Graphs of Motion

How to Read speed and distance from a motion graph

Use this to find speed, acceleration or distance from a distance–time or speed–time graph.

Before you start

  1. Reading coordinates off a graph
  2. Using the gradient formula (change in vertical) ÷ (change in horizontal)
  3. Areas of a triangle and a rectangle

When to use it

Use this to find speed, acceleration or distance from a distance–time or speed–time graph.

The steps

  1. Identify which graph you have: distance–time or speed–time.
  2. For a gradient, pick two clear points and compute (change in vertical) ÷ (change in horizontal).
  3. On a distance–time graph the gradient is speed; on a speed–time graph it is acceleration.
  4. For distance travelled on a speed–time graph, find the area under the line.
  5. Split the area into triangles and rectangles, then add them.

Worked example

A speed–time graph is a straight line from (0 s, 0 m/s) to (10 s, 30 m/s). Find the acceleration and the distance travelled.

  1. The graph plots speed against time, so it is a speed–time graph.
  2. Pick two clear points, (0, 0) and (10, 30). Gradient = (30 − 0) ÷ (10 − 0) = 3.
  3. On a speed–time graph the gradient is the acceleration, so acceleration = 3 m/s².
  4. For distance travelled, find the area under the line. The line and the axes form a triangle.
  5. The area is one triangle: ½ × base × height = ½ × 10 × 30 = 150.

A second example, with a twist

The journey now has two phases, so the area under the line splits into a triangle plus a rectangle. On a speed–time graph, a bus speeds up from 0 to 20 m/s in the first 5 s, then travels at a steady 20 m/s until 15 s.

Find the total distance travelled.

  1. The graph plots speed against time, so it is a speed–time graph.
  2. (The first phase gradient is (20 − 0) ÷ (5 − 0) = 4 m/s², the acceleration.)
  3. For total distance, find the area under the whole line.
  4. Split the area into a triangle for 0 to 5 s and a rectangle for 5 to 15 s.
  5. Triangle = ½ × 5 × 20 = 50; rectangle = 10 × 20 = 200; add them: 50 + 200 = 250.

Formulae you may need

Distance between two points (given in the exam)
Distance = √(x2−x1)2 + (y2−y1)2
Given in the exam
Average speed (given in the exam)
Average speed = Total distance / Total time
Given in the exam
Gradient (two points) (given in the exam)
m = (y2−y1)/(x2−x1)
Given in the exam
Gradient (intercepts) (given in the exam)
m = −(y-intercept)/(x-intercept)
Given in the exam

Formula pages

Practise this in a KBAT problem

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Frequently asked questions

How do I tell a distance–time graph from a speed–time graph?

Read the label on the vertical axis. If it shows distance (or displacement), it is a distance–time graph and the gradient gives speed.

If it shows speed (or velocity), it is a speed–time graph and the gradient gives acceleration. Always check the axis before you start.

Does area under the line mean anything on a distance–time graph?

No. Area under the line is only useful on a speed–time graph, where it gives the distance travelled.

On a distance–time graph you read distance straight off the vertical axis, and you use the gradient for speed. Do not find area on a distance–time graph.

What if the speed–time line slopes downward?

A downward slope means the object is slowing down, so the gradient is negative and represents deceleration. The area under the line is still the distance travelled and is always taken as positive.

Just work out the gradient the usual way; the minus sign shows the speed is dropping.

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