Reading Motion Graphs — Print · Grade 8 Science

Reading Motion Graphs

Learning goals

I can use a position–time graph to describe rest, direction, and relative speed.
I can calculate average velocity over an interval from position and time data.

Readings for this unit

A soft bumper stops a toy differently

Before this unit

Read axes, intervals and units on a graph.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in reading motion graphs?

Illustrative evidence case. Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points.

Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Notice and wonder

Does the final downward line prove decreasing speed? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

position: Location relative to a chosen origin.
slope: Change in vertical value divided by change in horizontal value.
average velocity: Displacement divided by elapsed time, including direction.
model: A representation that explains selected features and makes predictions; it leaves other details out.
evidence: Relevant observations or records used to support or challenge a claim.
system boundary: The chosen limit separating the objects being studied from their surroundings.
assumption: A condition taken as true for a model or calculation and needing to be checked against the case.

A position–time graph shows where an object is relative to a chosen origin. Put time on the horizontal axis and position on the vertical axis. The slope tells average velocity over an interval: change in position divided by elapsed time. A graph shows a relationship, not necessarily the physical path the object traveled.

A horizontal segment means position stays the same: the object is at rest during that interval. A straight rising segment means constant positive velocity. A falling segment means motion in the negative direction, not automatically slowing down. Greater steepness means greater speed; the sign of the slope indicates direction.

Choose two data points within the interval you are studying. Subtract both the positions and times. On a curved position–time graph the velocity is changing. Do not confuse position–time with speed–time: a horizontal speed–time line means constant speed, which may be nonzero. Always read the axes first.

Positive is right. The object travels right at 3 m/s for 2 s, then left at 2 m/s for 2 s. A rectangle’s time width times velocity height gives displacement: +6 m followed by −4 m. Signed areas add to +2 m displacement; their magnitudes add to 10 m distance. The instantaneous switch is an idealization; an actual reversal takes time. This velocity graph is a motion record, not a map of the object’s path.

On a velocity–time graph, slope represents acceleration. A horizontal segment can mean moving at constant velocity, not standing still. The area between the graph and time axis represents displacement; a section below the axis contributes negative displacement.

Always identify which quantity is plotted before interpreting the shape. Use two points separated by a useful interval to calculate slope as vertical change divided by horizontal change. A line on a graph is not a picture of a hill or the actual path of the object.

Build your understanding

Where this fits

This unit helps answer the chapter question: What evidence lets us describe and predict motion? Keep a notebook table with the question, evidence, explanation and one unresolved issue. Revise a previous model only when you can name the new evidence or idea that changes it.

Part 1 · Read axes before interpreting a line

A position-time graph shows position on the vertical axis and time on the horizontal axis. Its line is not a drawing of a hill or the object’s route. Position is measured relative to a stated reference, and negative position can simply mean the other side of that reference.

Read the visual model

What does a horizontal position-time section show?

Visual reasoning — after your attempt

Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

A graph rising from position 0 m to 6 m over 3 s describes increasing position. It does not mean an object physically climbed a diagonal track. Write a sentence explaining each axis before describing motion.

Pause and explain

What does a horizontal position-time section show?

Check your reasoning after answering

Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

Part 2 · Use position-time slope to find velocity

Slope compares change in position with change in time. A straight segment has constant velocity in the model. A steeper magnitude means greater speed; a negative slope indicates motion in the negative direction. Two different starting positions can have the same slope and therefore the same velocity.

Position-time graph for the supplied data, with straight segments assuming constant interval velocity. It is a graph of measurements and a model between them, not the shape of the route.

Read the visual model

Why must both coordinate changes be used for an interval slope?

Visual reasoning — after your attempt

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

A line changing from 2 m to 8 m between 1 s and 4 s has slope (8−2)/(4−1)=2 m/s. Using total position divided by total clock time would answer a different question unless both starting values are zero.

Pause and explain

Why must both coordinate changes be used for an interval slope?

Check your reasoning after answering

Slope uses differences between endpoints. Ignoring the initial position or time can give the wrong velocity for the selected interval.

