Average Velocity and Average Speed — Question 3

Question 3

Equal distances and equal times produce different means

Two travelers move only east along straight tracks. Each uses a steady speed of $3\ \mathrm{m/s}$ during one moving stage and $6\ \mathrm{m/s}$ during the other, with no stops. Idealized changes between stages take negligible time.

Traveler A covers 60 m at each speed. Traveler B travels for 10 s at each speed. Use total distance divided by total time, rather than assuming a rule for averaging speeds.

1. Find each traveler’s total distance, total time, and average speed. Explain why each average velocity has the same numerical value as the corresponding average speed.

2. Which traveler has average speed $(3+6)/2\ \mathrm{m/s}$? Explain the role of the time spent in the slower stage.

3. For two positive speeds $u$ and $w$, derive the average speed over two equal distances. Separately derive the result over two equal durations.

4. Prove that the equal-distance result is no greater than the equal-time result, and state the equality condition. Does this comparison alone establish which traveler finishes an arbitrary real journey first?

Question 3 - Solution

Equal distances and equal times produce different means

(a) Compute totals for each actual schedule.

A spends $60/3=20\ \mathrm{s}$ in the slower stage and $60/6=10\ \mathrm{s}$ in the faster stage. Thus $$D_A=120\ \mathrm{m},\quad T_A=30\ \mathrm{s},\quad\boxed{\bar s_A=4\ \mathrm{m/s}}.$$ B covers $3(10)+6(10)=90\ \mathrm{m}$ in 20 s, giving $$\boxed{\bar s_B=4.5\ \mathrm{m/s}}.$$ Both move only east, so displacement equals distance on the east-positive axis: $\bar v_A=+4$ and $\bar v_B=+4.5\ \mathrm{m/s}$.

(b) Equal durations permit the arithmetic mean.

$$\bar s_A=\frac23(3)+\frac13(6)=4\ \mathrm{m/s},\qquad\bar s_B=\frac12(3)+\frac12(6)=4.5\ \mathrm{m/s}.$$

B spends equal time at each speed, so the arithmetic mean applies. A spends twice as much time at the slower speed; that speed has twice the time weight. Equal distances are not equal time weights.

(c) Derive both expressions from distance and time.

For equal distances $d>0$, $$\bar s_d=\frac{2d}{d/u+d/w}=\boxed{\frac{2uw}{u+w}}.$$ For equal durations $\tau>0$, $$\bar s_t=\frac{u\tau+w\tau}{2\tau}=\boxed{\frac{u+w}{2}}.$$

(d) Compare the means algebraically, then respect the comparison’s scope.

$$\bar s_t-\bar s_d=\frac{(u+w)^2-4uw}{2(u+w)} =\frac{(u-w)^2}{2(u+w)}\geq0.$$ Equality holds exactly when $u=w$. Average speed alone does not rank arrival times for journeys of unspecified lengths or departure times. Here A and B cover different total distances, so a claim about who completes a common route first would require a common route and schedule.