Velocity & Acceleration — Print · Grade 12 Science

Chapter: Motion & Kinematics

Velocity & Acceleration

Learning goals

Readings for this unit

Before this unit

Rates and signed velocity.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Does negative acceleration always mean an object is slowing?

Illustrative evidence case. A cart’s velocity crosses from positive to negative while its acceleration stays negative.

Acceleration describes velocity change, not motion direction aloneAcceleration describes velocity change, not motion direction aloneCaseVelocityAccelerationMoving east, speeding up+6 m/s+2 m/s²Moving east, slowing+6 m/s−2 m/s²Moving west, speeding up−6 m/s−2 m/s²Same signs → increasing speed in one dimension.Opposite signs → decreasing speed, before reversal.
Selected instantaneous states, before any turnaround; signs use east positive.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Does negative acceleration always mean an object is slowing?

Your first model

Sketch separate velocity and acceleration arrows before and after the turnaround.

Build the language from the idea

average velocity
Displacement divided by the time interval: v_avg = Δx ÷ Δt. A vector, so it can be zero even after a long trip.
instantaneous velocity
The velocity at a single instant — the value on a speedometer paired with a direction; the slope of a position-time graph at that point.
average speed
Total distance divided by total time; a scalar that is never negative.
acceleration
The rate at which velocity changes: a = (v − v₀) ÷ t. A vector, measured in m/s².
free fall
Motion under the influence of gravity alone, with air resistance neglected.
g
The acceleration due to gravity near Earth's surface, about 9.8 m/s² directed downward, the same for every freely falling object regardless of mass.

Velocity is displacement over time, and the word 'average' hides an important subtlety. Average velocity is the total displacement divided by the total time, v_avg = Δx ÷ Δt, and because it is built on displacement it is a vector that can even come out to zero. A runner who circles a 400 m track and finishes back at the start has an average velocity of exactly zero for the lap, even though she was never once standing still — her displacement was zero, so her average velocity must be too. Average speed, by contrast, uses total distance instead of displacement, so for that same lap it is a healthy 8 m/s if the lap took 50 seconds.

Average velocity smooths a whole trip into one number, but often you want to know how fast something is moving *right now*. That is instantaneous velocity — the velocity at a single instant, which is what a car's speedometer shows (paired with the direction the car is pointing). Formally it is the slope of the position-time graph at that exact moment, the limit of Δx ÷ Δt as the time interval shrinks toward zero. On a road trip your average velocity might be 90 km/h, while your instantaneous velocity swings from 0 at a rest stop to 120 on the open highway.

Subtract signed velocities before dividing by timeSubtract signed velocities before dividing by timev₀ = −3 m/sv = +5 m/sΔt = 4 saavg = [5 − (−3)]/4 = +2 m/s²Velocity change is +8 m/s, not +2 m/s.
Original signed-velocity example; no within-interval history is supplied.

Acceleration is the next step up: the rate at which velocity changes, a = (v − v₀) ÷ t, measured in meters per second per second, or m/s². Because velocity is a vector, a change in *either* its size or its direction counts as acceleration. A car speeding from 12 m/s to 30 m/s in 6 seconds accelerates at (30 − 12) ÷ 6 = 3 m/s². The sign matters and it is not just 'plus for go, minus for stop' — a negative acceleration means the velocity is becoming more negative, which slows an object moving in the positive direction but speeds up one already moving negative.

A zero velocity can occur during nonzero accelerationA zero velocity can occur during nonzero accelerationVelocity east-positive (m/s)-2024601234Time (s)The graph crosses zero at a reversal, not at zero acceleration.
Calculated line for v=6−2t m/s; constant acceleration is an explicit assumption.

The most important acceleration in this unit is the one gravity gives for free. Near Earth's surface, every object in free fall — moving under gravity alone, with air resistance ignored — accelerates downward at the same rate, about 9.8 m/s², a value we call g. That phrase 'the same for every object' is the counterintuitive heart of it: a bowling ball and a marble, dropped together in a vacuum, gain speed at identical rates and land together. Galileo argued this four centuries ago, and astronauts have since dropped a hammer and a feather side by side on the airless Moon to prove it.

