Changing Units — Question 5

Question 5

Pace is the reciprocal of speed

A runner travels two consecutive segments of equal length, $1.50\ \mathrm{km}$ each. The pace on segment A is $4.00\ \mathrm{min/km}$ and on B is $6.00\ \mathrm{min/km}$. Pace is time per distance; speed is distance per time. Assume each segment is covered at constant speed, with no stop between them.

Use exact $1\ \mathrm{min}=60\ \mathrm{s}$, $1\ \mathrm{h}=60\ \mathrm{min}$, and $1\ \mathrm{km}=1000\ \mathrm{m}$. Treat the stated data as exact within the model.

1. Find the speed on each segment in $\mathrm{km/h}$ and $\mathrm{m/s}$. Explain why multiplying a pace by a conversion factor alone cannot turn its physical dimension into speed without taking a reciprocal.

2. Determine the overall average speed and average pace from total distance and total time.

3. A student averages the two speeds arithmetically. Another averages the two paces arithmetically and then takes the reciprocal. Which procedure works for this journey, and why?

4. If the runner instead spends equal times at the two stated paces, which averaging procedure gives overall average speed? Explain what changed.

Question 5 - Solution

Pace is the reciprocal of speed

(a) Reverse the ratio, then convert units.

$$v_A=\frac{1}{4.00}\ \mathrm{km/min}=15.0\ \mathrm{km/h} =\frac{25}{6}\ \mathrm{m/s}\approx4.17\ \mathrm{m/s},$$ $$v_B=\frac{1}{6.00}\ \mathrm{km/min}=10.0\ \mathrm{km/h} =\frac{25}{9}\ \mathrm{m/s}\approx2.78\ \mathrm{m/s}.$$ Changing units preserves a quantity’s dimension. Pace has time/length dimension, whereas speed has length/time dimension. They are different, reciprocal quantities; a unit change by itself does not invert the physical ratio.

(b) Combine the actual times and distances.

Each time is distance multiplied by pace. Thus $t_A=(1.50)(4.00)\ \mathrm{min}=6.00\ \mathrm{min}$ and $t_B=(1.50)(6.00)\ \mathrm{min}=9.00\ \mathrm{min}$. Hence $$\overline v=\frac{3.00\ \mathrm{km}}{15.00\ \mathrm{min}} =\boxed{12.0\ \mathrm{km/h}}=\frac{10}{3}\ \mathrm{m/s}, \qquad \overline p=\boxed{5.00\ \mathrm{min/km}}.$$

(c) Equal distance weights the paces equally.

$$\overline v=\frac{2d}{d/v_A+d/v_B}=\frac{2v_Av_B}{v_A+v_B}=\frac{2(15)(10)}{15+10}\ \mathrm{km\,h^{-1}}=12\ \mathrm{km\,h^{-1}}.$$

The arithmetic mean of speeds is $12.5\ \mathrm{km/h}$, which is too large. The runner spends more time at the slower speed, so the speeds must not receive equal time weights.

For two equal distances $d$, total time is $d p_A+d p_B$ and total distance is $2d$. Hence average pace is $(p_A+p_B)/2=5.00\ \mathrm{min/km}$. Its reciprocal gives the correct overall speed. Equivalently, average speed is the harmonic mean $2/(1/v_A+1/v_B)$.

(d) Equal time changes the weights.

For equal times $\tau$, total distance is $v_A\tau+v_B\tau$. Dividing by $2\tau$ gives $(v_A+v_B)/2=\boxed{12.5\ \mathrm{km/h}}$. The arithmetic mean of speeds now works. The arithmetic mean of the paces no longer does: the distances covered at the two paces are unequal.