Motion — Chapter quiz · Physical Science: Matter, Forces & Energy, Grade 8

Describe movement with measurements and graphs, then use Newton’s laws to explain changes.

Problem 1

An illustrative walk covers 30 m east and then 10 m west in 20 s. The starting position is the chosen origin, with east positive. Find distance, displacement, average speed and average velocity. Explain why the two rates differ.
Answer and explanation

Distance is 40 m; displacement is +20 m east. Average speed is 40/20=2 m/s; average velocity is +20/20=+1 m/s east. Distance counts the full path, while displacement compares final and initial positions with direction. These averages do not establish a constant speed at every instant.

Distance is 40 m; displacement is +20 m east. Average speed is 40/20=2 m/s; average velocity is +20/20=+1 m/s east. Distance counts the full path, while displacement compares final and initial positions with direction. These averages do not establish a constant speed at every 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 2

Why is m/s² appropriate for acceleration?
Answer and explanation

Velocity changes in m/s and is divided by seconds. The resulting unit describes how much velocity changes per second.

Velocity changes in m/s and is divided by seconds. The resulting unit describes how much velocity changes per second.

Response criteria

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

Problem 3

Independent check: Right is positive. A cart changes velocity from −2 to −8 m/s in 3 s. Calculate average acceleration and explain whether it speeds up or slows down.
Answer and explanation

(−8−(−2))/3 = −2 m/s². It speeds up toward the left because speed rises from 2 to 8 m/s.

Subtract signed velocities before dividing.

Compare velocity direction with acceleration direction.

Response criteria

  • Calculate −2 m/s².
  • Explain leftward motion.
  • Reject negative-always-slows.

Problem 4

Independent chapter check: Right is positive. A cart’s velocity changes from −3 to −9 m/s in 2 s. Find average acceleration and explain its speed change.
Answer and explanation

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

Subtract signed velocities.

Compare the magnitudes and directions instead of using a sign slogan.

Average acceleration = (−9−(−3))/2 = −3 m/s²; speed increases leftward from 3 to 9 m/s. Negative acceleration does not always mean slowing.

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 5

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 6

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

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

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

Response criteria

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

Problem 7

Independent check: 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 8

Independent chapter check: 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.

Problem 9

Illustrative carts of 2 kg and 4 kg each experience a 6 N horizontal net force. Other conditions are comparable. Calculate acceleration magnitudes, identify an equal-and-opposite interaction pair and explain why zero net force does not require rest.
Answer and explanation

Accelerations are 3 and 1.5 m/s² because a=Fnet/m. Equal interaction forces act on different objects, such as hand-on-cart and cart-on-hand. A zero net force means no acceleration in the simple inertial-frame model and permits constant nonzero velocity. Mass affects response, not an extra permanent force needed just to keep moving.

Accelerations are 3 and 1.5 m/s² because a=Fnet/m. Equal interaction forces act on different objects, such as hand-on-cart and cart-on-hand. A zero net force means no acceleration in the simple inertial-frame model and permits constant nonzero velocity. Mass affects response, not an extra permanent force needed just to keep moving.

Response criteria

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

Problem 10

Do equal interaction forces require equal accelerations?
Answer and explanation

No. Acceleration depends on each object’s mass and all forces acting on it. Equal opposite partner forces act on different objects.

No. Acceleration depends on each object’s mass and all forces acting on it. Equal opposite partner forces act on different objects.

Response criteria

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

Problem 11

Plan two comparisons from the records: one isolating net force, the other mass. Specify motion measurements, repeats, controls and what outcome would challenge the predicted relation. Explain braking separately from velocity direction.
Hints

Use the illustrated study titled Investigate force and mass separately. Keep the observations, model and conditions separate.

Answer and explanation

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Compare the 1 kg cases at 1 and 2 N to isolate net force; compare 1, 2 and 4 kg at 2 N to isolate mass. Match path, resisting conditions, timing, measuring procedure and initial state. Keep all repeated records. The supplied model gives a=Fnet/m; unexpected motion needs a boundary/force/method review rather than automatic rejection of conservation or hidden invented values. A leftward net force can slow rightward motion.