Free fall exposes a classic trap about the difference between velocity and acceleration. Throw a ball straight up and, at the very peak of its flight, its velocity is momentarily zero — for one instant it isn't moving up or down. But its acceleration there is *not* zero; it is still 9.8 m/s² downward, exactly as it is during the whole flight. Gravity never switches off, so even at the top the ball is changing its velocity, which is precisely why it doesn't hover but immediately begins to fall. Zero velocity and zero acceleration are completely different conditions.

Put together, these ideas let you describe motion with real precision. Average velocity captures the net result of a trip, instantaneous velocity captures a single moment, average speed tracks the total ground covered, and acceleration measures how the motion is changing — with g as the ready-made value for anything falling freely. Engineers designing everything from elevator ride-comfort to spacecraft re-entry work in exactly these terms, and a self-driving car's computer updates an object's instantaneous velocity and acceleration many times every second.

Model check: Negative acceleration need not mean slowing; compare its direction with velocity. a=(−8+2)/3=−2 m/s²; speed increases because velocity and acceleration point together.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Treat velocity and acceleration as separate vectors

Acceleration is the rate of change of velocity. Its direction need not match the current velocity. In one dimension a positive velocity and negative acceleration describe eastward motion that is slowing until it reaches zero; continuing the same acceleration then produces westward motion with growing speed. Negative acceleration therefore does not always mean slowing down. Compare both signs in the named frame, and check whether the interval includes a reversal.

Acceleration describes velocity change, not motion direction aloneAcceleration describes velocity change, not motion direction aloneCaseVelocityAccelerationMoving east, speeding up+6 m/s+2 m/s²Moving east, slowing+6 m/s−2 m/s²Moving west, speeding up−6 m/s−2 m/s²Same signs → increasing speed in one dimension.Opposite signs → decreasing speed, before reversal.
Selected instantaneous states, before any turnaround; signs use east positive.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

A westward-moving object has eastward acceleration. Is its speed initially increasing?

Visual reasoning — after your attempt

No. With east positive, velocity is negative and acceleration positive, so speed initially decreases.

Velocity describes position change per time; acceleration describes velocity change per time. An object can accelerate by changing speed, direction, or both. A nonzero velocity does not itself imply nonzero acceleration, and zero velocity at one instant does not require zero acceleration.

At the top of ideal vertical motion, vertical velocity can be zero while gravitational acceleration remains downward. The instant is not a pause in gravity.

Pause and explain

Can constant speed coexist with acceleration?

Check your reasoning after answering

Yes. Motion around a curve changes velocity direction, so acceleration can be nonzero even while speed stays constant.

Part 2 · Read signs through the coordinate convention

The velocity line falls by 2 m/s every second, so its slope is −2 m/s² throughout this constant-acceleration model. At 3 s the velocity is zero only for an instant; acceleration has not stopped. A graph’s crossing of a horizontal zero line concerns its plotted quantity, while slope concerns its rate of change. A negative graph value and a negative slope are distinct statements that must be read separately.

A zero velocity can occur during nonzero accelerationA zero velocity can occur during nonzero accelerationVelocity east-positive (m/s)-2024601234Time (s)The graph crosses zero at a reversal, not at zero acceleration.
Calculated line for v=6−2t m/s; constant acceleration is an explicit assumption.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

At 4 s, what are velocity, acceleration and the change in speed?

Visual reasoning — after your attempt

Velocity=−2 m/s; acceleration=−2 m/s²; speed is increasing after the reversal.

The sign of acceleration identifies its direction, not automatically whether an object speeds up or slows down. In one dimension, velocity and acceleration with the same sign increase speed; opposite signs initially reduce it. Reversal can change that relationship within an interval.

