Displacement from a Velocity–Time Graph: Keep the Signed Areas

The signed area between a velocity–time graph and the time axis gives displacement. For distance travelled, add the magnitudes of the positive and negative contributions instead of cancelling them.

Read the axes before the shape

A height on a velocity–time graph is a velocity, not a position. A horizontal segment means constant velocity. Its area over a time interval is velocity multiplied by duration, with units (m/s) × s = m. A sloping segment can form a triangle or trapezium whose area is found geometrically.

The slope of this graph answers a different question: acceleration. Area and slope use the same drawing but different operations and different units. Writing the requested quantity beside the graph helps you avoid using an acceleration calculation for a displacement question.

Worked graph described in words

For the first 4 s, velocity increases linearly from 0 to +6 m/s. It then remains at +6 m/s for 3 s. Finally it is −2 m/s for 2 s. Treat the described segments as an idealised piecewise graph.

The first area is a triangle: (1/2) × 4 × 6 = +12 m. The second is a rectangle: 3 × 6 = +18 m. The final rectangle lies below the axis and contributes 2 × (−2) = −4 m. Displacement is 12 + 18 − 4 = +26 m.

Distance is 12 + 18 + 4 = 34 m. Both calculations use the same regions, but the final negative region has different treatment. The graph’s idealised abrupt changes are mathematical assumptions, not a claim that real objects change velocity instantaneously.

What if a line crosses zero?

If a straight segment crosses the time axis, split it at the zero-velocity time before calculating distance. For displacement, signed contributions can cancel. For distance, they must be made nonnegative separately. Taking the absolute value only after summing the signed areas loses information about the return travel.

Common mistake

Use the width of each interval, not its ending time measured from zero. A segment from 4 s to 7 s has width 3 s. Also distinguish a speed–time graph, whose heights cannot be negative, from the velocity graph used here. Check that your final unit is metres rather than metres per second.

Velocity graph mein area displacement hai; distance mein negative region bhi positive jodo.

Three practice questions with worked answers

1. Velocity is +3 m/s for 5 s. Find displacement.

   The rectangular area is 3 × 5 = +15 m. The positive sign follows the chosen direction.

2. Velocity increases linearly from 0 to 8 m/s in 6 s. Find displacement.

   The triangular area is (1/2) × 6 × 8 = 24 m. This uses the stated linear increase.

3. Velocity is +4 m/s for 2 s and −3 m/s for 2 s. Find displacement and distance.

   Signed areas give 8 − 6 = +2 m displacement. Their magnitudes give 8 + 6 = 14 m distance.

Related study guides

• Motion Class 9 Notes: Equations of Motion, Graphs and Solved Numericals
• Equations of Motion with Constant Acceleration

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Velocity vs. Time Graphs. No textbook or exam-board affiliation is claimed.