Response criteria

  • Design two separately controlled comparisons.
  • Specify recorded motion, units and repeats.
  • Explain a challenge result and braking mechanism.

Problem 12

Independent check: The cart diagram shows 10 N right and 4 N left on a 2 kg cart. A student calculates 5 m/s² using the push alone. Correct the method and say whether the cart must currently move right.
Answer and explanation

Net force is 6 N right; acceleration is 3 m/s² right. The current velocity is not given: the cart could be moving left and slowing.

Subtract resistance from the applied push.

Divide net force by mass.

Keep acceleration distinct from current motion.

Response criteria

  • Calculate 6 N and 3 m/s².
  • Do not count the green net arrow again.
  • Avoid inferring current velocity.

Problem 13

Independent chapter check: A 2.5 kg cart has a 15 N applied push and 5 N resistance. Find acceleration, and explain why its equal force on the pusher does not cancel that acceleration.
Answer and explanation

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

Compute net force before dividing by mass.

Name the receiver of each interaction partner.

Net force is 10 N in the push direction and acceleration 10/2.5 = 4 m/s². The cart-on-pusher partner acts on the pusher, so it is not part of the cart’s net force.

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 14

Illustrative matched-cart tests show bumper thickness 15 mm before, 8 mm during and 15 mm afterward; peak force is 12 N rather than 24 N for a rigid stop. Explain the temporary deformation and compare peak values without declaring all collisions safe.
Answer and explanation

The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.

The bumper undergoes elastic deformation in this simplified record because it regains its thickness. The padded peak is half the rigid-stop peak under matched conditions. Deformation can allow a longer stopping interval, but the readings do not prove that every object is protected or give the entire force–time history. Energy is redistributed rather than erased.

Response criteria

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

Problem 15

What do these peak records support?
Answer and explanation

Lower measured peak force for this padded setup under the stated matched conditions. They do not certify every pad or establish identical average forces or injury protection.

Lower measured peak force for this padded setup under the stated matched conditions. They do not certify every pad or establish identical average forces or injury protection.

Response criteria

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

Problem 16

Independent chapter check: Two carts receive equal opposite 12 N contact forces in a stated horizontal interval. Their masses are 3 and 1.5 kg. Compare accelerations and explain whether equal force means equal velocity change over the same duration.
Answer and explanation

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

Use each receiver’s mass in Fnet/m under the supplied net-horizontal-force assumption.

Multiply acceleration by the common interval to compare velocity changes.

Accelerations are 4 and 8 m/s² in opposite directions. Over the same duration the lighter cart’s velocity change has twice the magnitude, despite the equal force pair.

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 17

Illustrative designs have A: mean peak 13 N, thickness 4 cm; B: mean peak 9 N, thickness 7 cm. Maximum permitted thickness is 5 cm. Choose a qualifying starting design and propose an evidence-based revision.
Answer and explanation

A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.

A qualifies under the thickness rule; B does not despite its lower mean peak. Start with A and investigate whether a more compressible arrangement within 5 cm lowers the peak. Repeat matched tests, record all results and evaluate permanent deformation as well as force. Lower mean peak alone is not universal protection.

Response criteria

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

Problem 18

Does combining useful features prove the new protector is better?
Answer and explanation

No. Features can interact and violate limits. The revised design requires new tests under the same evaluation rules.

No. Features can interact and violate limits. The revised design requires new tests under the same evaluation rules.

Response criteria

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

Problem 19

Independent chapter check: A design must stay below 30 N in every trial. Its supplied results are 22, 24 and 35 N. Does a mean of 27 N establish that it meets the criterion? Explain a proper next step.
Answer and explanation

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

Read the criterion as every trial, not the mean.

Use all records before recommending a tested revision.

No. The 35 N trial fails the every-trial limit even though the mean is 27 N. Preserve the result, investigate variation, revise a mechanism and retest under matched conditions.

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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