Distance–time graphs

Key idea: Distance–time gradient gives speed; a horizontal section shows rest.

  • SEC G1 Science 2027

Effects of Force · Lesson 4 of 5 · about 25–35 min

What you need to understand

Distance–time gradient gives speed; a horizontal section shows rest.

Definitions

distance–time graph
a graph showing how an object's distance changes with timeFor example: A horizontal section shows that distance is not changing.
gradient
how much the vertical quantity changes for each horizontal changeFor example: A steeper distance–time line represents a greater speed.

Explanation

A distance–time graph records how distance changes as time passes. Read the axis labels and units first: time is horizontal and distance is vertical. Plot each measured pair at its correct coordinates; join the points only in a way supported by the observations.

For a straight sloping section, divide the change in distance by the elapsed time. This gradient is the speed. Equal increases in distance in equal time intervals give a straight line; larger increases make it steeper. A horizontal section means no change in distance, so the object is at rest.

A curve that becomes steeper records increasing speed. Compare gradients, not the height of the graph: a high point tells you how much distance has been covered, not how fast the object is moving. This distance-travelled model cannot decrease; a falling line would need a different quantity or an error in the record.

Pause and say it: Distance–time gradient gives speed; a horizontal section shows rest.

Common mistake

Tempting wrong idea: A motion graph is a drawing of the path followed by the object.

Why it fails: The graph is not a picture of a hill or road. A horizontal line means unchanged distance, not a flat road travelled at constant speed.

Use this instead: Read the labelled quantities: time is horizontal and the plotted motion quantity is vertical.

Worked examples

Modelled example 1

Read distance–time sections

Core

Problem

Read the axes before interpreting the line. The accompanying data table gives the same pairs. These values describe an illustrative journey.

Distance covered during a twelve-second journey

Time in seconds runs horizontally from 0 to 12; distance in metres runs vertically from 0 to 30. Points are (0, 0), (5, 20), (9, 20) and (12, 26).

Scroll across the graph to read all labels.

Time in seconds runs horizontally from 0 to 12; distance in metres runs vertically from 0 to 30. Points are (0, 0), (5, 20), (9, 20) and (12, 26).Time in seconds runs horizontally from 0 to 12; distance in metres runs vertically from 0 to 30. Points are (0, 0), (5, 20), (9, 20) and (12, 26).
Straight segments join the plotted time and distance pairs.
Open full-size graph
View figure data
Values for Distance covered during a twelve-second journey
Time (s)Illustrative journey
00
520
920
1226

Calculate the speed from 0 to 5 s.

Study the worked solution
  1. Use the change in distance and time

    Method

    Divide the change in distance by the elapsed time.

    Reason

    The gradient measures distance covered per second, which is speed.

    Working

    From 0 to 5 s, distance increases by 20 m in 5 s. Speed is 4 m/s. The line is straight, so this speed is constant over that section.

Try it. Describe the next two sections and calculate their speeds. Which sloping section is steeper? Use changes between endpoints, not distance divided by time since the whole journey began.

Check the graph interpretation

From 5 to 9 s distance stays at 20 m: the object rests for 4 s, with zero speed. From 9 to 12 s it covers 26 − 20 = 6 m in 12 − 9 = 3 s: 2 m/s. The first sloping section is steeper and represents the greater speed. Graph height tells you distance covered, not speed.

Practical work

What to show: Read the axes and units, use changes in distance and time, and distinguish rest from constant and changing speed.

Before you finish: An adequate graph must place all pairs correctly; checking this description alone does not assess your drawing.

Practical: Distance–time record of a rolling trolley

How does a trolley's distance from its release point change with time as it rolls down a gentle ramp, along a level floor and into a cushion?

Safety: Use a low ramp and a soft stop. Keep the path clear of feet and bags, and never stand on the ramp or trolley.

Change
time since release, with a reading every 0.5 s
Measure
distance of the trolley's front edge from the release point, in metres
Keep the same
same trolley, ramp slope and floor surface, released from rest at the same mark without a push, the front edge is the reference point for every reading
View the apparatus, method and observation record, then use the record

Apparatus

dynamics trolley or toy car, board raised a few centimetres at one end as a gentle ramp, measuring tape at least 2 m long, cushion or beanbag as a stop, phone or camera on a stand to record video.

