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Jot down your answers, check against the worked out answers at the end of the page

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1.3 Drawing and interpreting distance - time graphs and calculations - problem solving

The gradient (slope) at any point on the graph gives you the speed at that point.

Since speed = distance ÷ time, then for graphical work

speed = (change in vertical y axis for distance) ÷ (change in horizontal x axis for time)

speed = ∆d / ∆t

The steeper the gradient the greater the speed

(1)

Distance - time graphs - acceleration - speeding up

Graph  curves upwards, showing increasing speed/velocity with time (acceleration), for each incremental time unit (e.g. minute or second) there is an ever increasing (larger) distance (∆d) covered in the same time (∆t).

speed = ∆d/∆t = the gradient of graph (∆d/∆t) is continuously increasing.

 

(2)

Distance - time graphs - motionless - stationary

Graph  is flat/horizontal, indicating zero speed/velocity, there is no increase in distance with time, object has stopped moving.

 

(3)

Distance - time graphs - deceleration - slowing down

Graph shows the curve is levelling off, steadily decreasing speed/velocity with time (deceleration), for each incremental time unit (e.g. minute or second) there is an ever decreasing (smaller) distance covered in the same time.

 

(4)

Distance - time graphs - constant speed

Graph is linear, showing constant speed/velocity, the distance covered in any equal time increment is the same.

 

The four graphs above illustrate of what you might see at any point on a distance-time graph, BUT, in reality, any distance-time graph will be more complicated than any of these specific graphs. So, you find a graph with all four types of gradient in just one distance-time graph - see the graph based questions below.

 

Examples of distance-time graph calculation questions

Moving on from the stylised 'types of graph' to full numerical graphs.

Graph (1) The graph lines for objects A and B are linear, indicating constant speed.

Using a convenient triangle to calculate the constant speeds of A and B

Speed of A = 500 / 10 = 50 m/s   and   speed of B = 300 / 25 =  12 m/s

Note the steeper the gradient, the greater the speed.

 

Graph (2) The graph line is horizontal, no change in distance, so object C is stationary at a distance of 300 m.

 

Graph (3) Lets call the object DE:

Its movement is first in a forward direction, represented by line D at a constant speed of 400 / 5 = 80 m/s.

Its second movement is in the opposite direction, line E, at a constant speed of 400 / 10 = 40 m/s.

(Strictly speaking, the line E represents a speed of -40 m/s, because of the reversed direction.)

Note that object DE has travelled a total distance of 800 m, BUT, its displacement is zero!

 

You get the speed from the gradient at any given point on the graph, which is easy if the graph at that point is linear (constant speed, like the three graphs above).

For example look at the linear portion between the two purple 'blobs' on the right-hand graph - that's a linear section and easy to measure the gradient (speed).

If the graph is curved you will need to draw a tangent to the curve at the point where you want to know the speed.

For example I've drawn a tangent at ~56 seconds and converted it to a purple triangle so you can measure the gradient - see Q2.1 (c) below.

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Questions based on distance-time graphs

Q2.1 The graph below shows part of a car journey.

Distance - time graph for a short car journey

Interpretation question using speed = distance / time

(a) Including calculating the average speed, describe and explain the motion of the car between:

(i) 0 and 20 seconds?

(ii) 20 and 40 seconds?

(iii) 40 to 82 seconds?

(iv) 82 to 90 seconds?

 

(b) What is the average speed of the car while it is moving?

(c) How can you find the specific speed of the car  at any point on the graph?

 

(d) What is the speed of car at 56 seconds?

 

Worked out ANSWERS to distance-time graph questions

 

Q2.2 The graph below summarises part of the journey of a train (somewhat delayed at some point!).

Interpretation question using speed = distance / time and giving your answers in km/hour and m/s

(a) What is the average speed between 1300 and 1400 hours?

(b) What is the average speed between 1530 and 1800 hours?

(c) How long was the train stopped for?

(d) What was the average speed for the whole journey?

Worked out ANSWERS to distance-time graph questions

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Velocity - Time graphs and acceleration are dealt with in detail in section 2

INDEX physics notes: Speed calculations and distance-time graphs

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Key points about speed and velocity - interpreting distance-time graphs

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE physics 9-1 level science examinations

A structured set of summary revision notes on interpreting distance-time graphs for moving objects, tailored to the GCSE/IGCSE physics specifications across major UK exam boards: WJEC, CCEA, CIE, AQA, Edexcel, and OCR.

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Interpreting Distance-Time Graphs

What Is a Distance-Time Graph?

• A distance-time graph shows how far an object travels over time.
• X-axis: Time (s, min, h)
• Y-axis: Distance (m, km)

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Key Graph Features

Graph Feature	Interpretation
Horizontal line	Object is stationary (no change in distance)
Straight diagonal line	Object moves at constant speed
Steeper diagonal line	Object moves faster
Shallower diagonal line	Object moves slower
Curved line (upwards)	Object is accelerating
Curved line (downwards)	Object is decelerating

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Calculating Speed from Graph

• Speed = Gradient of the line
• Gradient = Speed = Change in distance / Change in time
• Use a triangle on the graph to calculate gradient.

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Worked Example

A graph shows a straight line from 0 m to 100 m over 20 s.

• Gradient = 100 / 20 = 5 m/s
• The object moves at constant speed of 5 m/s

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Typical Exam Board Syllabus Content

Specification Focus
Motion – interpreting graphs, calculating speed from gradient
Motion – graphical analysis, speed from slope
Motion – includes uniform/non-uniform motion and graph interpretation
Forces – interpreting distance-time graphs and calculating speed
Describing Motion – gradient analysis, curved graphs
Motion – graphical interpretation, gradient and area under graphs

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Student Exam Tips

• Draw large gradient triangles to improve accuracy in speed calculations.
• Label axes and units carefully - examiners often test unit conversions.
• Look for changes in slope to identify acceleration or deceleration.
• Practice sketching graphs from motion descriptions and vice versa.
• Use real-life examples (e.g. walking, driving) to visualise motion types.

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Relative motion in real life

Let’s explore how concepts like relative motion and distance-time graphs show up in everyday life - they’re more than just exam topics; they help us understand and predict the world around us.

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Distance-Time Graphs in Real Life

Fitness Trackers

• When you go for a jog, your app plots distance vs. time. A straight line = steady pace; a curve = acceleration.
• You can spot breaks or slowdowns as flat segments.

Traffic Monitoring

• Traffic cameras or GPS systems track how vehicles move over time.
• A flat line might indicate a jam; a steep incline = fast movement.

Lift in a Building

• While riding an elevator, if you measure its position over time, the graph can show when it starts, stops, and changes speed.

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Why It Matters

Understanding these concepts helps with:

• Safer driving decisions
• Efficient travel planning
• Analysing performance in sports
• Interpreting motion in science and engineering
• Realistic simulations in games and animation

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Keywords, phrases and learning objectives for speed, motion and graphs

Be able to draw a distance-time graph from given data of a moving object.

Be able to interpret a distance - time graphs and associated calculations reading data from the graph.

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INDEX for physics notes on speed calculations and constructing and interpreting distance-time graphs

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INDEX physics notes: Speed calculations and distance-time graphs