Teacher page: kinematics (AA 5.9)

The one thing to do with the figure

Ask how far it went, and accept both answers.

Run to t = 4 and ask the class. Some say 4, some say 12. Both are right, to different questions, and the argument is the lesson arriving by itself.

Here every leg is exactly 4, so the contrast is as clean as it gets: out, all the way back, out again.

This is Standard Level here

Kinematics is SL content on Analysis and Approaches. On Applications and Interpretation it is Higher Level only.

If you teach both courses, keep that straight: an AI SL class should not be sitting through this, and an AA SL class must not skip it.

The answers

v = 3(t − 1)(t − 3)	Zero at t = 1 and t = 3, negative between.
s = t³ − 6t² + 9t	s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4.
Displacement, 0 to 4	4 m.
Total distance	12 m, three legs of 4.
Acceleration	a = 6t − 12, zero at t = 2.
1, 2, 3	4, 12, 2.

Where the marks go

1 markSolving v = 0.

1 markSplitting the integral there.

1 markEvaluating each leg and adding the sizes.

1 markAnswering the quantity asked for, with units.

What each wrong answer tells you

They give	What it means
4 for the distance	Did not split. The defining error of the sub-topic.
t = 1 or 3 for zero acceleration	Confused v = 0 with a = 0. Questions exploit this deliberately.
Negative speed	Speed is a magnitude. A free mark to lose.
8 for the distance	Only two legs. The third, from t = 3 to 4, was missed.

Other things they will say

"Can distance be less than displacement?" Never. Equal only if the direction never reverses, which is a useful check on their own answer.

"Is deceleration negative acceleration?" Not quite. Decelerating means speed is falling, which happens when a and v have opposite signs.

"Can I integrate the modulus on the calculator?" Yes, as a check. Show the split or the method marks go.

A possible order

	What is happening
1	The animation. Collect both answers for "how far".
2	s, v, a, and speed against velocity.
3	The two integrals and why the modulus means split.
4	The worked example with the leg table built live.
5	Questions 1 to 3.
6	The t = 2 acceleration trap.

Two things not to say

Do not use distance and displacement loosely while demonstrating. Students copy the looseness and then cannot tell which a question wants.

Do not let them integrate a modulus symbolically. Find the zeros, split, add the sizes.

Questions to set

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.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

VelocityFor s = t³ − 9t² + 24t, find v and the times at rest.
v = 3t² − 18t + 24 = 3(t − 2)(t − 4), so rest at t = 2 and t = 4.
PositionsFind s at those times.
s(2) = 20 and s(4) = 16.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

AccelerationFind a and evaluate at t = 2 and t = 4.
a = 6t − 18, so a(2) = −6 and a(4) = 6.
DisplacementFind the displacement from t = 0 to t = 5.
s(5) − s(0) = 20 − 0 = 20 metres.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

DistanceFind the total distance travelled from t = 0 to t = 5.
The particle reverses at t = 2 and t = 4, so sum the legs: |20 − 0| + |16 − 20| + |20 − 16| = 28 metres, against a displacement of 20.
Read the two signsAt t = 3.5 the particle has v < 0 and a > 0. Describe the motion in words.
Moving backwards and slowing down, because the acceleration opposes the velocity. Positive acceleration means speeding up only when the velocity is also positive.

Practicalities

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.