1. 0-2, 4-6
2. 2-3 & 5-6
3. ~0.67 s
4. 6 m/s, 0, -3 m/s
5. Change after 3 seconds
Distance: 18 m
Displacement: 6 m
Avg. Speed: 3 m/s
Avg. Velocity: 1 m/s North
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AREA UNDER A VELOCITY–TIME GRAPH — insertable slide block
Unit 1, Kinematics. Pairs with area_under_vt_graph.pdf (13 Q, 11 pp).
HOW IT RUNS:
SLIDES 1–3 set the activity up, before the worksheet goes out
SLIDES 4–6 how to draw the low / high / midpoint rectangles —
show when the class reaches that part of Q5
SLIDES 7–25 the summary: what they did and what they found.
One continuous run, AFTER the worksheet is finished.
Slides 7 on are a recap, not a reveal. They are written to be read back
to a class that has already done the work, in the order they did it.
Figures are the vt_*.svg files in this folder.
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Deliberately NOT titled "area under the curve." The title is the
question students are about to answer for themselves.
Warm-up, 2 minutes, everyone gets it. Do NOT point out that 24 is also
the rectangle's area — the worksheet walks them into that in Q1(c)–(d).
Say the last line out loud, then hand out the worksheet.
The next three slides are for the moment the class reaches the bracketing
in Q5. Everything after those waits until the worksheet is done.
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HOW TO DRAW THE RECTANGLES — show when the class reaches the
bracketing in Q5, to the whole room or one group at the board.
This is technique, not the answer: the demo line is not the Q5 line,
it starts above zero and its axes carry no numbers, so nothing here
can be copied into the table.
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This is the step the first group could not find on their own. They were
trying to draw a rectangle before deciding what its height should be.
Ask: "over this one slice, what is the slowest the cart ever goes?"
Nothing here reveals the answer — the demo line is not the Q5 line.
The point to land: you are not choosing a good estimate, you are choosing a
guaranteed-wrong one, twice, in known directions. That is what a bracket is.
Step 5 is worth saying twice: every rectangle in a round has the SAME width,
so the sum is (sum of heights) x delta-t. Groups that miss this multiply
each height separately and lose the thread in arithmetic.
The falling-line note is for Q6, and for Object A in Part C.
THE SUMMARY STARTS HERE — run slides 7 to 25 straight through once the
worksheet is done. If you collected group wordings from Q7, read one or
two aloud here before moving on; the rest of the run answers them.
This is the slide worth slowing down on. Two follow-ups for the room:
- Panels are laid out to match the worksheet: position on the left,
velocity on the right, time axes aligned so you can read straight across.
- Ask: is the x–t graph steepest where the v–t graph is lowest? (No —
steepest where v is HIGHEST. Let them catch it themselves.)
- Slope of each x–t segment = 8, 2, 5 m/s. The velocities, handed back.
Q3. This is where the sign rule was actually built, and it was built on
RECTANGLES on purpose — no sloping line, no changing velocity, nothing else
to think about at the same time.
The question to re-ask here: "your question-2 rule said find the areas and add
them up. Which of these three is a problem for that?"
Then the payoff on the left: +12 and -12 cancel, so at t = 5 s the cart is back
where it started, and BOTH graphs say so — the position graph touches zero,
the velocity graph has equal areas on opposite sides.
Displacement over the 8 s is +6 m; distance travelled is 30 m.
This is the step the sequence used to skip. Students had done the slicing
without ever being told why slicing helps.
The argument in one breath: the formula is fine, it just needs a constant v;
so manufacture one, slice by slice, and choose it so you know which way you
are wrong. Everything after this is bookkeeping.
Four slices — one row of the table they filled in. This comes BEFORE the
squeeze on purpose: settle what the three methods are at a single slice count,
then show what refining does to them.
Ask before you advance: why is the third one exact?
Because on each slice the rectangle pokes ABOVE the line on the left half by
exactly as much as it falls BELOW on the right half — congruent triangles,
which is only true because the graph is a straight line.
The worksheet's "Midpoint" column averaged the two bounds; this slide uses the
mid-slice velocity instead. For a straight line they agree, and both give 36.
What this slide is FOR: it is the payoff of the previous one. Each rectangle is
still wrong, but the amount it is wrong by shrinks with the slice.
Ask for the two observations before you say anything:
(1) the gap halves (2) the midpoint never moves
The honest footnote to slide 13: the straight line was the easy case.
Ask what is DIFFERENT here. The answer worth extracting: the midpoint is no
longer exact. On a straight line the bit poking above the curve and the bit
missing below were congruent triangles and cancelled perfectly. On a curve they
only NEARLY cancel — 21.50, then 21.375, then 21.344, closing on 21.33 without
ever landing on it. Good enough is not the same as exact.
AP 1 does not ask for quantitative work on non-uniform acceleration, but the CED
does expect students to analyse such situations qualitatively — this slide is
that, and it is where the idea goes in calculus.
If a group's Q6 wording dropped "time axis" or dropped the sign, this is
the moment to show why each phrase is there rather than asserting it.
The sloping-line version of *Backwards Counts Too*. They built the sign rule
there, on rectangles; here it has to survive a triangle and a zero crossing.
Ask what changed, and settle on "nothing: only how much region there is and
which side of the axis it sits on."
This is the distinction that decays fastest — plan to re-ask it cold on
the warm-up two days from now.
The hinge of the summary: everything from here is Part C, and the point is
that none of it needed a new rule. Ask what they had to do differently.
The honest answer is "nothing, except watch the sign."
Read it as a matrix, not four unrelated pictures:
- ROW picks the sign of the acceleration, and that sets which way
every curve in the row bends.
- COLUMN picks slowing vs. speeding, which is about the SIGNS MATCHING,
not about the sign of a on its own.
The velocity graph on the right is the whole row at once — left of the
dashed line it is slowing, right of it, speeding up. Same straight line.
# Sketch and Translate
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# Simplify and Diagram
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# Represent ➡️ Solve:
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REDRAWN versions of the three handwritten figures above — same
worked example, same four steps, rebuilt in the style of the
vt_*.svg figures from the area-under-a-VT-graph block.
Here for comparison: keep one set and delete the other three.
─────────────────────────────────────────────────────────────────
Linger on the middle panel. That shaded triangle is the displacement,
and it is the same area rule they built for themselves earlier —
handed back to them inside a problem they are now solving.
Left column is slope (v-t graph), right column is area (v-t graph).
Same graph, two different tools — the "Two Graphs, Two Tools" slide,
cashed out on a real problem.
Don't explain it yet. Put it up, name the five, and ask what is missing
from the list (nothing is — that is the point). The payoff lands on the
next few slides when they see each equation is a loop around four of them.
The greyed-out variable is the one the equation cannot help you with.
Ask it as a question: "which one is missing from this loop?" before
you show the equation underneath.
The right-hand one is the odd shape: a ring that goes around 'a'.
That is the trapezoid-area equation from the v-t graph — the one they
built themselves back in the area block. Worth naming out loud.
Known: v0, vf, dx. Wanted: dt. That is four quantities, and the one
left out is 'a' — so it is the ring equation, dx = 1/2(v0 + v)dt.
Let them find that from the cross rather than telling them.