9.3 More About Velocity–Time Graphs

A free GCSE Physics lesson for AQA (8463) from OBA School: measuring velocity changes, reading the gradient of a velocity–time graph, and finding distance from the area under it. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Describe how light gates or a motion sensor can be used to measure velocity and acceleration
• Use a velocity–time graph to tell whether an object is accelerating, decelerating or moving at constant velocity
• Calculate acceleration from the gradient of a velocity–time graph
• (Higher tier) calculate distance travelled from the area under a velocity–time graph, including by counting squares
• Convert speeds between km/h and m/s

Key words

• Velocity–time graph: A graph of velocity (y-axis) against time (x-axis).
• Light gate: A sensor that detects when a beam of light is blocked; linked to a data logger it times very short events.
• Motion sensor: A sensor that measures the distance to a moving object many times per second, so a computer can work out and plot its velocity.
• Gradient: Change in velocity ÷ change in time on a velocity–time graph: the acceleration.
• Area under the graph: (Higher tier) The region between the line and the time axis. It represents the distance travelled.

Measuring velocity changes

Hand timing is too slow for fast changes, so electronic sensors are used:

• A motion sensor sends out pulses (e.g. of ultrasound) and times the echoes. The computer works out the velocity many times per second and plots a velocity–time graph straight away.
• A light gate times how long a card of known length blocks the beam. Velocity = length of card ÷ time the beam is blocked. Two light gates give two velocities, and the data logger records the time between them.

Investigation: does a steeper ramp give a bigger acceleration?

Independent variable: height of the raised end of the ramp (or the angle). Dependent variable: acceleration of the trolley.

Control variables: same trolley and card, released from rest at the same point, same distance between the light gates.

1. Measure the length of the card. 2. Release the trolley; record the time the card blocks each gate and the time between the gates.

3. Velocity at each gate = card length ÷ blocking time. 4. Acceleration = (v₂ − v₁) ÷ time between gates.

5. Repeat three times for each height and find the mean. Then add a block to raise the ramp and repeat.

Safety: put a foam stop or box at the bottom so the trolley cannot fall off the bench onto feet.

Worked example: using two light gates

A 5.0 cm card blocks gate 1 for 0.050 s and gate 2 for 0.025 s. The card takes 0.80 s to travel from gate 1 to gate 2.
v₁ = 0.050 m ÷ 0.050 s = 1.0 m/s; v₂ = 0.050 m ÷ 0.025 s = 2.0 m/s
a = (2.0 − 1.0) ÷ 0.80

Answer: 1.25 m/s² (≈ 1.3 m/s²)

The results show that a steeper ramp gives a greater acceleration, so the line on the trolley’s velocity–time graph is steeper.

Reading a velocity–time graph

• A line sloping upwards: the velocity is increasing, so the object is accelerating.
• A horizontal line: constant velocity, so the acceleration is zero.
• A line sloping downwards: the object is decelerating (negative acceleration).
• A line along the time axis: the object is stationary.
• A steeper line means a greater acceleration. A straight sloping line means the acceleration is constant.

Key fact: The gradient of a velocity–time graph represents acceleration. acceleration = change in velocity ÷ time taken

Worked example: gradients from the cyclist graph

0 to 4 s: a = (8 − 0) ÷ (4 − 0) = 2.0 m/s²
10 to 14 s: a = (0 − 8) ÷ (14 − 10) = −2.0 m/s²

Answer: accelerates at 2.0 m/s², then decelerates at 2.0 m/s²

Exam tip: Check the y-axis label before you start. A horizontal line on a velocity–time graph means constant velocity, but on a distance–time graph it means stopped.

Distance from the area under the graph (Higher tier)

For an object moving at a steady 12 m/s for 10 s, the distance is 12 × 10 = 120 m: that is the area of a rectangle on the graph. This works for any shape:

Key fact: The area under a velocity–time graph represents the distance travelled (the displacement).

Worked example: total distance for the van

triangle (0–6 s): ½ × 6 s × 12 m/s = 36 m
rectangle (6–16 s): 10 s × 12 m/s = 120 m
triangle (16–20 s): ½ × 4 s × 12 m/s = 24 m
total = 36 + 120 + 24

Answer: 180 m; average speed = 180 m ÷ 20 s = 9.0 m/s

Counting squares for a curved graph (Higher tier)

If the line is curved, count the squares under it. Count a square if at least half of it is under the line; ignore it if less than half is. Then multiply by the distance one square represents.

Worked example: motorboat distance in the first 10 s

one square = 1 m/s × 1 s = 1 m
about 71 squares are under the line

Answer: distance ≈ 71 × 1 m ≈ 71 m

Exam tip: Always work out what one square is worth from the axis scales – it is often not 1 m. For example, 2 m/s × 5 s = 10 m per square.

Speeds in km/h

Road speeds are often given in kilometres per hour. Convert them to m/s before using an equation or plotting a graph:

1 km/h = 1000 m ÷ 3600 s ≈ 0.278 m/s, so speed in m/s = speed in km/h ÷ 3.6

and speed in km/h = speed in m/s × 3.6

Worked example: a motorway speed

108 km/h = 108 000 m ÷ 3600 s = 108 ÷ 3.6

Answer: 30 m/s

Common mistakes

• Mixing up the graphs: on a velocity–time graph the gradient is acceleration, not speed.
• Saying a downward-sloping velocity–time line means the object is moving backwards – it is slowing down.
• Forgetting the ½ when finding the area of a triangle.
• Assuming one square is worth 1 m without checking the axis scales.
• Using km/h in calculations without dividing by 3.6.

Revision questions

Try these, then check your answers below each one.

Question 1. On a velocity–time graph, a straight line sloping down towards the time axis shows an object that is: (Choose from: moving backwards, decelerating uniformly, stationary, moving at constant velocity.)

Answer: decelerating uniformly

Working: The velocity is decreasing at a steady rate, so the object is decelerating uniformly. It is still moving forwards until the line reaches zero.

Question 2. (Higher tier) A cyclist accelerates uniformly from rest to 20 m/s in 8.0 s. Use the area under the velocity–time graph to find the distance travelled.

Answer: 80 m

Working: distance = ½ × 8.0 s × 20 m/s = 80 m.

Question 3. A car is travelling at 72 km/h. Convert this speed into m/s.

Answer: 20 m/s

Working: 72 ÷ 3.6 = 20 m/s (72 000 m ÷ 3600 s).

Related lessons

• Previous lesson: 9.2 Velocity and Acceleration
• Next lesson: 9.4 Analysing Motion Graphs
• In this chapter: 9.1 Speed and Distance–Time Graphs

Continue in the full course

This lesson is section 9.3 (AGCPH0903) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.