Reading Motion Graphs — Practice · Grade 8 Science

Chapter: Motion

What this practice covers

By the end of Reading Motion Graphs you should be able to:

12 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: position, slope, average velocity, model, evidence, system boundary, assumption.

Problem 1

A horizontal position–time segment means…
  1. rest
  2. constant positive acceleration
  3. infinite speed
  4. 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.

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

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.
Answer and explanation

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.

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.

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

Read the position–time axesRead the position–time axesTime (s)Position (m)001.52344.5668Illustrative model • use the labels and stated assumptions
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.
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.

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.

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.

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