Velocity-Time Graphs –

Velocity-time graphs are tested on both Foundation and Higher GCSE Maths papers and link algebra, geometry, and real-world problem solving. Understanding how to read and interpret them is essential for questions involving speed, acceleration, and distance.

What Is a Velocity-Time Graph?

A velocity-time graph shows how the velocity (speed in a given direction) of an object changes over time. The horizontal axis represents time and the vertical axis represents velocity. Different sections of the graph tell you different things about the motion.

The gradient of a velocity-time graph gives the acceleration. A positive gradient means the object is speeding up, a negative gradient (deceleration) means it is slowing down, and a gradient of zero (horizontal line) means the object is travelling at a constant speed.

The area under a velocity-time graph gives the distance travelled. For straight-line sections, this area can be calculated using the formulas for rectangles and triangles. On the Higher tier, you may need to estimate the area under a curved section using trapeziums.

Key Formulas

Step-by-Step Method

1. Identify the sections of the graph: rising (acceleration), flat (constant speed), and falling (deceleration).
2. To find acceleration, calculate the gradient of the relevant section: rise / run = change in velocity / change in time.
3. To find distance, calculate the area under the graph between the required time values.
4. Break the area into rectangles, triangles, or trapeziums and calculate each separately.
5. Add the individual areas together to get the total distance.

Worked Example 1 — Foundation Level

Question: A car accelerates from rest to 20 m/s in 10 seconds, then travels at constant speed for 15 seconds. Find the total distance.

Working:

Step 1 — Acceleration phase: this is a triangle with base 10 and height 20. Area = ½ × 10 × 20 = 100 m.

Step 2 — Constant speed phase: this is a rectangle with width 15 and height 20. Area = 15 × 20 = 300 m.

Step 3 — Total distance = 100 + 300 = 400 m.

Answer: 400 metres

Worked Example 2 — Higher Level

Question: A cyclist accelerates from 5 m/s to 15 m/s in 8 seconds. What is the acceleration?

Working:

Step 1 — Change in velocity = 15 - 5 = 10 m/s.

Step 2 — Change in time = 8 seconds.

Step 3 — Acceleration = 10 / 8 = 1.25 m/s².

Answer: 1.25 m/s²

Worked Example 3 — Exam Style

Question: A vehicle travels for 30 seconds. It accelerates uniformly from rest to 24 m/s in 10 seconds, travels at 24 m/s for 12 seconds, then decelerates uniformly to rest in 8 seconds. Find the total distance travelled and the deceleration. (5 marks)

Working:

Step 1 — Acceleration phase (triangle): area = ½ × 10 × 24 = 120 m.

Step 2 — Constant speed phase (rectangle): area = 12 × 24 = 288 m.

Step 3 — Deceleration phase (triangle): area = ½ × 8 × 24 = 96 m.

Step 4 — Total distance = 120 + 288 + 96 = 504 m.

Step 5 — Deceleration = change in velocity / time = (0 - 24) / 8 = -3 m/s². The deceleration is 3 m/s².

Answer: Total distance = 504 m. Deceleration = 3 m/s².

Common Mistakes

• Confusing velocity-time graphs with distance-time graphs. On a velocity-time graph, a horizontal line means constant speed, not being stationary. On a distance-time graph, a horizontal line means stationary.
• Forgetting to halve the area for triangular sections. A triangular section is ½ × base × height, not base × height.
• Using the wrong units. Acceleration is measured in m/s² (metres per second squared), not m/s.

Exam Tips

• Label each section of the graph (acceleration, constant speed, deceleration) before calculating.
• Use a trapezium when the graph goes from one velocity to a different velocity over a time period — area = ½ × (v₁ + v₂) × t.
• If the question asks for deceleration, give a positive value. Deceleration is the magnitude of negative acceleration.

Practice Questions

Q1 (Foundation): An object accelerates from rest to 30 m/s in 6 seconds. Find the acceleration.

Q2 (Foundation): A train travels at 40 m/s for 20 seconds. Find the distance covered.

Q3 (Higher): A car accelerates from 10 m/s to 25 m/s in 6 seconds, then decelerates to rest in 10 seconds. Find the total distance.

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Related Topics

• Speed, Distance and Time
• Real-Life Graphs
• Compound Measures: Density and Pressure

Summary

• The gradient of a velocity-time graph gives the acceleration (m/s²).
• The area under a velocity-time graph gives the distance travelled (m).
• A horizontal line means constant speed; a rising line means acceleration; a falling line means deceleration.
• Break the area under the graph into triangles, rectangles, and trapeziums to calculate total distance.
• Always include correct units in your answer.