R14: Graphs and Rates of Change
Interpret the gradient of a straight line graph as a rate of change
📋 Key Concepts
Rate of Change: The gradient of a straight line graph represents the rate of change of y with respect to x. It tells us how much y changes for each unit change in x.
Gradient = (change in y) ÷ (change in x)
Gradient = (y₂ - y₁) / (x₂ - x₁)
Gradient = (y₂ - y₁) / (x₂ - x₁)
Interpreting Gradient
- Positive gradient: y increases as x increases
- Negative gradient: y decreases as x increases
- Steeper line = larger rate of change
📝 Calculating Gradient
Example 1
Find the gradient of the line passing through (2, 5) and (6, 17).
Solution:
Gradient = (y₂ - y₁) / (x₂ - x₁)
Gradient = (17 - 5) / (6 - 2)
Gradient = 12 / 4 = 3
Rate of change: 3 units of y per 1 unit of x
📝 Distance-Time Graphs
Interpretation: On a distance-time graph, the gradient represents speed.
- Gradient = Speed
- Steeper line = Higher speed
- Horizontal line = Stationary (speed = 0)
Example 2
A distance-time graph shows a line from (0, 0) to (2, 100). The distance is in km and time in hours. Calculate the speed.
Solution:
Gradient = 100 / 2 = 50 km/h
Speed = 50 km/h
Example 3
Find the speed between 1 and 4 hours from this data:
Time (h): 1, 4 | Distance (km): 40, 160
Solution:
Speed = (160 - 40) / (4 - 1) = 120/3 = 40 km/h
📝 Velocity-Time Graphs
Interpretation: On a velocity-time graph, the gradient represents acceleration.
- Gradient = Acceleration
- Positive gradient = Speeding up
- Negative gradient = Slowing down
- Horizontal line = Constant speed
Example 4
A velocity-time graph shows velocity changing from 10 m/s to 30 m/s over 8 seconds. Calculate the acceleration.
Solution:
Acceleration = (30 - 10) / 8 = 20/8 = 2.5 m/s²
📝 Real-World Contexts
| Graph Type | y-axis | x-axis | Gradient |
|---|---|---|---|
| Distance-Time | Distance | Time | Speed |
| Velocity-Time | Velocity | Time | Acceleration |
| Cost-Quantity | Cost | Quantity | Unit price |
| Height-Age | Height | Age | Growth rate |
Example 5
A graph shows the cost of hiring a van. The line passes through (0, 30) and (5, 105). Interpret the gradient.
Solution:
Gradient = (105 - 30) / (5 - 0) = 75/5 = £15
Interpretation: Cost increases by £15 per hour
The £30 intercept is the fixed charge.
🧮 Worked Examples — Exam Style
Rate = gradient. Distance-time → speed; cost graphs → price per unit; filling graphs → flow rate. The units tell you what the gradient means.
A delivery van's distance-time graph is horizontal from 10:00 to 10:45, then rises 30 km in 45 minutes. Describe the motion.
10:00–10:45: stationary (unloading). Then: speed = 30 ÷ 0.75 = 40 km/h
M1 for the 0.75 h conversion, A1 for 40. Real time-axes: 45 minutes = 0.75 hours, always.
A tank drains: volume falls from 60 L to 24 L in 6 minutes. Find the rate and describe the graph.
Rate = (24 − 60)/6 = −6 L/min — falling 6 litres per minute; a straight line sloping down.
M1 for the gradient calculation, A1 for −6 with units. Negative gradients mean decreasing — quote the minus.
Exam wording to expect: "Find the rate of change", "interpret the gradient", "which section is fastest?".
❓ Practice Questions
Q1: Find the gradient of the line through (3, 8) and (7, 24).
Q2: A distance-time graph shows points (0, 0) and (3, 180). Distance in miles, time in hours. Find the speed.
Q3: A velocity-time graph has gradient -4. What does this mean?
Q4: Cost of fuel: graph passes through (10, 14) and (25, 35). Units: litres and £. Find the cost per litre.
Q5: A car travels 150 miles in 3 hours at constant speed. What is the gradient of the distance-time graph?
Q6: A tap fills a tank at 12 litres per minute. Describe the volume-time graph.
Q7: A cooling graph falls steeply then levels off. Why?