Part 3 · Read velocity-time graphs differently

On a velocity-time graph, a horizontal line means constant velocity, not necessarily rest. Slope represents acceleration. The signed area under the graph gives displacement; distance requires adding the magnitudes of traveled portions. A negative velocity means motion in the negative direction under the chosen convention.

Read the visual model

Why is a flat velocity-time line not always a stopped object?

Visual reasoning — after your attempt

Its vertical value can be nonzero. A flat line at 3 m/s describes constant motion; only a line at zero velocity describes rest.

A velocity of −2 m/s maintained for 3 s gives displacement −6 m. The graph is flat but the object moves. Always name the graph type before applying a horizontal-line rule learned from position-time graphs.

For the supplied graph choose right positive. From 0–2 s, velocity is +3 m/s; the rectangular area is 2 s×3 m/s = +6 m. From 2–4 s, velocity is −2 m/s; its area is −4 m. Net displacement is +2 m, but total distance is 10 m. Area units are seconds multiplied by metres per second, leaving metres. The vertical switch is an idealized reversal; a real object needs time to reverse.

Pause and explain

Why is a flat velocity-time line not always a stopped object?

Check your reasoning after answering

Its vertical value can be nonzero. A flat line at 3 m/s describes constant motion; only a line at zero velocity describes rest.

Part 4 · Separate data from an interpolated model

A few measured points do not reveal every detail between readings. Joining points with straight lines is a modeling choice. Label measured values and explain any assumption of constant velocity between them. Avoid interpreting a smooth curve as more precise than the measurement supports.

The only supplied measurements are position 0 m at 0 s and position 4 m at 2 s (gold dots). The blue straight-line model has constant 2 m/s velocity. The red model, position = time² in these units, has increasing velocity and reaches 1 m at 1 s instead of 2 m. Both fit the endpoints and have the same average velocity. More measurements between the endpoints are needed to decide which account fits the actual motion. Intermediate curves are proposed models, not measured records.

Read the visual model

What would test a claim of constant velocity between two measurements?

Visual reasoning — after your attempt

Collect additional position-time readings within the interval and compare their slopes. Endpoint data alone establish an interval average, not every moment.

Positions measured at 0, 2, and 4 s can support average velocities for two intervals. If no intermediate readings exist, several different within-interval motions might fit the same points. More frequent measurements can test the assumed straight segments.

The gold endpoints in the new figure are the only actual supplied position records. The blue straight interpolation and red curved proposal both reach 4 m after 2 s, so they share average velocity 2 m/s. At 1 s they predict different positions. Measuring an intermediate position tests the model. Connecting records with a straight line should not silently turn missing data into proof of constant velocity.

Pause and explain

What would test a claim of constant velocity between two measurements?

Check your reasoning after answering

Collect additional position-time readings within the interval and compare their slopes. Endpoint data alone establish an interval average, not every moment.

Apply the whole unit · Make and interpret a motion graph

Illustrative positions: at 0, 2, 4, and 6 s a cart is at 0, 4, 4, and 0 m. Model each intervening interval as constant velocity, with positive position to the right.

Your task

Draw a labeled position-time graph, calculate velocity for each interval, and describe the motion in words. Find total distance and displacement, then identify how a new measurement could test the straight-segment assumption.

What your response should show

Use the supplied observations and measurements accurately; include units where needed.
Explain the science behind your answer, rather than giving only a number or label.
Show a labeled model, calculation, or evidence comparison as requested.
State an assumption or limitation and suggest a useful next measurement.

Response guidance — after your attempt

Velocities are +2, 0, and −2 m/s. The cart moves right 4 m, pauses, and returns left 4 m. Distance=8 m and displacement=0 m. Label time in seconds and position in meters. Intermediate readings would test constant velocity; the connected segments are a stated model rather than additional observations.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Use graph-reading sentences before calculations. Contrast a flat section on position-time and velocity-time axes. Ask the learner to separate measured dots from assumed connections. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation.

The key idea

On a position–time graph, slope describes velocity; a horizontal segment means rest.

Worked examples

Example 1

An object is at 2 m at 1 s and 10 m at 5 s. Find its average velocity.

Change in position = 10 − 2 = 8 m.
Elapsed time = 5 − 1 = 4 s.
Average velocity = 8 ÷ 4.

