Mathematics Standard • Year 11 • Module 1 • Lesson 7
Quadratic Formulas in Applied Contexts
Practise HSC-style writing on quadratic models — three multi-mark short answers and one extended response with marking criteria.
1. Short-answer questions
1.1 Use D = 0.01v² + 0.3v to estimate the stopping distance when v = 70 km/h. Show your substitution with brackets and give the answer in metres. 3 marks Band 3
1.2 Use A = s² to find the area of a square paving slab with side length s = 8.5 m. Round to 1 d.p. and include units. 3 marks Band 3-4
1.3 The supplied model for the height of a flare is h = 1.5 + 20t − 5t², where h is in metres and t in seconds.
(a) Calculate h when t = 3 s.
(b) State whether the flare is rising or falling at t = 3 s, using your value from (a) and an earlier value at t = 2 s (you may use h(2) = 21.5 m).
(c) Explain in one sentence why a constant-rate (linear) model would not describe a flare's height accurately. 4 marks Band 4
2. Extended response
2.1 A road-safety pamphlet uses two supplied models for vehicle stopping distance.
Reaction distance: R = 0.3v (metres), where v is speed in km/h.
Braking distance: B = 0.01v² (metres).
Total stopping distance: D = R + B.
(a) Calculate R, B and D for a driver travelling at v = 50 km/h.
(b) Calculate R, B and D for the same driver at v = 100 km/h.
(c) State by what factor each of R, B and D has changed from v = 50 to v = 100, and explain in one sentence why the factor for B is different from the factor for R.
(d) A car needs at least D = 80 m of clear road to stop safely. Use B = 0.01v² and R = 0.3v to set up an inequality 0.01v² + 0.3v ≤ 80, then find by trial the largest whole speed v (km/h) for which the car can still stop in 80 m. State your conclusion as a safe upper-speed recommendation. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — R = 15, B = 25, D = 40 m at v = 50.
Part (b) — 1 mark
• 1 mark — R = 30, B = 100, D = 130 m at v = 100.
Part (c) — 2 marks
• 1 mark — R changes by factor 2; B by factor 4; D by factor 3.25.
• 1 mark — explanation refers to v² in B (doubling v squares to 4 ×).
Part (d) — 3 marks
• 1 mark — writes the inequality 0.01v² + 0.3v ≤ 80.
• 1 mark — trials values (e.g. v = 80: D = 64 + 24 = 88 too far; v = 75: D = 56.25 + 22.5 = 78.75 OK).
• 1 mark — clear conclusion: largest safe whole speed is 75 km/h (or equivalent rounded answer).
Your response:
Stuck on (d)? Trial v = 80 first (you'll get D = 88, too far). Lower v in steps of 5 until D ≤ 80.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Stopping distance at 70 km/h (3 marks)
Sample response.
D = 0.01(70)² + 0.3(70) = 0.01(4900) + 21 = 49 + 21 = 70 m.
Marking notes. 1 mark — bracketed substitution. 1 mark — (70)² = 4900 evaluated correctly. 1 mark — final 70 m with units. A response missing brackets but reaching 70 m scores 2/3.
1.2 — Square paver area (3 marks)
Sample response.
A = s² = (8.5)² = 72.25 ≈ 72.3 m² (to 1 d.p.).
Marking notes. 1 mark — correct substitution. 1 mark — exact value 72.25. 1 mark — rounded to 72.3 m² with the correct square unit.
1.3 — Flare height model (4 marks)
(a) Sample response. h(3) = 1.5 + 20(3) − 5(3)² = 1.5 + 60 − 5(9) = 1.5 + 60 − 45 = 16.5 m.
(b) Sample response. h(2) = 21.5 m and h(3) = 16.5 m. The height has decreased between t = 2 and t = 3, so the flare is falling at t = 3 s.
(c) Sample response. A linear model has a constant rate of change, so it cannot capture a height that first rises and then falls — only a quadratic (with the −5t² term) can model the curve.
Marking notes. 1 mark — correct h(3). 1 mark — squaring done before multiplying. 1 mark — direction stated with evidence (compares to t = 2). 1 mark — explanation references quadratic / non-constant rate.
2.1 — Road safety stopping distance (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) v = 50 km/h.
R = 0.3(50) = 15 m. B = 0.01(50)² = 0.01(2500) = 25 m. D = 15 + 25 = 40 m. [1 mark.]
(b) v = 100 km/h.
R = 0.3(100) = 30 m. B = 0.01(100)² = 0.01(10 000) = 100 m. D = 30 + 100 = 130 m. [1 mark.]
(c) Comparison.
R goes 15 → 30 (factor 2). B goes 25 → 100 (factor 4). D goes 40 → 130 (factor 3.25). [1 mark — factors stated.]
The factor for B is 4 (not 2) because B contains v²: when v doubles, v² quadruples, so the braking distance grows much faster than the reaction distance. [1 mark — v² explanation.]
(d) Safe upper-speed search.
Set up: 0.01v² + 0.3v ≤ 80. [1 mark — correct inequality.]
Trial values:
v = 80: D = 0.01(6400) + 0.3(80) = 64 + 24 = 88 m (too far).
v = 75: D = 0.01(5625) + 22.5 = 56.25 + 22.5 = 78.75 m (under 80 — OK).
v = 76: D = 57.76 + 22.8 = 80.56 m (just over 80). [1 mark — trials shown.]
Conclusion: The largest whole-number speed that still stops within 80 m is 75 km/h. The pamphlet should recommend no more than 75 km/h on this stretch of road. [1 mark — clear contextual conclusion.]
Total: 7/7.
Band descriptors for marker.
Band 3: Substitutes correctly for one speed but does not split R and B, or omits units. ≈ 2 marks.
Band 4: All values in (a) and (b) correct; factors quoted in (c) but no v² explanation. ≈ 4 marks.
Band 5: Inequality set up and one trial shown, but conclusion vague or off by one (e.g. "80 km/h" without checking). ≈ 5-6 marks.
Band 6: Complete and correct, with conclusion tied back to a realistic speed recommendation. 7/7.