Distance-Time and Velocity-Time Graphs
Motion can be represented graphically using distance-time graphs, where gradient gives speed, and velocity-time graphs, where gradient gives acceleration and the area under the graph gives distance travelled. It covers reading and calculating speed and acceleration from graphs, distinguishing uniform and non-uniform motion, and using the equations speed = distance / time and acceleration = change in velocity / time. It also covers the equation of motion v^2 = u^2 + 2as, part of the extended IGCSE course beyond core GCSE, together with determining distance travelled from a velocity-time graph by calculating the area of its sections, including graphs with more than one section of motion.
Before you start
No specific prerequisites - this is a good place to start.
Method
- For a distance-time graph, identify the gradient of the line: a steeper positive gradient means faster speed, a horizontal (flat) section means the object is stationary, and a negative gradient means the object is moving back toward the start.
- Calculate speed from a distance-time graph using speed = gradient = change in distance / change in time, reading two convenient points on the line.
- For a velocity-time graph, identify the gradient: a positive gradient means acceleration, a negative gradient means deceleration, and a horizontal section means constant velocity.
- Calculate acceleration from a velocity-time graph using acceleration = gradient = change in velocity / change in time, or acceleration = (v - u) / t.
- Calculate distance travelled from a velocity-time graph by finding the area between the line and the time axis, splitting the area into rectangles and triangles if the graph has more than one section.
- For problems giving initial velocity, acceleration and distance but not time, use v^2 = u^2 + 2as, rearranging to find the unknown quantity.
- Check the sign of your answer: deceleration is often given as a negative acceleration, and distance travelled cannot be negative.
Worked example
A cyclist's velocity-time graph shows the cyclist accelerating uniformly from rest to 8.0 m/s in 4.0 s, then travelling at a constant 8.0 m/s for a further 6.0 s. Calculate the total distance travelled by the cyclist.
- Split the motion into two sections: the acceleration phase (0 to 4.0 s) forms a triangle under the graph, and the constant-speed phase (4.0 to 10 s) forms a rectangle.
- Calculate the area of the triangle (distance in the acceleration phase): area = 0.5 x base x height = 0.5 x 4.0 x 8.0.
- 0.5 x 4.0 = 2.0, then 2.0 x 8.0 = 16 m.
- Calculate the area of the rectangle (distance in the constant-speed phase): area = base x height = 6.0 x 8.0 = 48 m.
- Add the two areas together: total distance = 16 + 48.
- State the final answer with its unit: total distance = 64 m.
Practice questions
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Q1A distance-time graph shows a straight line from the origin to the point (5.0 s, 20 m). Calculate the speed.Show answer
Answer: 4.0 m/s (20/5.0).
Q2State what a horizontal section of a distance-time graph represents.Show answer
Answer: The object is stationary (not moving).
Q3A velocity-time graph shows velocity decreasing from 12 m/s to 0 m/s in 3.0 s. Calculate the deceleration.Show answer
Answer: 4.0 m/s^2 ((12-0)/3.0), i.e. an acceleration of -4.0 m/s^2.
Q4State what the area under a velocity-time graph represents.Show answer
Answer: The distance travelled (or displacement).
Q5A car accelerates uniformly from 5.0 m/s at 2.0 m/s^2 over a distance of 24 m. Use v^2 = u^2 + 2as to calculate its final velocity.Show answer
Answer: 11 m/s (v^2 = 5.0^2 + 2 x 2.0 x 24 = 25 + 96 = 121, so v = the square root of 121 = 11).
Q6A velocity-time graph shows constant velocity of 6.0 m/s for 10 s. Calculate the distance travelled.Show answer
Answer: 60 m (6.0 x 10, area of rectangle).
Q7Two runners start together. Runner A's distance-time graph has a steeper gradient than Runner B's. State what this means about their speeds.Show answer
Answer: Runner A is travelling faster than Runner B, since a steeper gradient on a distance-time graph means a greater speed.
Exam-style questions
Written in the style of a IGCSE Science exam paper, with a full mark scheme.
A ball is thrown so that it decelerates uniformly from 18 m/s to 6.0 m/s while travelling in a straight line, covering a distance of 24 m. Use v^2 = u^2 + 2as to calculate the deceleration of the ball.
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The velocity-time graph for a train journey shows the train accelerating uniformly from rest to 30 m/s in 60 s, travelling at a constant 30 m/s for 120 s, then decelerating uniformly to rest in 40 s. (a) Describe, using the shape of the graph, how you can tell when the train is accelerating, travelling at constant velocity, and decelerating. (3 marks) (b) Calculate the total distance travelled by the train during the whole journey. (3 marks)
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Free printable worksheet
Want more practice on paper? Download the distance-time and velocity-time graphs worksheet pack - 6 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
This topic is chapter 22 of IGCSE Science Workbook, the whole course as one free printable PDF.
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