For right-positive motion, v=−6 m/s and a=+2 m/s² initially describe leftward motion slowing. After reaching zero, continued positive acceleration produces rightward motion.

Pause and explain

Is positive acceleration always speeding up?

Check your reasoning after answering

No. It depends on the velocity direction; positive acceleration can reduce the speed of a left-moving object before it reverses.

Part 3 · Calculate a rate of velocity change

Average acceleration is Δv/Δt, so the numerator is final velocity minus initial velocity with signs retained. Here the change from −3 to +5 m/s is +8 m/s over 4 s, giving +2 m/s². This average does not by itself prove acceleration was constant within the interval. The units mean a velocity change of meters per second for every second, not a distance traveled per second squared.

Subtract signed velocities before dividing by timeSubtract signed velocities before dividing by timev₀ = −3 m/sv = +5 m/sΔt = 4 saavg = [5 − (−3)]/4 = +2 m/s²Velocity change is +8 m/s, not +2 m/s.
Original signed-velocity example; no within-interval history is supplied.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Would a different nonlinear velocity history with these endpoints have the same average acceleration?

Visual reasoning — after your attempt

Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.

Average acceleration is Δv/Δt with signed or vector velocity changes. Units m/s² mean a velocity change in m/s per second. Using a speed difference instead can miss acceleration caused solely by a direction change. State whether acceleration is assumed constant between observations.

Changing from +4 to −2 m/s in 3 s gives average acceleration −2 m/s². That average alone does not prove a constant acceleration history.

Pause and explain

What is lost if only speeds are recorded?

Check your reasoning after answering

Direction information is lost. Identical speeds before and after can conceal a changed velocity and therefore a nonzero average acceleration.

Part 4 · Check the physical model before predicting

An equation such as v=v₀+at requires constant a for the stated interval. Two histories can have the same start, finish and average acceleration but different intermediate velocities and displacements. Compare repeated velocity observations, sensor response and plausible force changes before choosing the simpler model. A model can be useful within an interval without being an exact description of an entire trip; state the interval and test discrepancies rather than hiding them.

Matching endpoints do not establish constant accelerationMatching endpoints do not establish constant accelerationVelocity (m/s)02460123Time (s)Constant slope modelVariable slope model
Illustrative piecewise-linear histories; areas, not final velocity alone, determine displacement.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Which displayed model has constant acceleration, and do the two have equal displacement?

Visual reasoning — after your attempt

The blue straight line has constant acceleration. Their areas differ, so their modeled displacements differ even though endpoints match.

Constant-acceleration equations require a constant acceleration over the interval in the chosen inertial frame. Gravity near Earth is often approximated by a constant downward value when height changes are small and drag is negligible. Real powered or resistive motion can require a changing model.

A single pair of readings supports an average. More time points can test whether increments of velocity are consistent across equal intervals.

Pause and explain

How would you test a constant-acceleration claim?

Check your reasoning after answering

Compare measured velocity changes across equal time intervals, with uncertainty and conditions considered; systematic changes in those increments challenge the model.

Apply the whole unit · Follow motion through a reversal

Illustrative one-dimensional model: v₀=−6.0 m/s, a=+2.0 m/s² for 5.0 s, right positive.

Your task

Calculate final velocity, reversal time, displacement, and distance. Explain why the acceleration sign does not mean speeding up for the entire interval.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. v=−6+2×5=+4 m/s; reversal occurs at t=3 s.

  2. Displacement=−6×5+½×2×25=−5 m.

  3. Travel before reversal is 9 m left; after reversal 4 m right, giving distance 13 m.

The object slows during its leftward motion, then speeds up rightward after crossing zero velocity. Negative net displacement and positive final velocity are compatible because they describe different features. Splitting at reversal is necessary for distance; adding signed displacements directly gives only the net −5 m.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask for a velocity-time sketch and the sign of acceleration before doing arithmetic. Require separate explanations of final motion direction and overall displacement. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

Free Fall: speed grows 9.8 m/s each secondt = 0 s0 m/st = 1 s9.8 m/st = 2 s19.6 m/st = 3 s29.4 m/s
Ideal near-surface, no-drag falling model. Downward velocity changes under approximately constantg; this is not a claim of identicalg everywhere.

