Distance-Time Graphs (Oxford AQA IGCSE Physics)

Revision Note

Leander Oates

Expertise

Physics

Distance-Time Graphs

  • distance-time graph is used to describe the motion of an object and calculate its speed

A Distance-Time graph of an object moving in a straight line

distance time graph for IGCSE & GCSE Physics revision notes
This distance-time graph shows an object moving in a straight line

Constant speed on a Distance-Time graph

  • A straight line represents constant speed

  • The slope of the straight line represents the magnitude of the speed:

    • A very steep slope means the object is moving at a fast speed

    • A shallow slope means the object is moving at a slower speed

    • A flat, horizontal line means the object is stationary (not moving)

The gradient of a Distance-Time graph

Distance -time graph, the steeper the line, the faster the speed. For IGCSE & GCSE Physics revision notes
This graph shows how the slope of a line is used to interpret the speed of moving objects. Both of these objects are moving at a constant speed because the lines are straight.

Calculating Speed from a Distance-Time Graph

  • The speed of a moving object can be calculated from the gradient of the line on a distance-time graph:

speed space equals space gradient space equals space fraction numerator increment y over denominator increment x end fraction

  • Where:

    • increment y is the change in y (distance) values

    • increment x is the change in x (time) values

Calculating the gradient of a straight line

Calculating the gradient of a distance-time graph for IGCSE & GCSE Physics revision notes
The speed of an object can be found by calculating the gradient of a distance-time graph
  • The gradient of a distance-time graph is equal to the object's speed

Worked Example

A distance-time graph is drawn below for part of a train journey. The train is travelling at a constant speed.

Distance time graph showing the train moving 8 kilometres over 6 minutes at a constant speed for IGCSE & GCSE Physics revision notes

Calculate the speed of the train.

Answer:

Step 1: Draw a large gradient triangle on the graph

  • The image below shows a large gradient triangle drawn with dashed lines

  • The increment y and increment x magnitudes are labelled, using the units as stated on each axes

Annotated distance-time graph showing that the rise is 8 kilometres and the run is 6 minutes

Step 2: Convert units for distance and time into standard units

  • The distance travelled = 8 km = 8000 m

  • The time taken = 6 mins = 360 s

Step 3: State that speed is equal to the gradient of a distance-time graph

  • The gradient of a distance-time graph is equal to the speed of a moving object:

speed space equals space gradient space equals space fraction numerator increment y over denominator increment x end fraction

Step 4: Substitute values to calculate the speed

speed space equals space gradient space equals space 8000 over 360

speed space equals space 22.2 space straight m divided by straight s

Worked Example

A man decides to take a stroll to the park. He finds a bench in a quiet spot and takes a seat, picking up where he left off reading his book on black holes. After some time reading, the man realises he lost track of time and runs home.

A distance-time graph for his trip is drawn below.

A chart showing three line segments. Segment A ascends steeply as the object travels 0 to 0.3 kilometres over 0 to 25 minutes. Segment B is horizontal as the object travels from 0.3 to 0.3 kilometres over the time interval 25 to 65 minutes.  Segment C ascends sharply again as the object travels 0.3 to 0.6 kilometres over 65 to 75 minutes

(a) How long does the man spend reading his book?

(b) There are three sections labelled on the graph, A, B and C. Which section represents the man running home?

(c) What is the total distance travelled by the man?

Answer:

Part (a)

  • The man spends 40 minutes reading his book

  • The flat section of the line (section B) represents an object which is stationary - so section B represents the man sitting on the bench reading

  • This section lasts for 40 minutes - as shown in the graph below

Annotated graph showing segment B (the man being stationary) for 65 minus 25 minutes equals 40 minutes

Part (b)

  • Section C represents the man running home

  • The slope of the line in section C is steeper than the slope in section A

  • This means the man was moving with a faster speed (running) in section C

Part (c)

  • The total distance travelled by the man is 0.6 km

  • The total distance travelled by an object is given by the final point on the line - in this case, the line ends at 0.6 km on the distance axis. This is shown in the image below:

Annotated graph showing the end segment C at 0.6 kilometres which is equal to the total distance travelled

Exam Tip

Use the entire line or line segment to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used - so remember to draw these directly on the graph itself!

Remember to check the units of variables measured on each axis. These may not always be in standard units - in these examples, the unit of distance was km and the unit of time was minutes. Double-check which units to use in your answer.

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Leander Oates

Author: Leander Oates

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.