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hscscience Maths Std · Y11
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Module 1 · L7 of 13 ~45 min ⚡ +75 XP available

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.

Today's hook, 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?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Think First, your gut answer first
+5 XP warm-up

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 calculatingwrite your gut explanation first.

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02
The formulas you need to use
+5 XP to read

Applied quadratic formulas are suppliedyour 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.

x y Vertex (2, 4) axis of symmetry x=0 x=4 y-int y = -(x-2)² + 4
$D = 0.01v^2 + 0.3v$ grows faster than linearly as $v$ increases
Squared means multiplied by itself
$v^2 = v \times v$. If $v = 40$, then $v^2 = 1600$. It is not $2 \times 40 = 80$.
Nonlinear growth
Doubling the input more than doubles the output when a squared term is present.
Round at the end
Keep full calculator values until the final step. Early rounding can shift comparison results.
03
What you'll master
Know

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.
Understand

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.
Can do

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.
04
Key terms
Quadratic formulaA formula that contains a squared variable, such as $v^2$ or $s^2$.
Squared term ($v^2$)The variable multiplied by itself: $v^2 = v \times v$. Not the same as $2v$.
Stopping distance ($D$)The total distance a vehicle travels from when the brakes are applied to when it stops. Measured in metres.
Nonlinear growthWhen output increases faster than a constant multiple of the input, caused by squared (or higher) terms.
SubstitutionReplacing a variable with its numerical value to calculate the formula output.
Square area ($A = s^2$)Area of a square equals side length multiplied by itself. Units are always square units (m², cm²).
05
Squared terms change the growth pattern
core concept

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.

Common error: $v^2$ means $v \times v$, not double $v$. Students sometimes write $40^2 = 80$ instead of $1600$.

Squared terms grow non-linearly: doubling the input more than doubles the output. In a stopping-distance formula like D = 0.01v² + 0.3v, the v² term dominates at high speeds. Calculate v² separately before substituting to avoid errors.

Pause, copy the non-linear growth property of squared terms (doubling the input more than doubles the output), the evaluation tip for D = 0.01v² + 0.3v (calculate 0.01v² and 0.3v separately then add), and note that the v² term dominates at high values of v into your book.

True or false: If $v = 6$, then $v^2 = 12$.

06
Round at the end
core concept

We just saw that squared terms grow non-linearly, in a stopping-distance formula like D = 0.01v² + 0.3v, doubling the speed more than doubles the stopping distance because v² dominates at high speeds. That raises a question: squared terms produce long decimals in intermediate calculations, when is the right time to round? This card answers it → round only at the final answer; rounding intermediate values introduces cumulative error that can push the final answer outside the required decimal accuracy.

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.

Rule: Calculate all intermediate steps exactly, then round the final answer to the required degree of accuracy.

Round only at the final answer. Rounding intermediate calculator values introduces cumulative error that can put the final answer outside the required decimal-place accuracy. Store exact values in memory or on screen throughout the calculation.

Pause, copy the round-at-the-end rule: store exact intermediate values in calculator memory or write full decimal answers at each step, and round only the final answer to the required number of decimal places, premature rounding compounds into your book.

Quick check: When should you round an answer when using a formula?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · STOPPING DISTANCE

Use $D = 0.01v^2 + 0.3v$ to estimate the stopping distance when $v = 50$ km/h.

1
$D = 0.01(50)^2 + 0.3(50)$
Substitute $v = 50$ into the formula. Replace every $v$ with 50.
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PROBLEM 2 · COMPARING TWO SPEEDS

Use $D = 0.01v^2 + 0.3v$ for $v = 40$ and $v = 80$. Does doubling speed double the stopping distance?

1
For $v = 40$: $D = 0.01(1600) + 0.3(40) = 16 + 12 = 28$ m
$40^2 = 1600$. Substitute into the formula.
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PROBLEM 3 · AREA WITH SQUARED TERM

A square has side length $s = 7.5$ m. Use $A = s^2$ to find its area.

1
$A = (7.5)^2$
Substitute $s = 7.5$ into $A = s^2$.
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Fill the gap: When $v = 60$ km/h, $v^2 =$ and $0.01 \times v^2 =$ .

Common errors · the 3 traps that cost marks

Trap 01
Treating $v^2$ as $2v$
The most common error: $40^2 \ne 80$. Squaring means multiplying by itself: $40^2 = 40 \times 40 = 1600$. Always check your squaring step first.
Trap 02
Rounding too early
If you round $0.01 \times 2500 = 25.0$ to $25$ mid-calculation that's fine, but rounding decimal intermediate values before squaring can shift the final answer. Keep full precision until the end.
Trap 03
Forgetting square units for area
$A = s^2$ gives an area, so units must be m², cm², etc. Writing just "56.25" without units loses a mark. Include the unit at every step.

Quick-fire practice · 4 calculations

1

Use $D = 0.01v^2 + 0.3v$ to estimate stopping distance at 60 km/h.

2

Compare stopping distance at 30 km/h and 60 km/h. Does it double?

3

Use $A = s^2$ to find the area of a square with side length 12 cm.

4

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?

10
Revisit the stopping-distance model

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.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer

Use supplied formulas and interpret the result.

ApplyBand 33 marks

Q1. Use $D = 0.01v^2 + 0.3v$ to estimate stopping distance when $v = 70$ km/h. Show full working. (3 marks)

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ApplyBand 32 marks

Q2. Use $A = s^2$ to find the area of a square with side length 8.5 m. Include units. (2 marks)

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AnalyseBand 42 marks

Q3. Explain why a formula containing $v^2$ may not behave like a constant-rate formula. (2 marks)

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📖 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].

01
Boss battle · Square-Term Sprint
earn bronze · silver · gold

For each formula, identify the squared term, substitute carefully, and say what the output means. Beat the boss to bank a tier.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering quadratic formula questions. Pool: lesson 7.

Mark lesson as complete

Tick when you've finished the practice and review.

Module overview · Maths Standard