Answer: 2 m/s in the positive direction.

Example 2 · Worked application

Position is 0 m at 0 s, 8 m at 2 s and 8 m at 4 s. Interpret the two graph segments.

From 0 to 2 s, position changes 8 m: slope=8/2=4 m/s.
From 2 to 4 s, position stays 8 m: slope=0.
Assuming straight segments, the object first moves at constant positive velocity, then stops.

Answer: 4 m/s for the first segment and 0 m/s for the second.

Check the reasoning

Read slope, not the vertical height, as velocity on a position–time graph.

Example 3

Position changes from 4 m at 2 s to 16 m at 6 s on a straight segment. Find its velocity.

Vertical change=16−4=12 m; time change=6−2=4 s.
Slope=12/4=3 m/s.

Answer: Constant velocity of +3 m/s over this segment.

Example 4 · Find and fix the mistake

Context: Position is 0 m at 0 s, 8 m at 2 s and 8 m at 4 s. Interpret the two graph segments. A student claims: “A horizontal position–time line means fast constant motion.” Evaluate the claim and write a corrected explanation.

Test the claim against the stated conditions: Read slope, not the vertical height, as velocity on a position–time graph.
Horizontal means position is constant, so velocity is zero. A horizontal velocity–time line has a different meaning.

Answer: Horizontal means position is constant, so velocity is zero. A horizontal velocity–time line has a different meaning.

Check the reasoning

Read slope, not the vertical height, as velocity on a position–time graph.

Example 5 · Displacement from signed areas

For the supplied velocity–time graph, motion is +3 m/s for 2 s then −2 m/s for 2 s. Find displacement, distance and average velocity.

First displacement = +3×2 = +6 m. Second displacement = −2×2 = −4 m.
Displacement = +2 m; distance = 6+4 = 10 m.
Average velocity = +2 m / 4 s = +0.5 m/s.

Answer: Displacement 2 m right; distance 10 m; average velocity 0.5 m/s right.

Check the reasoning

Average speed would be 10/4 = 2.5 m/s. It answers a different question.

Example 6 · Interpret the visual model

What does a horizontal position-time section show?

Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
A graph rising from position 0 m to 6 m over 3 s describes increasing position. It does not mean an object physically climbed a diagonal track. Write a sentence explaining each axis before describing motion.
Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

Answer: Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

Check the reasoning

Keep the model’s stated limits explicit: Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Avoid these mistakes

A downward line on a position graph does not mean the object moved downhill.

Check your understanding

Position changes from 4 m at 2 s to 16 m at 6 s on a straight segment. Find its velocity.

Discuss the answer

Constant velocity of +3 m/s over this segment.

How does a horizontal velocity–time line differ from a horizontal position–time line?

Discuss the answer

The slope of a position–time graph represents velocity. A horizontal segment means unchanged position, and a straight rising segment means constant positive velocity. A steep segment indicates a larger magnitude of velocity only when the axes and scales are comparable. On a velocity–time graph, slope represents acceleration. A horizontal segment can mean moving at constant velocity, not standing still. The area between the graph and time axis represents displacement; a section below the axis contributes negative displacement.

Return to the phenomenon

Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

The interval velocities are +3, 0 and −2 m/s. A level position graph means unchanged location; a level nonzero velocity graph means steady motion. The axes determine the interpretation. Straight segments describe the simplified interval model rather than measurements at every possible instant.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Supplied positions at 0, 2, 4 and 6 s are 0, 6, 6 and 2 m, with right positive. The graph connects these recorded points. Find average velocity for each interval and explain the difference between a level position graph and a level nonzero velocity graph. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

Use two relevant stated records or conditions.
Explain the mechanism rather than only repeat a definition.
Label observation and inference.
Name a changed model feature or a remaining uncertainty.

Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Does the final downward line prove decreasing speed? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

Answer the figure question with its stated labels.
Connect the visual representation to the scientific idea.
Identify a specific limitation from the caption.
Propose a useful change without misrepresenting supplied evidence.

