Graphs of Motion
Speed–time graph
A graph of speed against time; its gradient gives acceleration and the area beneath gives distance.
| English | Speed–time graph |
|---|---|
| Bahasa Melayu | Graf laju–masa |
| 中文 | 速率-时间图 |
How it is used
An object speeds up from 0 to 20 m/s in 4 s (a sloping line), then holds 20 m/s for 6 s. The gradient of the first part gives acceleration = 20 ÷ 4 = 5 m/s²; the area under the whole graph, ½ × 4 × 20 + 20 × 6 = 40 + 120 = 160 m, gives the distance.
Where it shows up in SPM
Central to the Graphs of Motion chapter (Form 5), mostly Paper 2. Typical tasks: find acceleration from a gradient, find distance or average speed from the area under the graph (often split into triangles and rectangles), and describe each stage of the motion.
Don't confuse it with
Open the chapter: Graphs of Motion →
Frequently asked questions
Why does the area under a speed–time graph give distance?
Because distance = speed × time, and area = height × width matches that: height is speed, width is time. For a constant 20 m/s over 6 s, the rectangle area is 20 × 6 = 120 m, exactly the distance travelled.
What does a line sloping downward mean?
It means the speed is decreasing, so the object is decelerating (slowing down). Its gradient is negative, for a drop from 20 m/s to 0 in 5 s, the gradient is (0 − 20) ÷ 5 = −4 m/s².
The object is still moving forward until the speed reaches zero.
How do I get average speed from a speed–time graph?
Find the total distance from the total area under the graph, then divide by the total time. If the area is 160 m over 10 s, average speed = 160 ÷ 10 = 16 m/s.
Do not simply average the top and bottom speeds unless the graph is one straight line.