Quadratic Formulas in Applied Contexts
Use supplied formulas with squared terms, calculate carefully, and interpret why outputs can grow faster than a constant-rate model.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A safety model estimates stopping distance with $D = 0.01v^2 + 0.3v$, where $v$ is speed in km/h. Why might doubling the speed more than double the stopping distance?
Without calculating — write your gut explanation first.
Applied quadratic formulas are supplied — your job is to substitute carefully and interpret the result. A squared term changes the growth pattern entirely.
Stopping distance: $D = 0.01v^2 + 0.3v$, where $D$ = stopping distance in metres and $v$ = speed in km/h.
Square area: $A = s^2$, where $A$ = area and $s$ = side length.
The key insight: $v^2$ means $v \times v$, not $2 \times v$. This is what makes the model nonlinear.
Key facts
- A quadratic formula contains a squared variable, such as $v^2$.
- Squaring is different from doubling.
- Rounding should usually happen at the end of a calculation.
Concepts
- Quadratic models can grow faster than linear models.
- The context determines whether the calculated value is reasonable.
- Technology or careful substitution can be used when formulas are supplied.
Skills
- Substitute into a formula containing a squared term.
- Calculate and compare outputs from a quadratic model.
- Interpret the result in context with suitable units.
A squared term means a value is multiplied by itself. This changes how quickly the output grows as the input increases.
If $v = 40$, then $v^2 = 40^2 = 1600$. It does not mean $2 \times 40$.
In the stopping-distance formula $D = 0.01v^2 + 0.3v$, the $0.01v^2$ part grows much faster than the $0.3v$ part as speed increases. This is why higher speeds are disproportionately dangerous.
What to write in your book
- Quadratic formula: contains a squared variable, e.g. $D = 0.01v^2 + 0.3v$.
- $v^2 = v \times v$ (not $2 \times v$). Example: $50^2 = 2500$, not 100.
- Stopping distance grows nonlinearly — the $v^2$ term makes it grow faster at higher speeds.
- Area of a square: $A = s^2$. Units must be square units.
True or false: If $v = 6$, then $v^2 = 12$.
When using a formula with decimals and squared terms, keep full calculator values until the final answer, unless the question gives a specific rounding instruction.
Rounding too early can shift the final answer, especially when comparing two model outputs.
What to write in your book
- Round only at the final step of a calculation.
- Early rounding can cause errors when comparing two outputs from the same model.
- Always include units in the final answer (e.g. metres for stopping distance, m² for area).
Quick check: When should you round an answer when using a formula?
Worked examples · 3 in a row, reveal as you go
Use $D = 0.01v^2 + 0.3v$ to estimate the stopping distance when $v = 50$ km/h.
Use $D = 0.01v^2 + 0.3v$ for $v = 40$ and $v = 80$. Does doubling speed double the stopping distance?
A square has side length $s = 7.5$ m. Use $A = s^2$ to find its area.
What to write in your book
- WE1: $v = 50$ km/h gives $D = 0.01(2500) + 15 = 40$ m.
- WE2: At 40 km/h, $D = 28$ m. At 80 km/h, $D = 88$ m. Doubling speed more than triples stopping distance.
- WE3: $A = (7.5)^2 = 56.25$ m². Area always uses square units.
Fill the gap: When $v = 60$ km/h, $v^2 =$ and $0.01 \times v^2 =$ .
Common errors · the 3 traps that cost marks
Quick-fire practice · 4 calculations
Use $D = 0.01v^2 + 0.3v$ to estimate stopping distance at 60 km/h.
Compare stopping distance at 30 km/h and 60 km/h. Does it double?
Use $A = s^2$ to find the area of a square with side length 12 cm.
Explain why $12^2$ is not the same as $2 \times 12$.
Match each value of $v$ to its correct $v^2$:
In your own words: Why does the stopping-distance model $D = 0.01v^2 + 0.3v$ grow faster than a linear model as speed increases?
Doubling speed can more than double stopping distance because the formula includes $v^2$. This makes the model nonlinear.
Earlier you predicted why doubling speed might more than double stopping distance. Check your answer against the worked examples.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Use supplied formulas and interpret the result.
Q1. Use $D = 0.01v^2 + 0.3v$ to estimate stopping distance when $v = 70$ km/h. Show full working. (3 marks)
Q2. Use $A = s^2$ to find the area of a square with side length 8.5 m. Include units. (2 marks)
Q3. Explain why a formula containing $v^2$ may not behave like a constant-rate formula. (2 marks)
📖 Answers (click to reveal)
Drill 1: $D = 0.01(3600) + 0.3(60) = 36 + 18 = 54$ m.
Drill 2: At 30 km/h: $D = 0.01(900) + 9 = 18$ m. At 60 km/h: $D = 54$ m. Ratio = $54 \div 18 = 3$. No — it triples, not doubles.
Drill 3: $A = 12^2 = 144$ cm².
Drill 4: $12^2 = 12 \times 12 = 144$. $2 \times 12 = 24$. Squaring means multiplying by itself, not by 2.
Q1 (3 marks): $D = 0.01(70)^2 + 0.3(70)$ [1] $= 0.01(4900) + 21$ [1] $= 49 + 21 = 70$ m [1].
Q2 (2 marks): $A = (8.5)^2 = 72.25$ m² [2]. (1 mark if correct calculation, second mark for units.)
Q3 (2 marks): The squared term $v^2$ grows faster than $v$ as $v$ increases [1], so doubling the input more than doubles the output — unlike a linear formula where output changes by a constant rate [1].
For each formula, identify the squared term, substitute carefully, and say what the output means. Beat the boss to bank a tier.
⚔ Enter the arenaClimb platforms by answering quadratic formula questions. Pool: lesson 7.
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