🚚 Rectangle: constant velocity
A horizontal line gives a rectangle under it.
area = height × width
The height is the velocity. The width is the time for that part of the journey.
Movement and Position » Distance from a Velocity-Time Graph
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 1.9
- Why the area between a velocity-time graph and the time axis is the distance travelled
- How to find the area of rectangles, triangles and shapes made from both
- How to find the total distance and the average speed for a journey with several stages
- How to count squares to estimate the distance when the graph is a curve
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Unlock This CourseWhy the area is the distance
A velocity-time graph hides a second piece of information. The area between the line and the time axis is the distance travelled.
To see why, picture a delivery van moving at a constant velocity of 15 m/s for 6 s. You already know how to find how far it goes:
distance = velocity × time = 15 × 6 = 90 m
Now picture the graph. It is a horizontal line at 15 m/s, running from 0 s to 6 s. The space under the line, down to the time axis, is a rectangle. The rectangle is 15 m/s high and 6 s wide.
area of rectangle = height × width = 15 × 6 = 90
It is the same sum. The height is a velocity and the width is a time, so multiplying them gives velocity × time, which is a distance. That is also why the unit works out: m/s × s = m.
This rule works for every velocity-time graph, whatever the shape of the line. Your job is to split the area under the line into simple shapes and add them up.
The two shapes you need
📐 Triangle: steady acceleration from rest
A straight line sloping up from the origin gives a triangle under it.
area = ½ × base × height
The base is the time. The height is the velocity reached.
Worked example 1: a triangle
A skier starts from rest and accelerates steadily down a slope. She reaches 10 m/s after 6 s. The graph is a straight line from (0 s, 0 m/s) to (6 s, 10 m/s).
The area under the line is a triangle with a base of 6 s and a height of 10 m/s.
distance = area = ½ × base × height
distance = ½ × 6 × 10
distance = 30 m
Why the half? The skier was not moving at 10 m/s for the whole 6 s. She started at 0 m/s and built up steadily, so her average velocity was 5 m/s, half of the top value. And 5 × 6 = 30 m, the same answer.
A triangle also appears when an object slows steadily to rest. The line slopes down to the time axis, and the area under it is still ½ × base × height.
A rectangle and a triangle together
Often an object is already moving when it starts to accelerate. The line then starts part of the way up the velocity axis, and the shape under it is not a simple triangle. Split it into two shapes with a horizontal dotted line drawn across from the starting velocity.
Worked example 2: splitting the shape
A taxi joins a fast road. It accelerates steadily from 10 m/s to 30 m/s in 8 s. The graph is a straight line from (0 s, 10 m/s) to (8 s, 30 m/s).
Draw a horizontal dotted line at 10 m/s. Below it is a rectangle. Above it is a triangle.
rectangle = height × width = 10 × 8 = 80 m
The height of the triangle is the change in velocity: 30 − 10 = 20 m/s.
triangle = ½ × base × height = ½ × 8 × 20 = 80 m
distance = 80 + 80 = 160 m
A journey with several stages
Most exam graphs show a journey in stages. Draw vertical dotted lines down to the time axis wherever the line changes direction. Find the area of each shape, then add them.
Worked example 3: a delivery drone
A drone flies in a straight line between two buildings. Its velocity-time graph has three straight sections:
| Stage | Time | What the line does | Shape under the line |
|---|---|---|---|
| A | 0 s to 10 s | Slopes up from 0 m/s to 12 m/s | Triangle |
| B | 10 s to 40 s | Horizontal at 12 m/s | Rectangle |
| C | 40 s to 60 s | Slopes down from 12 m/s to 0 m/s | Triangle |
Stage A: ½ × 10 × 12 = 60 m
Stage B: the width is 40 − 10 = 30 s, so 12 × 30 = 360 m
Stage C: the base is 60 − 40 = 20 s, so ½ × 20 × 12 = 120 m
total distance = 60 + 360 + 120 = 540 m
Once you have the total distance, you can find the average speed for the whole journey. Use the equation from the lesson Speed, Distance and Time.
Worked example 4: average speed for the drone
average speed = total distance ÷ total time
average speed = 540 ÷ 60
average speed = 9 m/s
This makes sense. The drone's top velocity was 12 m/s, but it spent time speeding up and slowing down, so its average must be less than 12 m/s.
Counting squares for a curved graph
If the acceleration is not steady, the line is a curve. You cannot split the area under a curve into exact rectangles and triangles, so you count squares on the graph paper instead.
- Work out what one square is worth. Multiply the time across one square by the velocity up one square.
- Count the whole squares between the curve and the time axis.
- Deal with the part squares. Count a square that is half or more under the curve as one square. Ignore any that are less than half under it.
- Multiply the number of squares by the distance for one square.
Worked example 5: a sprinter
A sprinter leaves the blocks. Her velocity-time graph is a curve that rises quickly at first and then levels off. On the graph paper, each square is 1 s wide and 2 m/s high.
one square = 1 × 2 = 2 m
For the first 5 s there are 11 whole squares under the curve, and 5 more squares that are at least half under it.
number of squares = 11 + 5 = 16
distance = 16 × 2 = 32 m
This is an estimate, so write "about 32 m". In an exam, an answer close to the expected value gets the marks.
Common mistakes
- Forgetting the half. For a triangle, base × height gives double the distance. Always write ½ × base × height.
- Finding the gradient instead. The gradient gives the acceleration. For distance you need the area.
- Using a time reading as the width. Stage B of the drone's journey ends at 40 s, but it only lasts 30 s. Subtract the start time from the end time.
- Using the wrong height for the triangle. When a triangle sits on top of a rectangle, its height is the change in velocity, not the top reading on the axis.
- Averaging the velocities to find average speed. Always use total distance ÷ total time.
- Counting squares as 1 m each. Check the scales on both axes and work out what one square is worth first.
Exam-style question
A kayaker paddles in a straight line across a lake. The velocity-time graph for the journey has three straight sections. The line goes from (0 s, 0 m/s) to (8 s, 4 m/s), then to (28 s, 4 m/s), then to (40 s, 0 m/s).
(a) State what the area between a velocity-time graph and the time axis represents. [1 mark]
(b) Calculate the distance the kayaker travels in the first 8 s. [2 marks]
(c) Determine the total distance travelled in the 40 s. [3 marks]
(d) Calculate the average speed of the kayaker for the whole journey. [2 marks]
Model answer
(a) The distance travelled. (1)
(b) distance = area of triangle = ½ × 8 × 4 (1)
distance = 16 m (1)
(c) middle section = 4 × (28 − 8) = 4 × 20 = 80 m (1)
last section = ½ × (40 − 28) × 4 = ½ × 12 × 4 = 24 m (1)
total distance = 16 + 80 + 24 = 120 m (1)
(d) average speed = total distance ÷ total time = 120 ÷ 40 (1)
average speed = 3 m/s (1)
Exam tip
In part (c), sketch the graph from the coordinates and draw dotted lines down from (8 s, 4 m/s) and (28 s, 4 m/s). You can then see one rectangle between two triangles. Write the width of each shape as a subtraction, such as 28 − 8, so you never use 28 s as the width by mistake. Part (b) is already one of the three areas, so reuse your 16 m.