Mathematics Standard • Year 11 • Module 1 • Lesson 7
Quadratic Formulas in Applied Contexts
Apply supplied quadratic formulas to real Australian contexts — driver stopping distances, paddock areas, projectile heights and a garden upgrade.
Problem 1 — School-zone stopping distances
Driver-education materials use D = 0.01v² + 0.3v, where v is the speed in km/h and D is the stopping distance in metres.
Set up: What are we solving for?
(i) Calculate D when v = 40 km/h (the school-zone limit). 2 marks
(ii) Calculate D when v = 50 km/h (suburban limit). 2 marks
(iii) A safety officer says "driving 10 km/h over the school-zone limit only adds 10 m to your stopping distance". Use your answers to (i) and (ii) to comment on whether this is reasonable. 2 marks
Stuck? Revisit lesson § Worked Example 2 — Compare two speeds. Compare the two D values, not the two v values.Problem 2 — Square paddock fence
A farmer is fencing a square paddock. The area model is A = s² (square metres) and the perimeter model is P = 4s (metres). Fencing costs $24 per metre.
Set up: What are we solving for?
(i) Calculate A when s = 25 m. 1 mark
(ii) Calculate the fencing cost for the same paddock. 2 marks
(iii) The farmer doubles the side length to s = 50 m. Calculate the new area and the new fence cost. State the factor by which each one increased. 3 marks
Stuck on (iii)? A = s² squares the side, so the area goes up by a factor of 4 when the side doubles, but the perimeter (and fence cost) only doubles.Problem 3 — Cricket ball height
A cricket coach uses the supplied model h = 1.5 + 20t − 5t², where h is the height in metres above the pitch and t is time in seconds after the ball leaves the bat.
Set up: What are we solving for?
(i) Calculate h when t = 1 s. 2 marks
(ii) Calculate h when t = 2 s. 2 marks
(iii) Calculate h when t = 4 s. State what your value tells you about where the ball is at that time (round to 1 d.p.). 2 marks
Stuck on (iii)? Square the 4 first, then multiply by −5. A negative or near-zero h means the ball has already landed.Problem 4 — Reaction vs braking distance
The full stopping distance has two parts: reaction distance R = 0.3v and braking distance B = 0.01v². The total is D = R + B (in metres) for v in km/h.
Set up: What are we solving for?
(i) Calculate R and B separately when v = 60 km/h. 2 marks
(ii) At v = 60 km/h, what percentage of the total stopping distance is the braking part? Round to the nearest %. 2 marks
(iii) Repeat (i) at v = 100 km/h (highway). State which part has grown more between 60 km/h and 100 km/h and explain why. 3 marks
Stuck on (iii)? B contains v², so braking distance grows faster than reaction distance as v increases.Problem 5 — Renovating a square courtyard
Mia is replacing pavers on a square courtyard with side length s = 4.6 m. New pavers cost $58 per square metre, and a fixed installation fee is $250.
C = 58 × s² + 250 (where C is total cost in dollars and s is side length in metres)
Set up: What are we solving for?
(i) Calculate C when s = 4.6 m. Round to the nearest dollar. 2 marks
(ii) Mia is considering a slightly larger courtyard with s = 5.0 m. Calculate C for that option and state the extra cost. 2 marks
(iii) Explain in one sentence why the extra cost in (ii) is not "$58 per extra metre of side length". 1 mark
Stuck? The cost depends on s² (area), not on s. Going from 4.6 m to 5.0 m adds more than 0.4 m² of pavers.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — School-zone stopping distance
Set up. Substitute each v into D = 0.01v² + 0.3v and compare.
(i) D(40) = 0.01(1600) + 0.3(40) = 16 + 12 = 28 m.
(ii) D(50) = 0.01(2500) + 0.3(50) = 25 + 15 = 40 m.
(iii) The actual increase is 40 − 28 = 12 m, slightly more than the "10 m" claim. The v² term means the extra 10 km/h adds more than just a linear amount, so 12 m is roughly in the right zone but the safety officer's rule is a slight under-estimate.
Problem 2 — Square paddock
Set up. Use A = s² for area and P = 4s for perimeter; fence cost = 24 × perimeter.
(i) A(25) = (25)² = 625 m².
(ii) Perimeter = 4(25) = 100 m. Cost = 24 × 100 = $2400.
(iii) New A(50) = (50)² = 2500 m² (4 × the old area). New perimeter = 200 m. New cost = 24 × 200 = $4800 (2 × the old cost). Doubling the side quadruples the area but only doubles the fence.
Problem 3 — Cricket ball height
Set up. Substitute t into h = 1.5 + 20t − 5t²; square first, then multiply.
(i) h(1) = 1.5 + 20(1) − 5(1) = 1.5 + 20 − 5 = 16.5 m.
(ii) h(2) = 1.5 + 20(2) − 5(4) = 1.5 + 40 − 20 = 21.5 m.
(iii) h(4) = 1.5 + 20(4) − 5(16) = 1.5 + 80 − 80 = 1.5 m. The ball is back near bat height — about to land (or has already in a real game).
Problem 4 — Reaction vs braking
Set up. R is linear in v; B is quadratic in v. Calculate both, then compare.
(i) At v = 60: R = 0.3(60) = 18 m; B = 0.01(60)² = 0.01(3600) = 36 m; D = 54 m.
(ii) Braking share = 36 / 54 × 100 ≈ 67 % of total.
(iii) At v = 100: R = 0.3(100) = 30 m; B = 0.01(100)² = 100 m. Reaction grew from 18 → 30 m (1.67 ×), but braking grew from 36 → 100 m (2.78 ×). The braking part grew more because it contains v², which scales faster than the linear reaction term.
Problem 5 — Courtyard upgrade
Set up. Substitute s into C = 58s² + 250.
(i) s² = (4.6)² = 21.16 m². C = 58(21.16) + 250 = 1227.28 + 250 = $1477.28 ≈ $1477.
(ii) s² = (5.0)² = 25 m². C = 58(25) + 250 = 1450 + 250 = $1700. Extra cost = 1700 − 1477 = $223.
(iii) Cost depends on the area (s²), not the side length. Going from 4.6 to 5.0 m adds (25 − 21.16) = 3.84 m² of pavers — far more than the 0.4 m of extra side would suggest.