Velocity–time graphs and travelled distance

FREE PHYSICS REVISION · UNIT 3 · B1.3.4

Pearson BTEC Level 3 AAQ Applied Science · Issue 7 (May 2026)

What you will learn

Describe v–t segments; displacement from signed area and distance from absolute area.

Specification mapping: B1.3.4, B1.3.5

Prerequisites: B1.3.3

Physical principles

A velocity–time graph directly shows velocity at each time. A horizontal line above zero represents constant positive velocity; below zero it represents constant negative velocity. A line crossing zero can represent a reversal. The gradient describes acceleration rather than the vertical height.

Signed area under the graph gives displacement because velocity multiplied by time has units metres. Areas below the time axis count negatively. Total distance is the sum of the magnitudes of all areas, splitting the interval at every zero crossing. When velocity remains non-negative, displacement and distance have the same numerical value; that restricted case explains the specification’s area shorthand.

Calculate rectangular, triangular and trapezium areas for straight segments. Curved segments may require counting squares or numerical approximation with stated assumptions. Do not add triangle areas without checking their signs. A velocity graph cannot show an instantaneous position jump: finite area over time gives a finite displacement.

Scientific diagram and interpretation

From 0 to 2 s velocity rises uniformly from 0 to 8 m s⁻¹; from 2 to 4 s it remains 8 m s⁻¹. Total area is 8 + 16 = 24 m.
From 0 to 2 s velocity rises uniformly from 0 to 8 m s⁻¹; from 2 to 4 s it remains 8 m s⁻¹. Total area is 8 + 16 = 24 m.

Equations and relationships

s = signed area under v–t; distance = sum of absolute areas

Worked examples and applications

Two stages

Velocity increases from 0 to 8.0 m s⁻¹ in 2.0 s, then stays at 8.0 m s⁻¹ for another 2.0 s. Distance = ½ × 2.0 × 8.0 + 2.0 × 8.0 = 24 m.

A reversal

Velocity is +3.0 m s⁻¹ for 4.0 s and −2.0 m s⁻¹ for 3.0 s. Displacement = +12 − 6 = +6.0 m. Total distance = 12 + 6 = 18 m.

Common misconceptions

Area below the axis contributes negative displacement but positive distance.

Examination technique

State whether you are finding displacement or total distance before combining areas.

For an explanation, connect the physical cause to its effect. For a calculation, state the model, rearrange before substitution, retain intermediate digits, and finish with units. For a data question, refer to the actual readings and distinguish a trend from proof of a cause.

Check your understanding

Five questions. Choose an option, then Check Answer. Read the feedback and use Try Again to reset. Use Tab between controls, arrow keys within radio groups, and Enter or Space on buttons. These local practice answers are not saved.

Question 1 of 5 · B1.3.4

Signed area under v–t gives…

Question 2 of 5 · B1.3.4

A horizontal v–t segment above zero means…

Question 3 of 5 · B1.3.4

A triangle of base 4 s and height 6 m s⁻¹ has area…

Question 4 of 5 · B1.3.4

Areas are +10 m and −4 m. Total distance is…

Question 5 of 5 · B1.3.4

Those same areas give displacement…