember:
✓ Write u, v, a, s, t before choosing an equation.
✓ “From rest” means u = 0.
✓ “Comes to rest” means v = 0.
✓ Gradient of a velocity-time graph = acceleration.
✓ Area under a velocity-time graph = displacement.
✓ For upward motion, gravity usually has a negative sign.
✓ Convert km/h to m/s by dividing by 3.6.
✓ Write u, v, a, s, t before choosing an equation.
✓ “From rest” means u = 0.
✓ “Comes to rest” means v = 0.
✓ Gradient of a velocity-time graph = acceleration.
✓ Area under a velocity-time graph = displacement.
✓ For upward motion, gravity usually has a negative sign.
✓ Convert km/h to m/s by dividing by 3.6.
16. Summary Sheet — One-page quick review
Simple explanation: Use this as your final scan before an exam: key facts, key equations, and the traps you are most likely to forget.
- Scalars: magnitude only, such as mass, time, speed, distance.
- Vectors: magnitude and direction, such as force, velocity, displacement, acceleration.
- SUVAT: Use only when acceleration is constant.
- Graphs: Velocity-time graph gradient = acceleration; area = displacement.
- Gravity: 9.8 m/s² downward; sign depends on chosen positive direction.
17. Checklist — Self-assessment
Simple explanation: Be honest with the checklist. It is much better to spot a weak area now than in the exam hall.
- I can define scalar, vector, displacement, velocity, and acceleration.
- I can choose the correct SUVAT equation.
- I can identify hidden data in exam wording.
- I can calculate gradient and area from a velocity-time graph.
- I can explain vertical motion using signs correctly.
- I can describe a light-gate experiment to measure acceleration.
18. Answers — Mark Schemes
Simple explanation: Study the mark scheme wording carefully. It shows what examiners reward, not just what students hope will be accepted.
Q1.
(a) v = u + at [1]
v = 15 + (2)(6) [1]
v = 27 m/s [1]
(b) s = ut + ½at² [1]
s = (15)(6) + ½(2)(6²) [1]
s = 90 + 36 = 126 m [1]
Q2.
Correct axes and labels [1]
Correct acceleration section from 0 to 20 s [1]
Correct constant speed section from 20 to 60 s [1]
Correct deceleration section from 60 to 70 s [1]
Area 1 = ½(20)(15) = 150 m [2]
Area 2 = (40)(15) = 600 m [2]
Area 3 = ½(10)(15) = 75 m [2]
Total distance = 825 m [2]
Q3.
At the highest point, the ball is momentarily stationary [1]
so velocity = 0 [1]
Gravity still acts on the ball [1]
gravity acts downward [1]
therefore acceleration = 9.8 m/s² downward [1]
so zero velocity does not mean zero acceleration [1]
(a) v = u + at [1]
v = 15 + (2)(6) [1]
v = 27 m/s [1]
(b) s = ut + ½at² [1]
s = (15)(6) + ½(2)(6²) [1]
s = 90 + 36 = 126 m [1]
Q2.
Correct axes and labels [1]
Correct acceleration section from 0 to 20 s [1]
Correct constant speed section from 20 to 60 s [1]
Correct deceleration section from 60 to 70 s [1]
Area 1 = ½(20)(15) = 150 m [2]
Area 2 = (40)(15) = 600 m [2]
Area 3 = ½(10)(15) = 75 m [2]
Total distance = 825 m [2]
Q3.
At the highest point, the ball is momentarily stationary [1]
so velocity = 0 [1]
Gravity still acts on the ball [1]
gravity acts downward [1]
therefore acceleration = 9.8 m/s² downward [1]
so zero velocity does not mean zero acceleration [1]