A15: Gradients & Areas Under Graphs

Foundation Higher AQAEdexcelOCREduqasCCEA

Calculate or estimate gradients and areas under graphs; interpret in distance-time and velocity-time contexts

Exam Board:

📋 Key Concepts

Gradient of a graph: The rate of change. Calculated by finding the change in y divided by the change in x.

Area under a graph: Can represent important quantities in real contexts (e.g., distance for velocity-time graphs).

📝 Finding Gradients

Formula:

Gradient = ^{change in y}⁄_{change in x} = ^Δy⁄_Δx

Example 1

Find the gradient between points (1, 3) and (4, 9) on a graph.

Solution:

Gradient = ^{9 - 3}⁄_{4 - 1} = ⁶⁄₃ = 2

📝 Gradients on Distance-Time Graphs

Interpretation: On a distance-time graph, the gradient represents SPEED.

Example 2

A distance-time graph shows a car travelling from 0 km to 150 km in 3 hours. Calculate the speed.

Solution:

Speed = gradient = ^{150 - 0}⁄_{3 - 0} = 50 km/h

Example 3

At a specific point on a distance-time graph, the gradient is 25. What does this mean?

Solution:

The instantaneous speed is 25 (units per hour/minute depending on scale).

📝 Gradients on Velocity-Time Graphs

Interpretation: On a velocity-time graph, the gradient represents ACCELERATION.

Example 4

A velocity-time graph shows velocity increasing from 0 m/s to 20 m/s in 5 seconds. Calculate acceleration.

Solution:

Acceleration = gradient = ^{20 - 0}⁄_{5 - 0} = 4 m/s²

Example 5

On a velocity-time graph, what does a horizontal line mean?

Solution:

Gradient = 0, so acceleration = 0. The object is moving at constant velocity.

📝 Area Under a Graph

Method:

Divide the area into shapes (rectangles, triangles, trapeziums)
Calculate the area of each shape
Add them together

Example 6

Find the area under the line y = 2x from x = 0 to x = 4.

Solution:

This forms a triangle with base 4 and height 8.

Area = ½ × 4 × 8 = 16 square units

📝 Area Under Velocity-Time Graphs

Interpretation: On a velocity-time graph, the area under the graph represents DISTANCE TRAVELLED.

Example 7

A car accelerates uniformly from 0 to 15 m/s in 10 seconds, then travels at constant speed for 20 seconds. Find total distance.

Solution:

Part 1 (triangle): Area = ½ × 10 × 15 = 75 m

Part 2 (rectangle): Area = 20 × 15 = 300 m

Total distance: 75 + 300 = 375 m

Example 8

A velocity-time graph shows:

0-5s: velocity increases from 0 to 10 m/s
5-15s: constant velocity of 10 m/s
15-20s: velocity decreases from 10 m/s to 0

Calculate total distance.

Solution:

Area 1 = ½ × 5 × 10 = 25 m

Area 2 = 10 × 10 = 100 m

Area 3 = ½ × 5 × 10 = 25 m

Total: 25 + 100 + 25 = 150 m

📝 Estimating Area Under Curves

Method: Count squares or use the trapezium rule for curved graphs.

Example 9

Estimate the area under a curve by counting squares. If 15 full squares and 8 half-squares are under the curve, and each square represents 2 units²:

Solution:

Total squares = 15 + 4 = 19 squares

Area ≈ 19 × 2 = 38 units²

🧮 Worked Examples — Exam Style

Graphs encode rates and totals: gradient = rate of change; area under = total accumulated (distance from velocity, work done from force).

A velocity-time graph is a straight line from (0, 0) to (20 s, 10 m/s). Find the acceleration and the distance travelled.

Acceleration = gradient = 10/20 = 0.5 m/s²

Distance = area = ½ × 20 × 10 = 100 m

M1 for the gradient, A1 for 0.5; M1 for the triangle area, A1 for 100. Triangular areas: ½ × base × height.

