Rising line
Speed is increasing: the object is accelerating.
Learning Objectives
- interpret sections of a speed–time graph;
- identify constant speed, acceleration and deceleration;
- calculate acceleration from gradient;
- calculate distance from area under the graph;
- solve multi-stage travel problems.
Introduction
This lesson develops Speed–Time Graphs from meaning and patterns before moving to calculations. Look for what the quantities, shapes or symbols represent, then connect that idea to the method.
Key Notes
Read the definitions and key facts below carefully. Each rule is most useful when you can explain why it fits the situation, not simply recall it.
🤔 Think About This...
A car increases its speed, travels steadily and then brakes.
Which parts of a speed–time graph would slope, and which part would be horizontal?
🇿🇼 Zimbabwe example: The same graph skills can model mobile-data use, ZESA electricity units, rainfall in Mutare, kombi journeys and school attendance.
Reading a Speed–Time Graph
Horizontal line
Speed is constant.
Falling line
Speed is decreasing: the object is decelerating.
On the time axis
Speed is zero: the object is stationary.
Gradient Gives Acceleration
acceleration = change in speed ÷ change in time
A car speeds up from 4 m/s to 16 m/s in 6 seconds.
1
Change in speed = 16 − 4 = 12 m/s.
2
Change in time = 6 s.
3
Acceleration = 12 ÷ 6 = 2 m/s².
Area Gives Distance
The area between a speed–time graph and the time axis represents
distance travelled.
A vehicle travels at 12 m/s for 8 seconds.
1
The area is a rectangle.
2
Distance = 12 × 8 = 96 m.
Speed rises uniformly from 0 to 20 m/s in 5 seconds.
1
The area is a triangle.
2
Distance = ½ × 5 × 20 = 50 m.
Compound Areas
A journey may require splitting the area into rectangles, triangles or trapezia, calculating each area and then adding them.
Do not use the final speed multiplied by the total time when speed is
changing. Use the area under the graph.
✅ Quick Check
What does a horizontal section represent?
Speed changes from 5 to 17 m/s in 4 s. Find acceleration.
Find the distance at 15 m/s for 10 s.
🧠 Let’s Think Together
Before calculating, identify the information given, the result required and the mathematical relationship that connects them. Predict whether the answer should be larger, smaller or unchanged, then use that prediction to check the result.
Practice Questions
Foundation- Describe the motion shown by rising, horizontal and falling sections.
- Find the acceleration from 0 to 24 m/s in 8 seconds.
- Find the distance travelled at 18 m/s for 12 seconds.
- A car accelerates from rest to 20 m/s in 5 s, travels at 20 m/s for 10 s, then stops uniformly in 4 s. Find the total distance.
- Calculate the deceleration during the final section.
- A trapezium under a speed–time graph has parallel sides 8 m/s and 18 m/s and width 6 s. Find the distance.
✅ Check Your Work
Attempt the questions before opening the solutions.
Show answers and working
Foundation
- Rising: accelerating. Horizontal: constant speed. Falling: decelerating.
- a=(24−0)÷8=3 m/s².
- d=18×12=216 m.
Developing
- Distance=½×5×20 + 10×20 + ½×4×20 = 50+200+40=290 m.
- Deceleration=(0−20)÷4=−5 m/s², magnitude 5 m/s².
Challenge
- Area=½(8+18)×6=78 m.
Common Mistakes
- Choosing a rule before identifying what the question describes.
- Skipping working or changing notation part-way through a solution.
- Accepting an answer without checking its size, sign or units.
Exam Focus
Show the mathematical relationship you are using, keep notation consistent and give a clear final answer. Use a quick estimate, diagram or inverse operation to check that the result is reasonable.
Summary
- Gradient on a speed–time graph gives acceleration.
- Area under the graph gives distance.
- Horizontal lines show constant speed.
- Falling lines show deceleration.