1.5 - Velocity-Time Graphs
Reading a speed-time graph
A speed-time graph shows how an object's speed changes as time passes. Speed has no direction, so the line never drops below the time axis.
- A flat section means a steady (constant) speed.
- An uphill section (going up to the right) means the speed is increasing.
- A downhill section (going down to the right) means the speed is decreasing.
- The steeper the line, the faster the speed is changing.
- A straight line means the speed is changing at a steady rate.
- A curve means the rate of change of speed is itself changing.

The gradient gives the rate of change of speed
- The gradient (steepness) of a speed-time graph equals the rate of change of speed - it's the change in speed divided by the time taken.
- For a straight section, divide the change in speed (the rise) by the time taken (the run).
The area gives the distance moved
- The area between the line and the time axis equals the distance moved in that time interval.
- Where the line is straight, split the area into rectangles and triangles, work out each area, and add them together.
Worked example - Rate of change of speed and distance
A speed-time graph shows a car's journey. The speed rises in a straight line from 0 to 24 m/s over the first 8 s, then stays constant at 24 m/s. Find the rate of change of speed over the first 8 s, and the distance the car moves in the first 14 s.

Step 1 - Find the change in speed and the time
Change in speed = 24 − 0 = 24 m/s
Time taken = 8 s
Step 2 - Divide the change in speed by the time
Rate of change of speed is the gradient over the first 8 s:
Rate of change of speed = 24 ÷ 8 = 3 m/s2
Step 3 - Find the distance as the area under the graph
The distance for the first 14 s is the area under the graph. Split it into a triangle (first 8 s) and a rectangle (next 6 s):
Triangle = ½ × base × height = ½ × 8 × 24 = 96 m
Rectangle = width × height = 6 × 24 = 144 m
Step 4 - Add the areas together
Distance = 96 + 144 = 240 m
The area under a velocity-time graph is the displacement
- The area between a velocity-time graph and the time axis equals the displacement in that time interval.
- Displacement is a vector: the straight-line distance from the start in a given direction, which isn't always the same as the total distance travelled.

Distance and displacement
- If the velocity stays positive (above the time axis), the displacement and the distance travelled have the same value.
- If part of the line drops below the time axis, the object has reversed direction. That area counts as a negative displacement, so it is:
- Subtracted when finding the overall displacement.
- Added when finding the total distance travelled.
Splitting the area into shapes
Where the acceleration is constant the line is straight, so the area underneath is made of simple shapes. Split it into rectangles and triangles, work out each area, and add them together:
- The area of a rectangle is width × height.
- The area of a triangle is ½ × base × height.
Counting squares for a curve
Where the line is curved the area isn't a simple shape. Estimate it by counting the squares under the line on the graph grid:
- Find the displacement one square represents by multiplying the width of a square (in seconds) by its height (in metres per second).
- Count the squares under the line, adding part-squares together to make whole ones.
- Multiply the number of squares by the displacement one square represents.
Worked example - Acceleration and distance from the graph
A velocity-time graph shows a car's journey. The velocity rises in a straight line from 0 to 24 m/s over the first 8 s, then stays constant at 24 m/s. Find the acceleration of the car over the first 8 s, and the distance it travels in the first 14 s.

Step 1 - Write down the change in velocity and the time
Change in velocity = 24 − 0 = 24 m/s
Time taken = 8 s
Step 2 - Divide the change in velocity by the time
Acceleration is the gradient over the first 8 s:
Acceleration = 24 ÷ 8 = 3 m/s2
Step 3 - Find the distance as the area under the graph
The distance for the first 14 s is the area under the graph. Split it into a triangle (first 8 s) and a rectangle (next 6 s):
Triangle = ½ × base × height = ½ × 8 × 24 = 96 m
Rectangle = width × height = 6 × 24 = 144 m
Step 4 - Add the areas together
Distance = 96 + 144 = 240 m