The key idea

Acceleration is velocity change per time. Near Earth, ideal free fall has approximately constant downwardg; negative acceleration does not always mean decreasing speed.

Average velocity

v_avg = Δx ÷ Δt

Δx is displacement (a vector) and Δt is the time interval; a round trip has Δx = 0, so its average velocity is 0 even though distance was covered.

Average speed

average speed = total distance ÷ total time

a scalar built from path length, not displacement; it is never negative.

Acceleration

a = (v − v₀) ÷ t

v is the final velocity and v₀ the initial velocity (both in m/s), t is time in s, and a comes out in m/s². A negative a means the velocity is decreasing.

Free-fall acceleration

g ≈ 9.8 m/s²

the constant downward acceleration of any object in free fall near Earth's surface, independent of the object's mass.

Worked examples

Example 1

A runner completes one 400 m lap of a track, finishing exactly where she started, in 50 s. Find (a) her average speed and (b) her average velocity.
  1. Average speed = total distance ÷ total time = 400 ÷ 50 = 8 m/s.
  2. Average velocity = displacement ÷ time. She returns to the start, so her displacement is 0.
  3. Average velocity = 0 ÷ 50 = 0.
Answer: (a) average speed = 8 m/s; (b) average velocity = 0 m/s

Example 2 · Worked application

Velocity changes from −6 to −2 m/s in 2 s. Find acceleration and whether speed rises.
  1. a=(−2−(−6))/2=+2 m/s².
  2. Speed decreases from 6 to 2 m/s.
  3. Positive acceleration opposes the initial negative motion here.
Answer: +2 m/s²; the object slows during this interval if the change is uniform.
Check the reasoning

Acceleration sign alone does not say whether speed increases.

Example 3

A stone is dropped from rest and falls freely. How fast is it moving after 3.0 s? (Use g = 9.8 m/s².)
  1. In free fall from rest, v₀ = 0 and the acceleration is g.
  2. v = v₀ + gt = 0 + (9.8)(3.0).
  3. = 29.4.
Answer: 29.4 m/s downward

Example 4 · Find and fix the mistake

Context: Velocity changes from −6 to −2 m/s in 2 s. Find acceleration and whether speed rises. A student claims: “Positive acceleration always means speeding up.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Acceleration sign alone does not say whether speed increases.
  2. Speed rises when velocity and acceleration point the same way; opposite directions can reduce speed.
Answer: Speed rises when velocity and acceleration point the same way; opposite directions can reduce speed.
Check the reasoning

Acceleration sign alone does not say whether speed increases.

Example 5 · Reason from the visual model

Subtract signed velocities before dividing by timeSubtract signed velocities before dividing by timev₀ = −3 m/sv = +5 m/sΔt = 4 saavg = [5 − (−3)]/4 = +2 m/s²Velocity change is +8 m/s, not +2 m/s.
Original signed-velocity example; no within-interval history is supplied.
Would a different nonlinear velocity history with these endpoints have the same average acceleration?
  1. Average acceleration is Δv/Δt, so the numerator is final velocity minus initial velocity with signs retained. Here the change from −3 to +5 m/s is +8 m/s over 4 s, giving +2 m/s². This average does not by itself prove acceleration was constant within the interval. The units mean a velocity change of meters per second for every second, not a distance traveled per second squared.
  2. Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.
Answer: Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.
Check the reasoning

Original signed-velocity example; no within-interval history is supplied.

Avoid these mistakes

Check your understanding

v changes−2 to−8 m/s in 3 s. Find acceleration and speed change.

Discuss the answer

a=(−8+2)/3=−2 m/s²; speed increases because velocity and acceleration point together.

