3.4: Units in the Numerator and the Denominator

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Learning Objectives

Given a quantity, convert from one set of units to another using dimensional analysis showing canceling on units. This includes compound units (eg: mph to kps).

Compound Units

There are many units that combine a unit in the numerator with a unit in the denominator. Common examples are miles per hour, meters per second, and miles per gallon. The "per" that appears when we talk about the compound unit or when we write it in text form translates to the divider between numerator and denominator when we write it symbolically. Therefore miles per hour would be written mi/h, meters per second would be written m/s, and miles per gallon would be written mi/gal.

For example, suppose we want to convert 0.0128 kilometers per minute to meters per hour.

Concept Map

Calculation

\[ \begin{align*} \dfrac{0.0128 \; \cancel{\rm{km}}}{1\; \cancel{\rm{min}}} \times \dfrac{60 \; \cancel{\rm{min}}}{1\; \rm{h}} \times \dfrac{1,000\; \rm{m}}{1\; \cancel{\rm{km}}} & = \dfrac{768\; \rm{m}}{\rm{h}} \\ &= 768 \; \rm{m/h}\ \end{align*}\]

Which we would write or say as 768 meters per hour.

Example \(\PageIndex{1}\): Unit Conversion

Convert 45 miles per gallon to kilometers per liter in one multi-step calculation.

Solution

Steps for Problem Solving	Unit Conversion
Identify the "given" information and what the problem is asking you to "find."	Given: 45 mi/gal Find: km/L
List other known quantities	\(1 km = 0.6214 mi \) \(1 gal = 3.785 L \)
Prepare a concept map.
Calculate.	\[ \begin{align} \dfrac{45 \; \cancel{\rm{mi}}}{1\; \cancel{\rm{gal}}} \times \dfrac{1 \; \cancel{\rm{gal}}}{3.785\; \rm{L}} \times \dfrac{1\; \rm{km}}{0.6214\; \cancel{\rm{mi}}} & = \dfrac{19\; \rm{km}}{\rm{L}} \\ &= 19 \; \rm{km/L}\ \end{align}\]

Exercise \(\PageIndex{1}\)

Convert 1.25 cm/s to in/min.

Answer: \(29.5\, in/min\)