POP

E1.12 Rates

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01 — Explanation

What are the key rate formulas I need?

A rate compares two different quantities measured in different units:

Rate	Formula	Typical units
Speed	$\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$	m/s, km/h
Density	$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}}$	g/cm³, kg/m³
Pressure	$\text{Pressure} = \dfrac{\text{Force}}{\text{Area}}$	N/m², N/cm²

02 — Explanation

How do I calculate density, speed or pressure from given quantities?

Substitute the given values directly into the relevant formula — the only real danger is mixing units (see below) before you substitute.

Worked Example

A solid cuboid measures $5\text{ cm} \times 4\text{ cm} \times 10\text{ cm}$ and has a mass of $840\text{ g}$ . Find its density in g/cm³.

Volume $= 5\times4\times10 = 200\text{ cm}^3$ .
Density $= \dfrac{\text{mass}}{\text{volume}} = \dfrac{840}{200} = 4.2\text{ g/cm}^3$ .

03 — Explanation

How do I convert a rate between compound units, like g/cm³ to kg/m³?

Convert the top and bottom of the rate separately, remembering that volume (a cubic unit) scales by the cube of the linear conversion factor, not the factor itself.

1\text{ m}^3 = 100^3\text{ cm}^3 = 1\,000\,000\text{ cm}^3, \qquad 1\text{ kg} = 1000\text{ g}

so multiplying a density in g/cm³ by $1000$ converts it directly to kg/m³:

4.2\text{ g/cm}^3 \times 1000 = 4200\text{ kg/m}^3

Examiner Tip: A very common slip is multiplying or dividing by $100$ (the linear factor) instead of $100^3$ when converting between cm³ and m³. Whenever a unit is squared or cubed, its conversion factor must be squared or cubed too.

04 — Explanation

How do I convert a speed between km/h and m/s?

This is the single most-tested rate conversion on the syllabus. Since $1\text{ km}=1000\text{ m}$ and $1\text{ h}=3600\text{ s}$ :

\text{km/h} \to \text{m/s: divide by } 3.6 \qquad \text{m/s} \to \text{km/h: multiply by } 3.6

Worked Example

A car travels at $108\text{ km/h}$ . Find its speed in m/s.

Convert km to m and h to s directly: $108\text{ km/h} = \dfrac{108\times1000}{3600}\text{ m/s}$ .
Evaluate: $\dfrac{108\,000}{3600} = 30\text{ m/s}$ .

05 — Explanation

What other types of rate might appear in an exam question?

The syllabus also expects you to apply rate reasoning to less common contexts — hourly pay rates, exchange rates, fuel consumption (e.g. km per litre), flow rates (e.g. litres per minute filling a tank), and population density (people per km²). Each works exactly like speed or density: identify which quantity is "per" which other quantity, and divide accordingly.

Examiner Tip: Whatever the context, the units in the question tell you the formula. "km per litre" means $\dfrac{\text{distance}}{\text{fuel used}}$ ; "people per km²" means $\dfrac{\text{population}}{\text{area}}$ — read the unit, don't try to memorise every named formula separately.

06 — Explanation

How do I rearrange a rate formula to find a different unknown?

If the question gives you the rate and one other quantity, rearrange the formula before substituting rather than trying to reverse the arithmetic afterwards. For example, from $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$ , finding a missing time means rearranging to $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$ first, then substituting the known values directly into that rearranged version.