Return to the phenomenon

Return to the initial case. Does negative acceleration always mean an object is slowing? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Speed falls before the reversal and rises afterward; negative identifies an axis direction, not a universal slowing rule.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Does negative acceleration always mean an object is slowing? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Speed falls before the reversal and rises afterward; negative identifies an axis direction, not a universal slowing rule.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A cart changes from +8 to −4 m/s in 6 s under constant acceleration. Find acceleration, reversal time and displacement.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

a=(−4−8)/6=−2 m/s². v=0 at 4 s. Displacement=8×6+½(−2)×6²=12 m; positive net displacement does not mean it moved east for the entire interval.

Words to know

average velocity
Displacement divided by the time interval: v_avg = Δx ÷ Δt. A vector, so it can be zero even after a long trip.
instantaneous velocity
The velocity at a single instant — the value on a speedometer paired with a direction; the slope of a position-time graph at that point.
average speed
Total distance divided by total time; a scalar that is never negative.
acceleration
The rate at which velocity changes: a = (v − v₀) ÷ t. A vector, measured in m/s².
free fall
Motion under the influence of gravity alone, with air resistance neglected.
g
The acceleration due to gravity near Earth's surface, about 9.8 m/s² directed downward, the same for every freely falling object regardless of mass.

Instantaneous velocity and the invention of calculus

Average velocity over a whole trip is easy, but 'how fast am I going at this exact instant?' is a surprisingly deep question — it asks for Δx ÷ Δt as the time interval shrinks toward zero, which is a division of nearly-nothing by nearly-nothing. Wrestling with exactly this problem led Isaac Newton and Gottfried Leibniz, working independently in the late 1600 s, to invent calculus. Instantaneous velocity is the derivative of position with respect to time, and instantaneous acceleration is the derivative of velocity. Your car's speedometer is, in effect, computing a derivative in real time — reading how fast position is changing at this very moment, not averaged over the trip.

Why g is the same for a bowling ball and a feather

It is genuinely strange that heavy and light objects fall at the same rate. A heavier object is pulled harder by gravity — but it also has more mass resisting that pull, and the two effects cancel exactly, leaving the same acceleration g for everything. This equivalence puzzled physicists for centuries until Einstein made it the cornerstone of general relativity. In everyday life air resistance hides the effect (a feather really does flutter down slowly), but remove the air and the trick vanishes: in a vacuum chamber, a feather and a bowling ball dropped together hit the bottom at the same instant, a demonstration you can watch physicists perform today.

The Leaning Tower legend and Galileo's real experiments

The famous story of Galileo dropping two balls from the Leaning Tower of Pisa is probably a myth, but the science behind it is real. Galileo couldn't time fast free fall with the clocks of his day, so he cleverly slowed gravity down by rolling balls down gentle ramps, where he could measure that the distance covered grew with the square of the time — the signature of constant acceleration. His conclusion overturned Aristotle's 2,000-year-old claim that heavier objects fall faster. Centuries later, in 1971, Apollo 15's David Scott closed the loop by dropping a hammer and a feather on the airless Moon: they landed together, on live television.

Negative acceleration is not the same as slowing down

A common misconception is that negative acceleration always means 'slowing down.' In truth, the sign of acceleration only tells you which way the velocity is changing, not whether speed is increasing or decreasing. What matters is how the acceleration's direction compares to the velocity's. A ball thrown upward has a negative (downward) acceleration the whole time — but on the way up that slows it down, while on the way down that speeds it up. An object speeds up whenever acceleration points the same way as velocity, and slows down when they point opposite ways, regardless of the raw sign. Careful sign bookkeeping, with a clearly chosen positive direction, is what keeps kinematics problems honest.

In the real world

A car's speedometer displays instantaneous speed; pairing it with the heading from a compass or GPS gives instantaneous velocity — exactly the quantity a self-driving car's computer updates dozens of times a second to decide when to brake or turn.

Exit check

v changes−2 to−8 m/s in 3 s. Find acceleration and speed change.

Check your explanation after answering

a=(−8+2)/3=−2 m/s²; speed increases because velocity and acceleration point together.

