O-Level Physics Motion Graphs: Read the Axis Before Finding a Gradient

In a public O-Level Physics discussion, a student asked how to improve, including Paper 2. Motion graphs provide a focused place to practise: a calculation can be perfectly executed yet answer the wrong question if the vertical axis is misread. This guide develops one concrete skill rather than predicting topics or grade thresholds.

Name the graph before doing anything

Write down the quantity and unit on each axis. On a displacement-time graph, the gradient gives velocity. On a velocity-time graph, the gradient gives acceleration, while the signed area between the graph and the time axis gives displacement. The 2026 SEAB Physics 6091 syllabus includes interpreting these graphs and finding displacement from velocity-time areas for uniform velocity or acceleration.

A horizontal segment therefore means different things. Constant displacement means the object is at rest. Constant non-zero velocity means it continues moving. Before writing “stationary”, ask what quantity is constant. The OpenStax position-time explanation is useful background for connecting slope with motion.

Work through an original journey

Imagine a trolley travelling along a straight track, with one direction defined as positive. Its velocity increases uniformly from 0 to 4 m/s over the first 2 seconds. It then stays at 4 m/s from 2 to 5 seconds. Finally, velocity decreases uniformly from 4 to −2 m/s between 5 and 8 seconds. Draw these three straight segments on velocity-time axes.

The first acceleration is (4 − 0) ÷ 2 = 2 m/s². The first displacement is the triangular area, ½ × 2 × 4 = 4 m. The middle displacement is 3 × 4 = 12 m. During the last interval, acceleration is (−2 − 4) ÷ 3 = −2 m/s². Calculate changes using the endpoints of the relevant segment, rather than dividing by the total time since the journey began.

Split an area when direction changes

During the final segment, velocity reaches zero at 7 seconds: it takes 2 seconds to decrease from 4 m/s to zero at −2 m/s². From 5 to 7 seconds, the positive triangular area is ½ × 2 × 4 = 4 m. From 7 to 8 seconds, the negative triangular area is −½ × 1 × 2 = −1 m.

The total displacement is 4 + 12 + 4 − 1 = 19 m. The total distance travelled is 4 + 12 + 4 + 1 = 21 m. Distance adds how far the trolley travels in both directions; displacement retains direction. Taking the absolute value of the final displacement does not recover the distance travelled after a reversal.

Explain what a negative value means

Negative velocity means motion in the chosen negative direction. Negative acceleration means acceleration in that direction; it does not always mean slowing down. From 5 to 7 seconds, the positive velocity decreases towards zero, so speed decreases. From 7 to 8 seconds, velocity becomes more negative, so speed increases even though acceleration remains negative.

Describe motion interval by interval. The trolley is momentarily at rest at 7 seconds, but it is not stationary throughout the final interval. Likewise, crossing the time axis is a change in direction only if velocity changes sign. These distinctions help turn a graph description into an account of the actual journey.

Check your answer with units and a second representation

Velocity divided by time has units m/s². Velocity multiplied by time has units metres. If an “acceleration” answer has units metres, review whether you used area instead of gradient. Sketching displacement against time also helps: its slope should flatten to zero at the turning instant and become negative afterwards. See OpenStax on velocity and speed for the underlying distinction.

Select a graph question from the Secondary Pure Physics collection. Before calculating, state the axis quantity, interval and requested result. Check the paper's syllabus and year against your course. After marking, redo just the interval where your interpretation failed; a clear explanation of one negative segment is more useful than repeating several unexplained calculations.