Stand Up & Math · Teacher Answers
The Getaway Car
Teacher Answers
Not for distribution · AI HL 5.5 · 5.8 · Full worked solutions + facilitation notes
"Your team are forensic analysts. Two nights ago, a car fled a scene and led police on a 60-second chase before it was boxed in at an intersection. The black box survived. It recorded the car's speed every 10 seconds, nothing else, no distance, no GPS. The court has one question for you: how far did this car actually travel? You don't get a formula. You get seven speed readings and your own thinking."
Give RECT01 only once a group has both a left total and a right total, and can explain out loud why the two differ, not just written on the board. Give TRAP02 once a group has found the trapezoid total and noticed, before you tell them, that it sits exactly halfway between their two Phase 1 totals; if they haven't spotted it, prompt with "look at your three numbers together" rather than stating the relationship. Give ZERO03 once a group can describe in their own words what happens as slice width shrinks to zero, before they see the formal integral notation confirmed. Give CASE04 once every group has evaluated the integral on their own GDC using their own regression equation; small variation in coefficients between groups is expected and fine.
Give RECT01 once both totals are found and the gap is explained aloud, ~15 min on VNPS
| t (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| v (m/s) | 0 | 12 | 22 | 28 | 26 | 18 | 8 |
Average of the seven readings: \(\dfrac{0+12+22+28+26+18+8}{7} = \dfrac{114}{7} \approx 16.29\) m/s. Multiplied by 60 s gives roughly 977 m.
This method quietly treats every reading as equally representative of the whole 60 seconds, but the two endpoint readings (\(t=0\) and \(t=60\)) really only "belong" to one 10-second slice each, while each interior reading is shared between two neighbouring slices. A correctly weighted time-average speed (see Task 2d) works out closer to \(1100 \div 60 \approx 18.33\) m/s, noticeably higher. The instantaneous readings alone, averaged naively, undercount the interior speeds that matter twice as much.
Using the left (earlier) reading of each 10-second slice as a constant height:
Using the right (later) reading of each slice:
Right > Left here. On a rising slice the right-hand height overestimates and the left-hand height underestimates; the reverse happens on a falling slice. There are exactly three rising slices (\(0\to10\to20\to30\)) and three falling slices (\(30\to40\to50\to60\)), but the rises (+12, +10, +6) outweigh the falls (\(-2, -8, -10\)) in total, so the right sum comes out ahead.
Bonus pattern worth drawing out: because every interior reading cancels between the two sums, \(\text{Right} - \text{Left} = \Delta t\,(v_6 - v_0) = 10 \times (8-0) = 80\), exactly matching \(1140-1060\). This telescoping fact generalizes to any data set and previews why the gap must shrink as \(\Delta t \to 0\) (Task 3a).
With a reading every 1 second there would be 60 rectangles instead of 6, each only 1 second wide. Every individual rectangle's error (the sliver of mismatch between the flat top and the true curve) shrinks, and because the left/right gap is \(\Delta t(v_n-v_0)\), a smaller \(\Delta t\) directly shrinks the gap between the two totals. This is the seed of Phase 3: push \(\Delta t \to 0\) and the gap disappears entirely.
Give TRAP02 once the halfway relationship is noticed, ~12 min on VNPS
Slice-by-slice average heights: 6, 17, 25, 27, 22, 13 (m/s). Distances (×10 s each): 60, 170, 250, 270, 220, 130 m.
The shape under each slice is now a trapezoid: two parallel vertical sides of different heights (the two readings) joined by a straight slanted top, instead of a flat rectangle top.
1100 sits exactly halfway between 1060 and 1140: \(\frac{1060+1140}{2}=1100\). This is the key "aha" of the phase, worth letting groups discover before confirming.
With \(\Delta t = 10\) and readings \(v_0,\dots,v_6\):
Adding the two and dividing by 2:
which is exactly the sum of the six slice-by-slice trapezoid areas from Task 2a, since every interior reading \(v_1,\dots,v_5\) is shared between two neighbouring trapezoids and so is counted twice, while the two endpoints \(v_0\) and \(v_6\) are each counted once. This holds for any readings, any \(n\), not just this data set.
General form for \(n\) equal slices of width \(\Delta t\):
Every interior reading is doubled; only the two endpoints are counted once. Checked against our data: \(T=\frac{10}{2}[0+2(106)+8]=5(220)=1100\) m, matching Task 2a exactly.
Give ZERO03 once the shrinking-slice idea is verbalized, ~20 min, GDC needed
As \(\Delta t \to 0\) and the number of slices \(n \to \infty\), the left and right sums both converge to the same exact value, since their gap, \(\Delta t (v_n-v_0)\), shrinks to zero along with \(\Delta t\). Every method (left, right, trapezoid) becomes exact in this limit; the accuracy is bounded only by how well the assumed shape (flat top, sloped top) matches the true curve inside a vanishingly small slice.
Accept any answer along the lines of: "add up \(v(t)\) times an infinitesimally thin slice of time \(dt\), for every instant \(t\) from 0 to 60, using infinitely many such slices." This is the Riemann sum of Phase 1, taken to its limit. Push groups to say it without leaning on the word "area" here; that framing gets tested directly in Task 4d.
Quadratic regression on the seven \((t,v)\) points gives:
Small variation between groups' coefficients (rounding, calculator model) is expected and fine; the shape and \(R^2\) should be consistent.
Using the unrounded regression coefficients, \(\displaystyle\int_0^{60}v(t)\,dt \approx 1131\) m. Groups using the rounded equation above will typically land within about ±10 m of this. This is the team's best available estimate of the true distance.
Give CASE04 once every group has evaluated their own integral, ~13 min
Left (1060 m) falls short of 1100 m by 40 m. Right (1140 m) clears it by 40 m. The trapezoid total (1100 m) lands exactly on the threshold, genuinely ambiguous on its own. The integral estimate (≈1131 m) clears it by roughly 31 m.
The point to draw out: a single rough method can sit right on top of a real decision threshold and give no confident answer either way. The more refined, technology-supported estimate is the one a court could actually rely on.
\(R^2\approx0.99\) means the quadratic explains about 99% of the variation in the readings, an excellent but imperfect fit. The model disagrees most at \(t=0\) (predicting a slightly negative, physically impossible speed) and near \(t=30\) (predicting about 26.6 m/s against an actual reading of 28). Real cars don't accelerate and brake along a perfectly smooth curve: gear changes, road surface, driver reaction time, and panic braking all introduce small kinks a single quadratic can't capture.
If the driver had braked and accelerated repeatedly inside a single 10-second gap, a method sampling only every 10 seconds would never "see" that behaviour, since it assumes the speed changes smoothly (or not at all) between readings. The estimate could be badly wrong in either direction. Trusting a numerical estimate for a genuinely erratic function needs either a much shorter \(\Delta t\) (far more frequent sampling) or a continuously logged source, exactly what the engine control module supplied in Phase 3.
Model answer: the integral is adding up infinitely many infinitesimally small pieces, each one a speed \(v(t)\) multiplied by an infinitesimally thin slice of time \(dt\), across every instant from 0 to 60. Because every piece's width is allowed to shrink all the way to zero while the number of pieces grows without bound, none of the leftover error from a finite-width rectangle or trapezoid survives; the sum becomes exact rather than approximate. This is the core idea the whole activity was built to surface, look for it in students' own words, not the phrase itself.