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.

45 min Algebra Formulas and equations Lesson 7 of 13

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Use the printable version for substitution into quadratic formulas, rounding, and interpreting nonlinear growth.

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Think First

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?

Type your first explanation below.

Write your first explanation in your book.

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Saved

Know

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

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

Substitute into a formula containing a squared term.
Calculate and compare outputs from a quadratic model.
Interpret the result in context with suitable units.

q

Applied Quadratic Models

$D = 0.01v^2 + 0.3v$

$D$ = stopping distance in metres, $v$ = speed in km/h

$A = s^2$

$A$ = square area, $s$ = side length

Core Content

1. Squared Terms Change the Growth Pattern

A squared term means a value is multiplied by itself.

If $v = 40$, then $v^2 = 40^2 = 1600$. It does not mean $2 \times 40$.

Common error: $v^2$ means $v \times v$, not double $v$.

Worked Example 1

Use a stopping-distance formula

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

$D = 0.01(50)^2 + 0.3(50)$

$D = 0.01(2500) + 15$

$D = 25 + 15 = 40$

Answer: The estimated stopping distance is 40 m.

Worked Example 2

Compare two speeds

Use $D = 0.01v^2 + 0.3v$ for $v = 40$ and $v = 80$.

For $v = 40$: $D = 0.01(1600) + 12 = 28$ m.

For $v = 80$: $D = 0.01(6400) + 24 = 88$ m.

Doubling speed from 40 to 80 more than doubles the stopping distance.

Interpretation: The squared term makes the model grow faster at higher speeds.

Worked Example 3

Use an area formula with a squared term

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

$A = (7.5)^2$

$A = 56.25$

Answer: The area is 56.25 m2.

Unit habit: Area is measured in square units, such as m2 or cm2.

2. Round at the End

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.

Activity

Applied Quadratic Practice

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 $2 \times 12$.

Complete the applied quadratic practice in your book.

Revisit the Stopping-Distance Model

Doubling speed can more than double stopping distance because the formula includes $v^2$. This makes the model nonlinear.

Explain the squared term in your book.

Assessment

MC

Multiple Choice

Random questions from the lesson bank - feedback appears immediately.

SA

Short Answer

Use supplied formulas and interpret the result.

1. Use $D = 0.01v^2 + 0.3v$ to estimate stopping distance when $v = 70$ km/h. 3 MARKS

Answer in your book.

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

Answer in your book.

3. Explain why a formula containing $v^2$ may not behave like a constant-rate formula. 2 MARKS

Answer in your book.

Square-Term Sprint

For each formula, identify the squared term, substitute carefully, and say what the output means.

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