Position and Displacement — Question 9

Question 9

A repeating position display can hide the direction of a net change

A cart travels on a straight east-positive track. Its display gives only the repeating coordinate $q$, with $0\leq q<10\ \mathrm{m}$. The actual position has the form $$X=10n\ \mathrm{m}+q,\qquad n\text{ an integer}.$$ Thus the display returns to zero at every 10 m boundary and does not report which 10 m interval contains the cart. The display is a position code, not an odometer.

The initial true position is independently known to be $X_i=8\ \mathrm{m}$, and the final display reads $q_f=2\ \mathrm{m}$. A separate measurement establishes $|\Delta X|<7\ \mathrm{m}$. Treat all data as exact in this model; the path is continuous.

1. Derive all possible signed displacements before applying the bound. Then identify every candidate satisfying $|\Delta X|<7\ \mathrm{m}$.

2. Why does the raw display difference $2-8=-6\ \mathrm{m}$ not by itself establish westward net motion? Give a simple continuous path for each bounded candidate.

3. An independent odometer now reports a total traveled distance of exactly 6 m. Does this select one candidate if reversals are allowed? Construct a path of that length for each candidate, preserving the known initial position and final display.

4. If it is additionally known that the cart never reverses during the run, which candidate is selected by the 6 m distance? State precisely which additional information makes this conclusion possible.

Question 9 - Solution

A repeating position display can hide the direction of a net change

(a) Keep the unknown integer interval.

$$-7<-6+10n<7\quad\Rightarrow\quad-0.1<n<1.3\quad\Rightarrow\quad n\in\{0,1\}.$$

The possible final positions are $X_f=2\ \mathrm{m}+10n\ \mathrm{m}$, so $$\boxed{\Delta X=(-6+10n)\ \mathrm{m},\qquad n\in\mathbb Z}.$$ The bound requires $-7<-6+10n<7$, or $-0.1<n<1.3$. The only integers are $n=0$ and $n=1$, giving $$\boxed{\Delta X=-6\ \mathrm{m}\quad\text{or}\quad+4\ \mathrm{m}}.$$

(b) A wrapped coordinate difference loses the interval count.

The route $8\to2\ \mathrm{m}$ gives a $-6\ \mathrm{m}$ displacement. The route $8\to12\ \mathrm{m}$ gives $+4\ \mathrm{m}$, with the display passing through its reset at 10 m and ending at 2 m. Both are continuous and satisfy the independent displacement bound. Subtracting the displayed numbers assumes the same interval label at both endpoints, which is not established.

(c) Total distance still permits both net changes if reversals are allowed.

$$D_{8\to13\to12}=5+1=6\ \mathrm{m},\qquad\Delta X=4\ \mathrm{m}.$$

For the negative candidate, direct westward travel $8\to2\ \mathrm{m}$ has distance 6 m. For the positive candidate, a route $$8\ \longrightarrow\ 13\ \longrightarrow\ 12\quad(\mathrm{m})$$ has distance $5+1=6\ \mathrm{m}$, displacement $+4\ \mathrm{m}$, and final display 2 m. Thus both candidates remain possible. The odometer measures path length; it does not supply a signed net change when opposing travel occurs.

(d) A no-reversal constraint turns length into displacement magnitude.

On a straight line with no reversal, including possible pauses, $D=|\Delta X|$. The measured $D=6\ \mathrm{m}$ then selects $$\boxed{\Delta X=-6\ \mathrm{m},\qquad X_f=2\ \mathrm{m}}.$$ The $+4\ \mathrm{m}$ candidate would have distance 4 m if there were no reversal, contradicting the odometer. The selection uses the repeating-code model, the independent displacement bound, the traveled length, and the no-reversal condition together. No one display subtraction contains all that information.