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How do motion graphs let us read off speed, acceleration and distance?

Interpreting distance-time and velocity-time graphs, finding speed from the gradient of a distance-time graph, finding acceleration from the gradient of a velocity-time graph, and finding distance from the area under a velocity-time graph.

A CCEA GCSE Double Award Science (Physics Unit P1) answer on reading motion graphs: finding speed from the gradient of a distance-time graph, acceleration from the gradient of a velocity-time graph, and distance travelled from the area under a velocity-time graph.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Distance-time graphs
  3. Velocity-time graphs
  4. Finding distance from the area
  5. Reading curved lines
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA Double Award wants you to read distance-time and velocity-time graphs: to find speed from the gradient of a distance-time graph, acceleration from the gradient of a velocity-time graph, and distance from the area under a velocity-time graph. Graph questions reward careful axis labels and unit work, so read what each axis shows before you start.

Distance-time graphs

A distance-time graph plots distance (vertical axis) against time (horizontal axis).

To find the speed, work out the gradient: divide the change in distance (rise) by the change in time (run).

Velocity-time graphs

A velocity-time graph plots velocity (vertical axis) against time (horizontal axis). It carries two pieces of information at once.

Finding distance from the area

To find the distance travelled, work out the area between the line and the time axis. Break the shape into simple parts: triangles (area =12×base×height= \tfrac{1}{2} \times \text{base} \times \text{height}) and rectangles (area =base×height= \text{base} \times \text{height}), then add them.

Reading curved lines

Not every motion graph is made of straight lines. On a distance-time graph, a curve that gets steeper means the object is speeding up (the speed, which is the gradient, is increasing), while a curve that levels off means it is slowing down. To find the speed at one instant on a curve, draw a tangent to the curve at that point and find the gradient of the tangent.

On a velocity-time graph, a curve means the acceleration is changing. A falling object reaching terminal velocity gives a line that is steep at first (large acceleration) and then curves to horizontal (zero acceleration) as air resistance grows to balance the weight.

Examples in context

Example 1. A bus route
A distance-time graph for a bus has sloping sections (driving) separated by flat sections (waiting at stops). The steepest slope shows where the bus was fastest, and the flat sections show where it was stationary at stops.
Example 2. A dropped ball
On a velocity-time graph a falling ball gives a straight upward line whose gradient is the acceleration of free fall, until air resistance starts to curve it towards terminal velocity.
Example 3. Comparing two cars
If two cars are shown on the same velocity-time graph, the one whose line is steeper has the greater acceleration, and the one whose line encloses the larger area has travelled the greater distance. Reading both the gradient and the area lets you compare motion at a glance, which is exactly what longer CCEA graph questions ask you to do.

Try this

Q1. What does the gradient of a distance-time graph tell you? [1 mark]

  • Cue. The speed.

Q2. What does the area under a velocity-time graph represent? [1 mark]

  • Cue. The distance travelled.

Q3. An object moves at a steady 15 m/s15\ \text{m/s} for 8.0 s8.0\ \text{s}. Find the distance using the area under the graph. [2 marks]

  • Cue. Rectangle: 15×8.0=120 m15 \times 8.0 = 120\ \text{m}.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA-style3 marksA velocity-time graph shows an object accelerating uniformly from 0 to 12 m/s in 4.0 s, then travelling at 12 m/s for 6.0 s. Calculate the total distance travelled.
Show worked answer →

The distance is the area under the velocity-time graph.

The first phase is a triangle: area =12×4.0×12=24 m= \tfrac{1}{2} \times 4.0 \times 12 = 24\ \text{m}.

The second phase is a rectangle: area =12×6.0=72 m= 12 \times 6.0 = 72\ \text{m}.

Total distance =24+72=96 m= 24 + 72 = 96\ \text{m}.

Markers reward area as triangle plus rectangle, the two areas (24 m and 72 m), and the total 96 m.

CCEA-style3 marksDescribe how you would find the acceleration of an object from its velocity-time graph, and state what a horizontal line on such a graph means.
Show worked answer →

Find the gradient of the line: divide the change in velocity (rise) by the time taken (run).

A steeper line means a greater acceleration. A horizontal line means the velocity is constant, so the acceleration is zero.

Markers reward gradient as change in velocity over time, and a horizontal line meaning constant velocity (zero acceleration).

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