AQA GCSE Combined Science (8464), Higher tier · Physics › Forces › Forces and motion › Describing motion along a line
Practise The distance–time relationship. 15 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
A distance–time graph shows how far an object is from a point as time goes on. You need to draw them from measurements, describe motion from their shape, find speed from the gradient and, on the Higher tier, find the speed of an accelerating object by drawing a tangent.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Describe motion from a distance–time graphA horizontal line means stationary; a straight sloping line means constant speed.
4
Plot a distance–time graph from dataTime on the x-axis, distance on the y-axis, labelled axes with units and sensible scales.
5
Calculate speed from the gradientSpeed = change in distance ÷ change in time for a straight section.
6
Calculate average speed from a graphRead the total distance and total time, then divide.
7
Recognise acceleration and deceleration on the graphA curve getting steeper shows acceleration; a curve getting less steep shows deceleration.
8
Find speed at an instant using a tangentDraw a tangent to the curve at that time and calculate its gradient.
Notes
Reading a distance–time graph
Distance is on the y-axis and time is on the x-axis.
A horizontal line means the object is stationary: its distance is not changing.
A straight sloping line means the object is moving at a constant speed. The steeper the line, the greater the speed.
If the graph shows distance from a starting point, a line sloping back down towards the time axis means the object is moving back towards the start.
Speed from the gradient
Speed = gradient = change in distance ÷ change in time.
Use two points far apart on a straight section and draw a large triangle.
Example: from (2 s, 10 m) to (6 s, 34 m), speed = (34 − 10) ÷ (6 − 2) = 24 ÷ 4 = 6 m/s.
Average speed for a whole journey = total distance ÷ total time, read from the start and end of the graph.
Curved lines: changing speed
A curve that gets steeper shows that the object is accelerating.
A curve that gets less steep shows that the object is decelerating.
Speed at an instant grade 8+
If an object is accelerating, you can find its speed at a particular time by drawing a tangent to the curve at that time.
A tangent is a straight line that just touches the curve at that point, with the same slope as the curve there.
Draw the tangent long, then calculate its gradient using a large triangle. This gradient is the speed at that instant.
Drawing the graph
Put time on the x-axis and distance on the y-axis. Label both axes with units.
Choose scales that use more than half the grid, plot the points accurately, then draw a line of best fit (a straight line or a smooth curve).
Cheatsheet
Distance–time graph: gradient = speed
Horizontal line = stationary
Straight sloping line = constant speed
Steeper line = greater speed
Curve getting steeper = accelerating; getting less steep = decelerating
Average speed = total distance ÷ total time
Speed at an instant = gradient of the tangent at that time grade 8+
How to answer each type of question
Describe the motion shown by a graph
2 to 3 marks3
Describe each section in turn.
Use the words stationary, constant speed, accelerating or decelerating.
Compare sections: 'faster than', 'slower than'.
Example. A distance–time graph for a jogger has three sections. A: a steep straight line. B: a horizontal line. C: a straight line that is less steep than A. Describe the motion of the jogger in each section.
Show the model answer
A: constant speed (1) B: stationary (1) C: constant speed, slower than in A (1)
Calculate speed from the gradient
2 marks5
Read two points on the straight line, far apart.
Speed = change in distance ÷ change in time.
Give the unit (usually m/s).
Example. The distance–time graph for a cyclist is a straight line from 0 m at 0 s to 450 m at 90 s. Calculate the speed of the cyclist.
Show the model answer
speed = 450 ÷ 90 (1) speed = 5.0 m/s (1)
Calculate an average speed from a graph
3 marks6
Read the total distance and the total time from the graph.
Convert the units if needed.
Average speed = total distance ÷ total time.
Example. A distance–time graph for a walk shows that the walker is 1800 m from home after 25 minutes, including a 5-minute rest. Calculate the average speed of the walker over the 25 minutes, in m/s.