Exam wording to expect: "Find the acceleration", "estimate the area under the graph", "What does the gradient represent?".

❓ Practice Questions

Q1: On a distance-time graph, what does the gradient represent?

Q2: On a velocity-time graph, what does the gradient represent?

Q3: On a velocity-time graph, what does the area under the graph represent?

Q4: A car travels 200 miles in 4 hours. Calculate the gradient of the distance-time graph.

Q5: A velocity-time graph shows a triangle with base 10s and height 30 m/s. Calculate the distance travelled.

Q6: A car accelerates from 10 m/s to 30 m/s in 8 seconds. Calculate the acceleration.

Q6: A velocity-time graph rises in a straight line from 0 to 12 m/s over 6 seconds. Find the distance travelled.

Q7: On a curved graph, how do you estimate the gradient at a point?

✅ Answers

Speed
Acceleration
Distance travelled
50 miles per hour
150 m
2.5 m/s²

📋 Model Answers — Full Working

Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.

Q1. Distance-time gradient = speed

Q2. Velocity-time gradient = acceleration

Q3. Area under velocity-time = distance travelled

Q4. 200 ÷ 4 = 50 mph

Q5. Triangle: ½ × 10 × 30 = 150 m

Q6. Acceleration = 12/6 = 2 m/s²; distance = ½ × 6 × 12 = 36 m

Q7. Draw a tangent at the point and find its gradient (two grid-friendly points on the tangent)

📌 Key Facts to Memorise

Gradient of distance-time = speed; of velocity-time = acceleration
Area under velocity-time = distance; count grid squares for estimates (area of one square × squares)
Straight sections: constant rate; curves: changing rate (tangent for instantaneous)
Area below the x-axis counts as NEGATIVE distance (moving back)
Units of area under: (m/s) × (s) = m

🗺️ In the Exam

Gradients and areas: 3–5 marks on Papers 2/3. Estimating curved areas by counting trapeziums/squares is the AO3 skill; interpreting gradients in context is AO2.

Links to: instantaneous rates (R15), real-life graphs (A14), compound measures (R1).

⚡ Targeted Drills

Cover the answers, work each one, then mark yourself. If any takes over a minute, revisit the matching section above.

Gradient between (1, 4) and (5, 16) → 12/4 = 3
Velocity-time: 0 → 8 m/s in 4 s → acceleration 2 m/s², distance ½ × 4 × 8 = 16 m
Area under: trapezium bases 6 and 10, width 3 → ½(6 + 10) × 3 = 24 m
Distance-time gradient 0 → stationary · velocity-time gradient 0 → constant speed
Curve gradient at a point → draw a tangent, rise ÷ run

🧭 Board Focus — OCR

Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.

Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.

🎯 Exam Tips

Distance-time: gradient = speed, area = no meaning
Velocity-time: gradient = acceleration, area = distance
Split area into triangles and rectangles for calculations
For curves, count squares or estimate
Always check units in your answer

🧠 Problem-Solving Strategies

Problem-Solving

Distance-time: gradient = speed. Velocity-time: gradient = acceleration, area = distance. Split curved areas into trapeziums or count squares. For instantaneous rate, draw a tangent and find its gradient.

Multi-Step Problem

A velocity-time graph shows: 0-4s: velocity increases from 0 to 12 m/s; 4-10s: constant velocity 12 m/s; 10-14s: velocity decreases from 12 to 0 m/s. (a) Find the acceleration for each section. (b) Find total distance. (c) Find average speed.

Solution:

(a) 0-4s: ¹²⁄₄ = 3 m/s². 4-10s: 0 m/s². 10-14s: ^-12⁄₄ = -3 m/s²

(b) Triangle: ½×4×12 = 24. Rectangle: 6×12 = 72. Triangle: ½×4×12 = 24. Total = 120 m

⚠️ Common Errors

Watch Out!