Method

  1. Set up a gentle ramp about 0.8 m long leading onto a smooth level floor. Lay the tape along the path from the release point and place the cushion 2.00 m from that point.
  2. Place the camera level with the tape, facing it square-on, so the whole path is in view. Start recording.
  3. Hold the trolley with its front edge at 0 m, then let go without pushing.
  4. Play the video back. Take time = 0 at the frame where the trolley starts to move, then read the position of its front edge on the tape every 0.5 s until it has rested against the cushion for at least 1 s.
  5. Repeat for three runs from the same mark and calculate the mean distance at each time.
  6. Plot mean distance (m) on the vertical axis against time (s) on the horizontal axis.
Illustrative readings for three runs (not real class data; record your own readings in the same layout)
Time / sRun 1 / mRun 2 / mRun 3 / mMean / m
0.00.000.000.000.00
0.50.050.060.050.05
1.00.200.210.190.20
1.50.450.460.440.45
2.00.800.810.790.80
2.51.191.211.201.20
3.01.601.611.591.60
3.52.002.002.002.00
4.02.002.002.002.00
4.52.002.002.002.00

Use the record

  1. Find how far the trolley moves in each 0.5 s interval using the mean distances. During which times is it speeding up, moving at a steady speed and at rest?

    Check your answer

    From 0 to 2.0 s it moves 0.05, 0.15, 0.25 and 0.35 m in successive intervals, so it is speeding up on the ramp. From 2.0 to 3.5 s it moves 0.40 m in every interval, so its speed is steady on the floor. After 3.5 s the distance does not change, so it is at rest against the cushion.

  2. Describe the shape of your distance–time graph in each of those three sections.

    Check your answer

    A curve that gets steeper from 0 to 2.0 s, a straight sloping line from 2.0 to 3.5 s, and a horizontal line after 3.5 s.

  3. Calculate the trolley's steady speed along the floor.

    Check your answer

    Speed = gradient = (2.00 − 0.80) m ÷ (3.5 − 2.0) s = 1.20 m ÷ 1.5 s = 0.80 m/s.

  4. Write a conclusion that answers the question.

    Check your answer

    On the ramp the trolley covered more distance in each equal time interval, so it was speeding up. On the floor it covered equal distances in equal times, a steady speed of 0.80 m/s. Once it reached the cushion its distance stopped changing, so it was at rest. The steady section lasted only 1.5 s, so a longer floor would be needed to test whether the speed stays steady for longer.

Limitation: A reading can be misjudged if the camera views the tape at an angle or if the trolley is blurred in the paused frame.

Improvement: Keep the camera square-on to the middle of the tape, use a higher frame rate or slow-motion mode, and read the same edge of the trolley every time.

Guided practice

Try it with support

Use the graph in the worked example. Find the speed during the first sloping section and describe the horizontal section.

  1. Use changes on both axes.
Check the guided answer

Answer: The first speed is 20 m ÷ 5 s = 4 m/s. From 5 s to 9 s, distance stays at 20 m, so the object is at rest.

Check: Graph height is distance; gradient gives speed.

Practise and continue

Practise this

Plot distance / m against time / s for these pairs: (0, 0), (2, 6), (4, 12), (6, 12), (8, 16). Describe each section and find its speed.

Need a hint?
  • Label both axes and use uniform scales.
Check your answer

Answer: From 0–4 s the straight line has gradient 3 m/s. From 4–6 s it is horizontal: rest. From 6–8 s its gradient is 2 m/s.

Check: An adequate graph must place all pairs correctly; checking this description alone does not assess your drawing.

Think like a scientist

A learner sketches a distance-travelled graph that falls after the trolley stops. Find the first mistake.

Check the reasoning

Distance already travelled cannot decrease. The line should become horizontal at the stopping distance.

Remember: Do not treat the graph as a picture of the trolley’s route.

One-minute check

  1. Hide the page and explain distance–time graphs in your own words.
  2. Give a new example that is different from the worked example.
  3. Correct this common mistake: “A motion graph is a drawing of the path followed by the object.”

Next small step

Lesson 5: Turning effects

For the same perpendicular force, a greater distance from the pivot gives a greater turning effect.

Continue to lesson 5

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G1 Science
Edition
SEC G1 Science 2027