Hint

Response guidance — discuss after your attempt

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude. Recorded positions give interval slopes +3, 0 and −2 m/s. A horizontal position segment means unchanged position, while a downward segment means a negative velocity under the chosen positive direction. Straight connections are a simplified interval model, not a record at every intermediate instant.

Words to know

position: Location relative to a chosen origin.
slope: Change in vertical value divided by change in horizontal value.
average velocity: Displacement divided by elapsed time, including direction.
model: A representation that explains selected features and makes predictions; it leaves other details out.
evidence: Relevant observations or records used to support or challenge a claim.
system boundary: The chosen limit separating the objects being studied from their surroundings.
assumption: A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Exit check

Does the final downward line prove decreasing speed? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A horizontal position–time segment means…

rest
constant positive acceleration
infinite speed
motion backward

Answer and explanation

rest

Position does not change.

Problem 2

Position changes from 3 m to 15 m in 4 s. Find average velocity.

Answer and explanation

3 m/s

(15 − 3) ÷ 4 = 3.

Problem 3

Does a falling position–time line always mean slowing down?

Answer and explanation

No. It means position decreases. A straight falling line represents constant negative velocity.

Problem 4

Position changes from 4 m at 2 s to 16 m at 6 s on a straight segment. Find its velocity. Explain the reasoning in your own words.

Answer and explanation

Constant velocity of +3 m/s over this segment.

Vertical change=16−4=12 m; time change=6−2=4 s.

Slope=12/4=3 m/s.

Response criteria

Use the stated conditions.
Show the evidence or calculation.
Explain the conclusion and any stated limitation.

Problem 5

A straight position–time segment goes from 12 m at 1 s to 4 m at 5 s. Find its velocity. Show the reasoning, not just the conclusion.

Hints

Read slope, not the vertical height, as velocity on a position–time graph.

Answer and explanation

(4−12)/(5−1)=−2 m/s; negative indicates motion toward decreasing position.

Problem 6

Can a position–time graph tell you the road's hills directly? Explain.

Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

No. The vertical axis is position along a chosen coordinate, not necessarily road elevation. Interpret the labeled axes.

Problem 7

Does the final downward line prove decreasing speed?

Answer and explanation

No. A straight downward position line has constant negative velocity in the simplified interval. Its direction differs from its speed magnitude.

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 8

Answer and explanation

Response criteria

Use the supplied conditions and relevant evidence.
Explain the scientific mechanism or calculation.
State a meaningful limitation or testable next step.

Problem 9

Read the supplied +3 m/s then −2 m/s graph, each interval lasting 2 s. Calculate distance, displacement and average speed.

Answer and explanation

Distance 10 m; displacement +2 m; average speed 2.5 m/s.

Compute signed areas +6 and −4 m.

Add magnitudes for distance and divide by 4 s for average speed.

Response criteria

Use signed area for displacement.
Use magnitudes for distance.
State average-speed units.

Problem 10

Position is measured as 0 m at 0 s and 4 m at 2 s. Does this prove constant 2 m/s velocity? Use the straight and curved proposals to suggest a test.

Answer and explanation

No. It establishes average velocity 2 m/s. The straight proposal predicts 2 m at 1 s and the curved proposal 1 m; an intermediate measurement can distinguish them.

Compare two models fitting the same endpoints.

Use a discriminating intermediate prediction.

Response criteria

Distinguish average from constant velocity.
Supply both intermediate predictions.
Propose an actual measurement.

Problem 11

For this unit’s illustrated case, what does a horizontal position-time section show?

Answer and explanation

Position remains unchanged during that interval, so the object is at rest relative to the chosen reference, provided the graph accurately represents the motion.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Problem 12

A velocity–time record is +4 m/s for 3 s followed by −1 m/s for 2 s. Calculate distance, displacement and average velocity over the full five seconds.

Answer and explanation

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Use signed rectangle areas for displacement.

Add path magnitudes for distance and divide displacement by total time for average velocity.

Distance 12+2 = 14 m; displacement 12−2 = 10 m right; average velocity 10/5 = 2 m/s right.

Response criteria

Use the named labels, directions or supplied quantities accurately.
Connect the visual evidence to the scientific mechanism or calculation.
State the relevant model conditions or limitation.

Browse all courses