Two answers from the same curve, and the split that is the whole method.
Ask how far it travelled, and accept both answers.
Run it to t = 4 and ask the class how far the particle has gone. Some will say 1.33, some 4. Both are right, to different questions, and that argument is the lesson arriving on its own.
The green and red shading does the rest: red counts backwards for displacement and forwards for distance, and the two readouts separate as soon as the particle turns.
Kinematics is not set at Standard Level on this course, which is why the SL pages avoid it entirely and say so.
If you teach a mixed group, be explicit about who needs this. SL students sitting through it is time they do not have.
| v = t² − 4t + 3 | = (t − 1)(t − 3), so at rest at t = 1 and t = 3, negative between them. |
| Displacement, 0 to 4 | ∫v dt = 4/3 ≈ 1.33 m. |
| Total distance, 0 to 4 | ∫|v| dt = 4 m. Each of the three legs is 4/3. |
| Acceleration | a = 2t − 4, zero at t = 2, and a(0) = −4. |
| 1. Smaller time at rest | 1. |
| 2. Which statement | B. Displacement 4/3, distance 4. |
| 3. a at t = 0 | −4 m per second squared. |
1 markSolving v = 0 to find where the motion reverses.
1 markSplitting the integral there, which is the method mark on any distance question.
1 markEvaluating each leg and adding the sizes.
1 markAnswering the quantity actually asked for, with units.
| They give | What it means |
|---|---|
| Distance = 4/3 | Did not split. They integrated v over the whole interval and got the displacement. The single commonest error here. |
| t = 2 (Q1) | Confused zero acceleration with zero velocity. Worth separating explicitly; questions exploit it. |
| "Displacement is zero" (Q2) | Stopped thinking at t = 3, where it genuinely is zero, and did not carry on to t = 4. |
| 3 (Q3) | Gave v(0), not a(0). Differentiate before substituting. |
| Negative speed | Speed is a magnitude. Free mark to lose. |
"Can distance be less than displacement?" Never. Distance is at least the size of the displacement, and equal only if the direction never reverses. A useful sanity check on their own answer.
"Is deceleration just negative acceleration?" Not quite. Decelerating means speed is falling, which happens when a and v have opposite signs. At t = 0 here, a is negative while v is positive, so it is slowing.
"Can I just integrate the modulus on the calculator?" Yes and it is a good check, but show the split or the method marks go.
| What is happening | |
|---|---|
| 1 | Run the animation. Collect both answers for "how far" and let the disagreement stand. |
| 2 | s, v, a and the two directions between them. Speed against velocity. |
| 3 | The two integrals, and why the modulus means split. |
| 4 | The worked example with the leg table built on the board. |
| 5 | Questions 1 to 3. |
| 6 | Acceleration, and the t = 2 trap. |
Do not use the words distance and displacement interchangeably while demonstrating. Students copy the looseness and then cannot tell which one a question wants.
Do not let them integrate a modulus symbolically. Find the zeros, split, add the sizes: that is the method and it is what is marked.
Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.
The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.
The same skill inside a real situation, where the first job is working out what is being asked.
Reasoning, working backwards, or spotting an error. These are where the top grades are decided.
Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.