1. Wrong: Area under a distance-time graph gives distance Correct: Area under a distance-time graph has no physical meaning — use gradient for speed

2. Wrong: Gradient of a velocity-time graph gives speed Correct: Gradient gives acceleration (rate of change of velocity)

3. Wrong: The area under a curved graph can be found exactly using a triangle formula Correct: For curves, estimate using trapeziums or counting squares — it's an approximation

✍️ 6-Mark Exam Question

Extended Answer

6 marks: A car accelerates uniformly from rest to 25 m/s in 10 seconds, maintains 25 m/s for 30 seconds, then decelerates uniformly to rest in 5 seconds. (a) Calculate the acceleration. (b) Calculate the total distance. (c) If the speed limit is 60 km/h, is the car breaking the limit? Show working.

(a) Acceleration = ²⁵⁄₁₀ = 2.5 m/s²

(b) Area 1: ½ × 10 × 25 = 125 m. Area 2: 30 × 25 = 750 m. Area 3: ½ × 5 × 25 = 62.5 m. Total = 937.5 m

Mark scheme: (a) 1 mark. (b) 3 marks. (c) 2 marks for conversion and conclusion.

📊 AO3: Reason & Interpret

Reasoning and Interpretation

A runner's velocity-time graph shows a curve that gradually flattens.

(a) What does the flattening curve tell you about the runner's acceleration?

(b) How would you estimate the total distance from this graph?

(c) Explain why a straight-line graph would be unrealistic for this situation.

Answers: (a) The runner's acceleration is decreasing — they speed up more slowly. (b) Count squares under the curve, or divide into trapeziums and sum areas. (c) A straight line means constant acceleration — runners cannot keep accelerating at the same rate; they tire and approach a maximum speed.

📝 Exam Questions by Topic

AQA Past Papers - Topic Questions -- AQA
Edexcel Past Papers - Topic Questions -- Edexcel
OCR Past Papers - Topic Questions -- OCR
Eduqas Past Papers - Topic Questions -- Eduqas
CCEA Past Papers - Topic Questions -- CCEA

A15: Gradients & Areas Under Graphs

📋 Key Concepts

📝 Finding Gradients

📝 Gradients on Distance-Time Graphs

📝 Gradients on Velocity-Time Graphs

📝 Area Under a Graph

📝 Area Under Velocity-Time Graphs

📝 Estimating Area Under Curves

🧮 Worked Examples — Exam Style

❓ Practice Questions

✅ Answers

📋 Model Answers — Full Working

📌 Key Facts to Memorise

🗺️ In the Exam

⚡ Targeted Drills

🧭 Board Focus — OCR

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

⚠️ Common Errors

Watch Out!

✍️ 6-Mark Exam Question

Extended Answer

📊 AO3: Reason & Interpret

Reasoning and Interpretation

📝 Exam Questions by Topic

🎬 Video Resources

Ready to ace your GCSE Mathematics exams?

🧠 Flashcards (Spaced Repetition)

📝 Exam Questions by Topic

🎯 Target Tests (Auto-Graded)

🎓 Smart Lesson (Guided)

A15: Gradients & Areas Under Graphs

📋 Key Concepts

📝 Finding Gradients

📝 Gradients on Distance-Time Graphs

📝 Gradients on Velocity-Time Graphs

📝 Area Under a Graph

📝 Area Under Velocity-Time Graphs

📝 Estimating Area Under Curves

🧮 Worked Examples — Exam Style

❓ Practice Questions

✅ Answers

📋 Model Answers — Full Working

📌 Key Facts to Memorise

🗺️ In the Exam

⚡ Targeted Drills

🧭 Board Focus — OCR

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

⚠️ Common Errors

Watch Out!

✍️ 6-Mark Exam Question

Extended Answer

📊 AO3: Reason & Interpret

Reasoning and Interpretation

📝 Exam Questions by Topic

🎬 Video Resources

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Ready to ace your GCSE Mathematics exams?

🧠 Flashcards (Spaced Repetition)

📝 Exam Questions by Topic

🎯 Target Tests (Auto-Graded)

🎓 Smart Lesson (Guided)