Show the model answer
t = 25 × 60 = 1500 s (1) average speed = 1800 ÷ 1500 (1) = 1.2 m/s (1)
Draw a tangent to find the speed at an instant
3 to 4 marks8
Draw a straight line that just touches the curve at the time given.
Choose two points far apart on the tangent.
Gradient of the tangent = speed at that time.
Example. A car accelerates from rest and its distance–time graph is a curve. A student draws a tangent to the curve at t = 4.0 s. The tangent passes through (2.0 s, 0 m) and (6.0 s, 48 m). Determine the speed of the car at t = 4.0 s.
Show the model answer
tangent drawn touching the curve at 4.0 s (1) gradient = (48 − 0) ÷ (6.0 − 2.0) (1) speed = 12 m/s (1)
Shortcuts and memory tricks
Flat = stopped. Steep = fast.
Gradient = rise ÷ run = distance ÷ time = speed.
Use a big triangle for gradients: it reduces reading errors.
Tangent tip: tilt your ruler until the gaps between the ruler and the curve look the same on both sides of the point.
Where marks are lost
Mixing up distance–time and velocity–time graphs: on a distance–time graph a horizontal line means stationary.
On a curve, dividing the distance by the time at one point (which gives an average speed) instead of drawing a tangent.
Using a time axis in minutes but giving the gradient in m/s without converting.
Drawing a 'tangent' that cuts through the curve instead of just touching it.
Exam technique
Draw your gradient triangle on the graph and write down the values you read.
When describing motion, use the key words stationary, constant speed, accelerating and decelerating, and compare sections.
Check the axis units: time may be in minutes or hours, and distance in km.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Suggest one advantage of this method over a student using a stopwatch and metre rule to record distance manually at intervals.
It records far more, and more precise, data points than manual timing could.
A student walks at a constant speed and covers 350 m in 250 s. Calculate their speed.
1.4 m/s
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy3 marks
(a) On a distance–time graph, what does a steeper gradient represent? Tick (✓) one box.[1]
A greater distance
A greater speed
A longer time
A smaller speed
(b) A student walks at a constant speed and covers 350 m in 250 s. Calculate their speed.[2]
Show the answer and mark scheme
(a)Answer: A greater speed
(b)Answer: 1.4 m/s
v = 350 ÷ 250
1.4 (m/s)
Question 2Medium4 marks
A student wants to obtain a distance–time graph for a toy car as it moves across the floor.
(a) Describe how the student could use an ultrasonic position sensor connected to a data logger to obtain a distance–time graph for the car.[3]
(b) Suggest one advantage of this method over a student using a stopwatch and metre rule to record distance manually at intervals.[1]
Show the answer and mark scheme
(a)Answer: Point a position sensor at the car; the data logger records distance at frequent regular intervals and plots it against time automatically.
set up the position sensor so that it faces the car and can detect the car throughout its motion
the sensor (and data logger) automatically records the car’s distance from the sensor at very short, regular time intervals
the data logger (or connected computer) plots this distance and time data directly as a distance–time graph
(b)Answer: It records far more, and more precise, data points than manual timing could.
it can record many more data points, much more frequently and precisely, than would be possible by hand, giving a smoother and more detailed graph
Question 3Hard6 marks
Explain how you could use a distance–time graph to determine: (i) whether an object is stationary, moving at a constant speed, or accelerating; (ii) the object’s speed at one instant, if the graph is a curve rather than a straight line.[6]
Show the answer and mark scheme
a horizontal line (zero gradient) shows the object is stationary
a straight, sloping line (constant, non-zero gradient) shows the object is moving at a constant speed
a curved line, where the gradient is changing, shows the object is accelerating (speeding up) or decelerating (slowing down)
speed at any point equals the gradient of the graph at that point
for a straight-line section, the gradient (and so the speed) can be found directly from any two points on the line: change in distance ÷ change in time
for a curved section, a tangent must be drawn to the curve at the instant of interest, and the gradient of that tangent (found using a large triangle) gives the speed at that instant
Marked with levels of response: the full level descriptors are in the app.