riaology
Unit 4 · Lesson 4 of 5

Displacement can cancel while distance accumulates.

A cart reverses direction. Its signed displacement and total distance answer different questions.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its signed displacement.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase end time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Motion illustration: position -1.5. Velocity direction shown where applicable.Position -1.5 · arrow: velocity direction
Time (s) versus Velocity (m/s). Signed velocity. Selected signed displacement: -1.5 m. Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.-200224Velocity (m/s)Time (s)
Signed velocity

signed displacementΔs=−1.5 m,d=2.5 m\Delta s=-1.5\,\mathrm m,\quad d=2.5\,\mathrm m

Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.

Assumptions, units, and limits

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

v(t)=t−2,Δs=∫0Tv dt=T22−2T,d=∫0T∣v∣ dtv(t)=t-2,\quad\Delta s=\int_0^Tv\,dt=\frac{T^2}2-2T,\quad d=\int_0^T|v|\,dt
02

Keep the reasoning visible

distance=∫∣v∣ dt≥∣∫v dt∣\text{distance}=\int|v|\,dt\ge\left|\int v\,dt\right|

Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.

03

What makes this model useful?

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

Negative signed area is physically meaningful for net change. It is not negative traveled distance.

A worked example

Calculate the starting scenario

  1. Use end time = 3 s.
  2. Δs=−1.5 m,d=2.5 m\Delta s=-1.5\,\mathrm m,\quad d=2.5\,\mathrm m
  3. Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal. The checked quantity is signed displacement.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the signed displacement using end time 3 s. State the meaning of its sign and units.

Checked quantity: signed displacement · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

Explore the visual solution
Motion illustration: position -1.5. Velocity direction shown where applicable.Position -1.5 · arrow: velocity direction
Time (s) versus Velocity (m/s). Signed velocity. Selected signed displacement: -1.5 m. Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.-200224Velocity (m/s)Time (s)
Signed velocity

signed displacementΔs=−1.5 m,d=2.5 m\Delta s=-1.5\,\mathrm m,\quad d=2.5\,\mathrm m

Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.

Assumptions, units, and limits

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

v(t)=t−2,Δs=∫0Tv dt=T22−2T,d=∫0T∣v∣ dtv(t)=t-2,\quad\Delta s=\int_0^Tv\,dt=\frac{T^2}2-2T,\quad d=\int_0^T|v|\,dt
Calculate and explain

Apply the idea to a new scenario

End the motion at T=4T=4 s. Find displacement and compare with total distance.

Checked quantity: signed displacement · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

Explore the visual solution
Motion illustration: position 0. Velocity direction shown where applicable.Position 0 · arrow: velocity direction
Time (s) versus Velocity (m/s). Signed velocity. Selected signed displacement: 0 m. Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.-200224Velocity (m/s)Time (s)
Signed velocity

signed displacementΔs=0 m,d=4 m\Delta s=0\,\mathrm m,\quad d=4\,\mathrm m

Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.

Assumptions, units, and limits

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

v(t)=t−2,Δs=∫0Tv dt=T22−2T,d=∫0T∣v∣ dtv(t)=t-2,\quad\Delta s=\int_0^Tv\,dt=\frac{T^2}2-2T,\quad d=\int_0^T|v|\,dt
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Motion illustration: position -1.5. Velocity direction shown where applicable.Position -1.5 · arrow: velocity direction
Time (s) versus Velocity (m/s). Signed velocity. Selected signed displacement: -1.5 m. Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.-200224Velocity (m/s)Time (s)
Signed velocity

signed displacementΔs=−1.5 m,d=2.5 m\Delta s=-1.5\,\mathrm m,\quad d=2.5\,\mathrm m

Area below the axis counts negatively for displacement but positively for distance. Split the integral at the reversal.

Assumptions, units, and limits

One-dimensional motion reverses at 2 s. Distance uses speed and is never negative.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Integrate signed velocity for displacement. For distance, integrate its absolute value and split at zeros where direction changes. The same distinction applies to net inflow versus total throughput in other systems.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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