Accumulation can cancel without erasing motion
Why can displacement be zero while distance traveled is positive?
Entering midway? Velocity is a signed rate. Multiplying a constant rate by time gives a change; an integral accumulates a changing rate.
Start with a concrete question
A one-dimensional object has velocity v(t)=t−turn. It can begin moving backward, stop, then move forward.
Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.
Go directly to the experiment · Review prerequisite mathematics and return here
What do you expect?
At turn=2 and end time=4, predict displacement and total distance separately.
Change one thing. Inspect what happens.
- Split the default graph at its zero crossing and calculate both triangular areas.
- Compare end times before, at and after the direction change.
- Set end time to zero and verify zero accumulation.
- Explain how differentiating the position-change formula recovers velocity.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Velocity crosses zero; accumulation retains its sign
Horizontal: Time (s). Vertical: Velocity (m/s).
- 1. Velocity (solid line)
- 2. Zero velocity (dashed line)
Axes use the displayed ranges.
Read plotted values as a table
| Series | Time (s) | Velocity (m/s) |
|---|---|---|
| Velocity | 0 | -2 |
| Velocity | 0.06667 | -1.933 |
| Velocity | 0.1333 | -1.867 |
| Velocity | 0.2 | -1.8 |
| Velocity | 0.2667 | -1.733 |
| Velocity | 0.3333 | -1.667 |
| Velocity | 0.4 | -1.6 |
| Velocity | 0.4667 | -1.533 |
| Velocity | 0.5333 | -1.467 |
| Velocity | 0.6 | -1.4 |
| Velocity | 0.6667 | -1.333 |
| Velocity | 0.7333 | -1.267 |
| Velocity | 0.8 | -1.2 |
| Velocity | 0.8667 | -1.133 |
| Velocity | 0.9333 | -1.067 |
| Velocity | 1 | -1 |
| Velocity | 1.067 | -0.9333 |
| Velocity | 1.133 | -0.8667 |
| Velocity | 1.2 | -0.8 |
| Velocity | 1.267 | -0.7333 |
| Velocity | 1.333 | -0.6667 |
| Velocity | 1.4 | -0.6 |
| Velocity | 1.467 | -0.5333 |
| Velocity | 1.533 | -0.4667 |
| Velocity | 1.6 | -0.4 |
| Velocity | 1.667 | -0.3333 |
| Velocity | 1.733 | -0.2667 |
| Velocity | 1.8 | -0.2 |
| Velocity | 1.867 | -0.1333 |
| Velocity | 1.933 | -0.06667 |
| Velocity | 2 | 0 |
| Velocity | 2.067 | 0.06667 |
| Velocity | 2.133 | 0.1333 |
| Velocity | 2.2 | 0.2 |
| Velocity | 2.267 | 0.2667 |
| Velocity | 2.333 | 0.3333 |
| Velocity | 2.4 | 0.4 |
| Velocity | 2.467 | 0.4667 |
| Velocity | 2.533 | 0.5333 |
| Velocity | 2.6 | 0.6 |
| Velocity | 2.667 | 0.6667 |
| Velocity | 2.733 | 0.7333 |
| Velocity | 2.8 | 0.8 |
| Velocity | 2.867 | 0.8667 |
| Velocity | 2.933 | 0.9333 |
| Velocity | 3 | 1 |
| Velocity | 3.067 | 1.067 |
| Velocity | 3.133 | 1.133 |
| Velocity | 3.2 | 1.2 |
| Velocity | 3.267 | 1.267 |
| Velocity | 3.333 | 1.333 |
| Velocity | 3.4 | 1.4 |
| Velocity | 3.467 | 1.467 |
| Velocity | 3.533 | 1.533 |
| Velocity | 3.6 | 1.6 |
| Velocity | 3.667 | 1.667 |
| Velocity | 3.733 | 1.733 |
| Velocity | 3.8 | 1.8 |
| Velocity | 3.867 | 1.867 |
| Velocity | 3.933 | 1.933 |
| Velocity | 4 | 2 |
| Zero velocity | 0 | 0 |
| Zero velocity | 6 | 0 |
Follow the calculation
- Velocity v(t)=t−2 m/s on 0≤t≤4; negative velocity means movement in the negative direction.
- An accumulated position change is F(t)=t²/2−2t.
- Signed integral = F(4)−F(0) = 8−8 = 0 m.
- Area below zero has magnitude 2; area above zero 2.
- Distance adds magnitudes: 2+2=4; displacement subtracts them.
An integral of velocity gives signed displacement. Integrating speed |velocity| gives distance. Cancellation in displacement does not mean no movement occurred.
Captures keep your inputs, prediction and observation. Results are recalculated from those inputs. Resetting inputs keeps your captures and challenge history.
Explain the mechanism
Start with the meaning
Velocity is a signed rate. Multiplying a constant rate by time gives a change; an integral accumulates a changing rate.
A one-dimensional object has velocity v(t)=t−turn. It can begin moving backward, stop, then move forward.
A worked explanation
An integral is signed accumulation. Areas below the horizontal axis subtract; areas above add.
With v(t)=t−2 over zero to four seconds, the negative triangle has magnitude 2 m and the positive triangle 2 m. Displacement is zero; total distance is 4 m.
The antiderivative F(t)=t²/2−2t has derivative v(t). Evaluating F(end)−F(start) connects instantaneous rate with total change: the fundamental theorem of calculus in this continuous example.
Symbols and relationship
Displacement=∫₀ᵀ(t−a)dt=T²/2−aT. Distance=∫₀ᵀ|t−a|dt. a and T are seconds; velocity is m/s under the chosen unit acceleration.
Use the graph and trace together
The graph and calculations use your current inputs. Read axis labels and units before comparing positions or slopes. Change one input, predict the effect, and use the numerical trace to verify the plotted behavior. A displayed curve is an illustration; the formula and its assumptions determine which conclusions are justified.
Scope and edge cases
- Programmed constant acceleration in one dimension; no force or collision model.
- Signed area is not the same as geometric area; distance requires splitting at a sign change.
Try a fresh case
Velocity is v(t)=2t−2 m/s from t=0 to t=3 s. What is signed displacement?
This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.
Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Compute distance and displacement from triangular regions.
College investigation
Extend to a piecewise rate and verify the fundamental theorem on each continuous interval.
Workplace investigation
Audit a rate dashboard that mistakes signed net change for total activity.
Keep the evidence, choose your next step
Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.