✅ Answers
- 4
- 60 mph
- Deceleration of 4 m/s² (slowing down)
- £1.40 per litre
- 50 (miles per hour)
📋 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. (24 − 8)/(7 − 3) = 16/4 = 4
Q2. 180 ÷ 3 = 60 mph
Q3. Gradient −4 → deceleration of 4 m/s²
Q4. (35 − 14)/(25 − 10) = 21/15 = £1.40 per litre
Q5. 150 ÷ 3 = 50 (mph) — constant speed gives a constant gradient
Q6. Straight line rising from (0, 0) with constant gradient 12 (litres per minute)
Q7. Cooling is fastest when hottest: big temperature difference at first (steep), then the rate drops as it approaches room temperature (levels off)
📌 Key Facts to Memorise
- Gradient = rate of change — read the UNITS to name it
- Straight line = constant rate; curve = changing rate; horizontal = zero rate
- Negative gradient = decreasing (deceleration, draining, cooling)
- Steepness compares rates between sections
- Time axes: minutes → hours (÷ 60) before speeds in per-hour units
🗺️ In the Exam
Rates from graphs: 2–4 marks on Papers 2/3 — journeys, filling, cooling, costs. The interpretation sentence ("the gradient means…") is the examined AO2 line.
Links to: gradients/areas (A15), real-life graphs (A14), compound measures (R1).
🧭 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
- Gradient = (change in y) / (change in x)
- Distance-time: gradient = speed
- Velocity-time: gradient = acceleration
- Always state units in your answer
- Read axes carefully to identify what the gradient represents
🧠 Problem-Solving Strategies
Problem-Solving
The gradient of a distance-time graph gives the speed. The gradient of a speed-time graph gives the acceleration. The area under a speed-time graph gives the distance travelled. For curved graphs, use a tangent to find the gradient at a specific point (instantaneous rate of change).
Multi-Step Problem
A speed-time graph shows a car accelerating uniformly from 0 to 25 m/s in 10 seconds, then travelling at constant speed for 20 seconds, then decelerating uniformly to rest in 5 seconds. Find the total distance travelled.
Solution: Phase 1: area = ½ × 10 × 25 = 125 m. Phase 2: area = 25 × 20 = 500 m. Phase 3: area = ½ × 5 × 25 = 62.5 m. Total = 125 + 500 + 62.5 = 687.5 m.
⚠️ Common Errors
Watch Out!
1. Wrong: Reading the speed from a distance-time graph directly off the y-axis at a point Correct: Speed = gradient of the distance-time graph, not the value on the axis. A flat line means zero speed regardless of the distance value.
2. Wrong: Calculating the area under a distance-time graph to find distance Correct: Area under a speed-time graph gives distance. Area under a distance-time graph has no useful physical meaning.
3. Wrong: Forgetting that a negative gradient on a speed-time graph means deceleration, not going backwards Correct: A negative gradient on a speed-time graph means the object is slowing down. The object only reverses when speed goes negative.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A car accelerates uniformly from rest, reaching a speed of v m/s in 8 seconds. It then travels at this constant speed for 12 seconds before decelerating uniformly to rest in 4 seconds. The total distance travelled is 560 m. (a) Sketch the speed-time graph. (b) Find the value of v. (c) Calculate the acceleration and deceleration.
(a) Graph: triangle (0 to 8s), rectangle (8 to 20s), triangle (20 to 24s).
(b) Total area = ½ × 8 × v + 12 × v + ½ × 4 × v = 4v + 12v + 2v = 18v. 18v = 560, so v = 560/18 = 31.1 m/s (1 d.p.).
(c) Acceleration = 31.1/8 = 3.89 m/s². Deceleration = 31.1/4 = 7.78 m/s².
Mark scheme: M1 sketch with correct shape, A1 labels, M1 total area expression, A1 v = 31.1, M1 a = v/t, A1 both values
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two cyclists A and B race along the same 10 km route. Cyclist A starts from rest, accelerates to 20 km/h in 2 minutes, then cycles at 20 km/h for the rest. Cyclist B cycles at a steady 18 km/h from the start.
(a) Who completes the 10 km first?
(b) A student says "A's average speed must be more than 18 km/h since A reaches 20 km/h." Is this necessarily true?
(c) What assumption have we made about A's acceleration phase?
Answers: (a) A: distance during acceleration ≈ ½ × 20 × (2/60) = 0.333 km. Remaining: 9.667 km at 20 km/h = 0.483 hr ≈ 29 min. Total ≈ 31 min. B: 10/18 = 0.556 hr ≈ 33.3 min. A wins. (b) Not necessarily — A's average speed depends on the full journey including the slow acceleration phase. In this case A's average = 10/(31/60) ≈ 19.35 km/h, which IS above 18, but a longer acceleration phase could change this. (c) We assumed uniform acceleration during the first 2 minutes.
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