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

Problem 1

A sprinter covers a 100 m straight-line displacement in 12.5 s. What is her average velocity?
Answer and explanation

8 m/s

Average velocity = displacement ÷ time = 100 ÷ 12.5 = 8 m/s.

Problem 2

A car speeds up from 10 m/s to 22 m/s in 4.0 s. Find its acceleration.
Answer and explanation

3 m/s²

a = (v − v₀) ÷ t = (22 − 10) ÷ 4 = 12 ÷ 4 = 3 m/s².

Problem 3

At the very top of its path, a ball thrown straight up momentarily has zero velocity. What is its acceleration at that instant?
  1. 0 m/s², since it is momentarily at rest
  2. 9.8 m/s² directed downward
  3. 9.8 m/s² directed upward
  4. It cannot be determined
Answer and explanation

9.8 m/s² directed downward

Gravity acts the whole time, so the acceleration stays 9.8 m/s² downward even when the velocity is instantaneously zero.

Problem 4

A rock is dropped from rest and falls freely for 5.0 s. How fast is it moving at the end of that time? (g = 9.8 m/s²)
Answer and explanation

49 m/s

v = v₀ + gt = 0 + 9.8 × 5.0 = 49 m/s.

Problem 5

A car moving at 28 m/s brakes smoothly to 4 m/s in 3.0 s. Find its acceleration, including the sign.
Answer and explanation

-8 m/s²

a = (v − v₀) ÷ t = (4 − 28) ÷ 3 = −24 ÷ 3 = −8 m/s². Here the acceleration is opposite the stated positive velocity, so speed decreases during the interval. A negative acceleration does not mean slowing in every motion case.

Problem 6

Explain the difference between average velocity and instantaneous velocity, using a road trip as your example.
Answer and explanation

Average velocity is the total displacement divided by the total time for the whole trip — it smooths everything into one number, like driving 270 km north in 3 hours for an average velocity of 90 km/h north. Instantaneous velocity is how fast (and which way) you are moving at one specific moment, like the 120 km/h your speedometer reads on the open highway or the 0 km/h at a rest stop. The average can hide big swings in the instantaneous value.

Problem 7

Velocity changes from 3 to 15 m/s in 4 s. Find average acceleration. Show the reasoning, not just the conclusion.
Hints

Acceleration sign alone does not say whether speed increases.

Answer and explanation

(15−3)/4=3 m/s².

Problem 8

Can average velocity be zero while average speed is positive?
Hints

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

Answer and explanation

Yes. A return trip has zero displacement but nonzero traveled distance.

Problem 9

Subtract signed velocities before dividing by timeSubtract signed velocities before dividing by timev₀ = −3 m/sv = +5 m/sΔt = 4 saavg = [5 − (−3)]/4 = +2 m/s²Velocity change is +8 m/s, not +2 m/s.
Original signed-velocity example; no within-interval history is supplied.
Interpret the supplied visual. Would a different nonlinear velocity history with these endpoints have the same average acceleration?
Answer and explanation

Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.

Average acceleration is Δv/Δt, so the numerator is final velocity minus initial velocity with signs retained. Here the change from −3 to +5 m/s is +8 m/s over 4 s, giving +2 m/s². This average does not by itself prove acceleration was constant within the interval. The units mean a velocity change of meters per second for every second, not a distance traveled per second squared.

Yes. The endpoint average is still +2 m/s², while instantaneous acceleration could vary.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A cart changes from +8 to −4 m/s in 6 s under constant acceleration. Find acceleration, reversal time and displacement.
Answer and explanation

a=(−4−8)/6=−2 m/s². v=0 at 4 s. Displacement=8×6+½(−2)×6²=12 m; positive net displacement does not mean it moved east for the entire interval.

a=(−4−8)/6=−2 m/s². v=0 at 4 s.

Displacement=8×6+½(−2)×6²=12 m; positive net displacement does not mean it moved east for the entire interval.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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