Two students look at the same speed–time graph. One calculates its gradient; the other calculates its area. They can both be doing correct mathematics while answering different physics questions.
The first decision is what you need to find. The gradient describes how quickly speed changes. The area under a speed–time graph gives distance travelled. Keeping these meanings separate is more useful than memorising an isolated triangle formula.
This guide uses an original, idealised journey. Check your Core or Extended requirements against the syllabus for your examination year on the Cambridge Physics 0625 page.
Read the axes before doing any arithmetic
Suppose the horizontal axis is time in seconds and the vertical axis is speed in metres per second. A point at (4, 8) means that the object is travelling at 8 m/s after 4 seconds. It does not mean the object has travelled 8 metres.
A horizontal line at 8 m/s means constant speed. A line along the time axis means zero speed. A downward-sloping speed line means the object is slowing down; it does not, by itself, mean the object is travelling backwards.
Find acceleration from the gradient
In our example, a trolley starts from rest and reaches 8 m/s uniformly in 4 seconds. Its acceleration during that interval is:
acceleration = change in velocity ÷ time = (8 − 0) ÷ 4 = 2 m/s².
Here the trolley moves in a fixed direction, so the increase in speed represents the change in velocity. The straight sloping segment tells us that the acceleration is constant over that interval.
Use two points on the segment you are investigating. Dividing the final speed by the total journey time would average over parts of the journey that may have different accelerations.
Find distance from the area
After accelerating, the trolley keeps travelling at 8 m/s for another 3 seconds. The graph now contains a triangle followed by a rectangle.
The triangular area is 1/2 × 4 × 8 = 16 metres. The rectangular area is 3 × 8 = 24 metres. Total distance is 16 + 24 = 40 metres.
The units explain why area works: seconds multiplied by metres per second gives metres. During the first four seconds, the average speed is 4 m/s because the speed increases uniformly from 0 to 8 m/s. Multiplying 4 by 4 gives the same 16-metre distance.
Do not confuse distance and displacement
A speed–time graph uses non-negative speed, so its area gives distance. A velocity–time graph can contain negative values. Its signed area gives displacement, while the sum of the magnitudes of the areas gives distance travelled.
If your question says velocity rather than speed, read the direction convention. Do not discard a minus sign before deciding what the question is asking.
Try a new journey
A cyclist moves in one direction, accelerating uniformly from 2 m/s to 10 m/s over 4 seconds. The cyclist then continues at 10 m/s for 6 seconds.
Find the acceleration in the first interval and the total distance in 10 seconds. Sketch the graph before calculating. The first region is a trapezium, not a triangle, because the initial speed is not zero.
Check your answer
Acceleration is (10 − 2) ÷ 4 = 2 m/s². Distance in the first interval is the average speed multiplied by time: ((2 + 10) ÷ 2) × 4 = 24 metres. The next interval contributes 10 × 6 = 60 metres. Total distance is 84 metres.
If you found 80 metres, check whether you used a triangle and lost the distance travelled at the initial speed. If you found 100 metres, you treated the cyclist as travelling at the final speed throughout.
Open the StudyForA Physics 0625 course and practise another motion example. Label each calculation “gradient” or “area” until the distinction becomes clear.