Average Velocity and Average Speed — Question 5

Question 5

Sparse position records cannot certify an average speed

A sensor reports positions on a straight line only at three instants: The recorded $(t,x)$ pairs are $(0\ \mathrm{s},0\ \mathrm{m})$, $(4\ \mathrm{s},12\ \mathrm{m})$ and $(10\ \mathrm{s},6\ \mathrm{m})$, shown as isolated points below. The observations are exact, the path is continuous, and nothing is known about motion between observations. In particular, connecting the samples by straight graph segments is not a measured fact. There is no supplied speed limit.

1. Find the exact average velocity over 0–10 s. Find the strongest lower bound on average speed implied by all three samples, and exhibit a path attaining it.

2. Explain why using only the initial and final samples gives a weaker lower bound.

3. Construct a path matching all three observations but having average speed exactly $3.0\ \mathrm{m/s}$. Specify any additional turning-point position and a compatible time for it.

4. Is there a finite upper bound on average speed from these data alone? Explain. What extra path information would make the lower bound from part (a) the exact answer?

Question 5 - Solution

Sparse position records cannot certify an average speed

(a) Endpoint displacement is known; the intervening distance is not.

The exact average velocity is $$\boxed{\bar v=(6-0)/10=+0.60\ \mathrm{m/s}}.$$ Any path must travel at least 12 m before the 4 s observation and at least 6 m afterward. Thus $D\geq18\ \mathrm{m}$ and $$\boxed{\bar s\geq1.80\ \mathrm{m/s}}.$$ A path traveling directly from 0 to 12 m during 0–4 s and directly back to 6 m during 4–10 s attains this bound. Its rates need not be constant to give the same distance.

(b) The middle observation proves otherwise hidden opposing travel.

Initial and final positions alone give $D\geq|6-0|=6\ \mathrm{m}$, or $\bar s\geq0.60\ \mathrm{m/s}$. This weaker bound ignores the required visit to 12 m and the subsequent return toward 6 m.

(c) Add a compatible excursion.

Use the timed route $$(t,x):\quad(0,0)\to(2,18)\to(4,12)\to(10,6),$$ with straight position–time segments between these events. It is continuous and matches every observation. Its distance is $18+6+6=30\ \mathrm{m}$, yielding $\boxed{\bar s=3.0\ \mathrm{m/s}}$. Its average velocity is still $+0.60\ \mathrm{m/s}$.

(d) Unobserved excursions need a bound of their own.

$$\bar s(M)=\frac{2M-6\ \mathrm{m}}{10\ \mathrm{s}}\longrightarrow\infty\quad(M\longrightarrow\infty).$$

Replacing the 18 m excursion by a peak $M\geq12\ \mathrm{m}$ at 2 s gives distance $M+(M-12)+6=2M-6\ \mathrm{m}$. This can grow without bound while preserving every sampled position and time. Each finite $M$ admits finite stage rates; no supplied constraint excludes them.

If independent information establishes no reversal within either sampling interval, the sum of the two endpoint separations is the actual distance. Then the lower bound $1.80\ \mathrm{m/s}$ becomes exact. Merely drawing straight connectors is an assumption, not evidence of this condition.