Changing Units — Question 1

Question 1

A conversion factor is one, but orientation still matters

A speed is $72\ \mathrm{km/h}$. Use the exact relations $1\ \mathrm{km}=1000\ \mathrm{m}$ and $1\ \mathrm{h}=3600\ \mathrm{s}$. Three proposed conversion chains are $$\begin{aligned} \mathrm{A}:&\quad 72\ \frac{\mathrm{km}}{\mathrm{h}} \frac{1000\ \mathrm{m}}{1\ \mathrm{km}} \frac{1\ \mathrm{h}}{3600\ \mathrm{s}},\\[4pt] \mathrm{B}:&\quad 72\ \frac{\mathrm{km}}{\mathrm{h}} \frac{1\ \mathrm{km}}{1000\ \mathrm{m}} \frac{3600\ \mathrm{s}}{1\ \mathrm{h}},\\[4pt] \mathrm{C}:&\quad 72\ \frac{\mathrm{km}}{\mathrm{h}} \frac{1000\ \mathrm{m}}{1\ \mathrm{km}} \frac{3600\ \mathrm{s}}{1\ \mathrm{h}}. \end{aligned}$$

1. Retain all units while simplifying each chain. Which produces a number directly multiplying $\mathrm{m/s}$? Explain why writing $\mathrm{m/s}$ after the numerical results of B or C is incorrect.

2. Each inserted ratio is a ratio of equal physical quantities and therefore equals one. How can B and C still fail to achieve the requested form without changing the underlying speed?

3. Convert the successful result back to $\mathrm{km/h}$ as a check. Explain why its numerical value in $\mathrm{m/s}$ is smaller, even though metres are smaller than kilometres.

Treat the initial speed as exact for this exercise; the issue is conversion logic, not measurement uncertainty.

Question 1 - Solution

A conversion factor is one, but orientation still matters

(a) Cancel symbols before discarding units.

$$72\ \frac{\mathrm{km}}{\mathrm h}\frac{1000\ \mathrm m}{1\ \mathrm{km}}=72000\ \frac{\mathrm m}{\mathrm h},\qquad\frac{72000}{3600}\ \frac{\mathrm m}{\mathrm s}=20\ \frac{\mathrm m}{\mathrm s}.$$

The three simplified expressions are $$\begin{aligned} \mathrm{A}:&\quad \boxed{20\ \mathrm{m/s}},\\ \mathrm{B}:&\quad 259.2\ \frac{\mathrm{km^2\,s}}{\mathrm{m\,h^2}},\\ \mathrm{C}:&\quad 259\,200\,000\ \frac{\mathrm{m\,s}}{\mathrm{h^2}}. \end{aligned}$$ In A, kilometre and hour cancel in the intended directions. The mixed units left in B and C are not the unit $\mathrm{m/s}$. Replacing them by that label would change the quantity rather than merely its representation.

(b) An unfinished conversion is not multiplication by a false identity.

All the ratios equal one, including the inverted ratios. Thus all three full expressions still represent the original speed. B and C simply leave inconvenient mixed units. For example, in C, $$259\,200\,000\ \frac{\mathrm{m\,s}}{\mathrm{h^2}} =\frac{259\,200\,000}{3600^2}\ \mathrm{m/s}=20\ \mathrm{m/s}.$$ B also gives $259.2(1000^2/3600^2)\ \mathrm{m/s}=20\ \mathrm{m/s}$ after its remaining units are reduced. Unit ratios should be oriented to cancel the starting unit and leave the desired one efficiently.

(c) Reverse the operation and explain the numerical scale.

$$20\ \frac{\mathrm{m}}{\mathrm{s}} \frac{1\ \mathrm{km}}{1000\ \mathrm{m}} \frac{3600\ \mathrm{s}}{1\ \mathrm{h}} =\boxed{72\ \mathrm{km/h}}.$$ Both distance and time units changed. A metre per second is $3.6\ \mathrm{km/h}$, a larger speed unit than a kilometre per hour, so fewer of these units are needed to express the same speed.