A14: Real-Life Graphs
Plot and interpret graphs including reciprocal and non-standard functions in real contexts
📋 Key Concepts
Real-life graphs: Graphs that represent real-world situations and data. Common types include distance-time graphs, conversion graphs, and filling/emptying containers.
📝 Distance-Time Graphs
Key features:
- Horizontal line: Stationary (not moving)
- Upward sloping line: Moving away from start
- Downward sloping line: Returning towards start
- Gradient: Speed (steeper = faster)
Example 1
A distance-time graph shows:
- 0-10 mins: line goes up from 0 km to 5 km
- 10-20 mins: horizontal line at 5 km
- 20-40 mins: line goes down from 5 km to 0 km
Interpretation:
- 0-10 mins: Travelling at 5⁄10 = 0.5 km/min
- 10-20 mins: Stationary (stopped)
- 20-40 mins: Returning at 5⁄20 = 0.25 km/min
Example 2
Find the speed from a distance-time graph where the distance changes from 0 m to 300 m in 20 seconds.
Solution:
Speed = distance⁄time = 300⁄20 = 15 m/s
📝 Speed-Time Graphs
Key features:
- Horizontal line: Constant speed
- Upward sloping line: Accelerating
- Downward sloping line: Decelerating
- Area under graph: Distance travelled
Example 3
A car accelerates from 0 to 20 m/s in 10 seconds, travels at constant speed for 20 seconds, then decelerates to 0 in 5 seconds. Calculate total distance.
Solution:
Area 1 (triangle): ½ × 10 × 20 = 100 m
Area 2 (rectangle): 20 × 20 = 400 m
Area 3 (triangle): ½ × 5 × 20 = 50 m
Total distance: 100 + 400 + 50 = 550 m
📝 Conversion Graphs
Conversion graphs: Used to convert between different units (e.g., currency, temperature, measurements).
Example 4
A conversion graph shows pounds (£) against euros (€). If the line passes through (0, 0) and (£10, €12):
a) Convert £25 to euros
b) Convert €36 to pounds
Solution:
Rate: €12/£10 = €1.20 per £1
a) £25 × 1.20 = €30
b) €36 ÷ 1.20 = £30
📝 Filling and Emptying Containers
Graphs showing filling containers:
- Height increases as container fills
- Shape of graph depends on container shape
- Wider parts fill more slowly (shallower gradient)
- Narrower parts fill more quickly (steeper gradient)
Example 5
Water is poured into a cylindrical container at a constant rate. Sketch the height-time graph.
Solution:
Since cylinder has constant width, height increases at constant rate.
Graph is a straight line (constant gradient).
Example 6
Water is poured into a cone (point down) at constant rate. Describe the graph.
Solution:
Bottom is narrow, so fills quickly at first (steep gradient).
Top is wide, so fills more slowly later (shallower gradient).
Graph gets less steep over time.
❓ Practice Questions
Q1: On a distance-time graph, what does a horizontal line mean?
Q2: A car travels 120 miles in 3 hours. What is its speed?
Q3: On a speed-time graph, what does the area under the graph represent?
Q4: A conversion graph shows 1 kg = 2.2 pounds. Convert 5 kg to pounds.
Q5: Water fills a vase that is narrow at the bottom and wide at the top. How does the gradient of the height-time graph change?
Q6: On a distance-time graph, which section shows fastest speed: a steep section or a shallow section?
✅ Answers
- Stationary (not moving)
- 40 mph
- Distance travelled
- 11 pounds
- Gradient starts steep (fills quickly at narrow bottom) and becomes shallower (fills slowly at wide top)
- Steep section (steeper = faster)
🎯 Exam Tips
- Distance-time: gradient = speed, flat = stationary
- Speed-time: gradient = acceleration, area = distance
- Conversion graphs: find the gradient and multiply or divide
- Container graphs: narrow = steep, wide = shallow
- Always read axes carefully and check units
🧠 Problem-Solving Strategies
Problem-Solving
On distance-time graphs: gradient = speed, flat = stationary, uphill = away, downhill = return. On speed-time graphs: gradient = acceleration, area under = distance. For containers: narrow = steep, wide = shallow gradient.
Multi-Step Problem
Tom leaves home at 9am, drives at 60 km/h for 2 hours, rests for 1 hour, then drives back at 40 km/h. (a) Draw a distance-time graph. (b) How far from home does he get? (c) What time does he get home?
Solution:
(a) Graph: line up from 0 to 120 km (9-11am), flat at 120 km (11am-12pm), line down to 0 (takes 3 hours at 40 km/h)
(b) Maximum distance = 60 × 2 = 120 km
(c) Return takes 120 ÷ 40 = 3 hours. Home at 3pm.
⚠️ Common Errors
Watch Out!
1. Wrong: On a distance-time graph, a downward line means the object is going downhill Correct: A downward line means returning towards the starting point
2. Wrong: The gradient of a speed-time graph gives distance Correct: The gradient gives acceleration; the AREA under gives distance
3. Wrong: A steeper line on a distance-time graph means a slower speed Correct: A steeper line means a FASTER speed (greater gradient = greater speed)
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A speed-time graph shows: 0-5s: acceleration from 0 to 20 m/s; 5-15s: constant speed 20 m/s; 15-20s: deceleration from 20 to 0 m/s. (a) Calculate the acceleration in the first 5 seconds. (b) Calculate total distance. (c) Calculate the average speed for the whole journey.
(a) Acceleration = 20-0⁄5 = 4 m/s²
(b) Area = triangle (½ × 5 × 20) + rectangle (10 × 20) + triangle (½ × 5 × 20) = 50 + 200 + 50 = 300 m
(c) Total time = 20s. Average speed = 300⁄20 = 15 m/s
Mark scheme: (a) 1 mark. (b) 3 marks (1 for each area + total). (c) 2 marks for method and answer.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two cars travel the same route. Car A: steady 50 mph. Car B: 70 mph for first half (time), then 30 mph for second half.
(a) Which car finishes first over 100 miles?
(b) Sketch both on the same distance-time graph.
(c) Explain why Car B doesn't arrive earlier despite travelling faster at first.
Answers: (a) Car A: time = 100/50 = 2 hours. Car B: average speed = (70+30)/2 = 50 mph, so time = 2 hours — same time! (b) Car A: straight line. Car B: steeper line then shallower line, both starting and ending at same points. (c) The slow second half exactly cancels the fast first half — the average speed equals Car A's speed.
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🎬 Video Resources
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