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Physics / Chapter 2: Motion Along a Straight Line / Section 2-2

Position and Displacement — Question 8

Section 2-2 · Physics pages 166–167

Question 8 A range is not a signed position A point target moves continuously along a straight line through a fixed beacon at x = 0. East is positive. An ideal range sensor reports only r = |x|, in metres, with no side information. Its initial and final readings are ri = 3 m, rf = 5 m. x (m) −5 −3 0 3 5 beacon East is positive. The range sensor reports distance to the beacon but not which side the target occupies. (a) List every possible ordered pair (xi, xf) and the corresponding signed displacement. Does subtracting the ranges, 5 −3, determine the displacement? (b) Now learn that the target starts west of the beacon. What two endpoint/displacement possibilities remain? (c) Add an independent bound that the total traveled distance is strictly less than 8 m. Which of those two possibilities is excluded, and why? Give one compatible path for the surviving possibility. (d) Instead of the path-length bound, suppose a complete crossing log records exactly one passage from one side of the beacon to the other. With the initial side still known, determine the final side and displacement. Could this crossing log and the less-than-8-m bound both hold under the same model?
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Question 8 - Solution A range is not a signed position x (m) −5 −3 0 3 5 beacon (a) Restore the signs that the range sensor omits. 3 →5 : ∆x = +2 m, 3 →−5 : ∆x = −8 m, −3 →5 : ∆x = +8 m, −3 →−5 : ∆x = −2 m. Thus all four signed displacements are possible. The range change +2 m is not generally the signed position change; it equals that displacement only for the first listed endpoint pair. (b) Initial side information removes two candidates. xi = −3 m ⇒∆x ∈{−2, +8} m. Starting west fixes xi = −3 m. The remaining possibilities are xf = +5 m with ∆x = +8 m, and xf = −5 m with ∆x = −2 m. (c) Path length cannot be shorter than endpoint separation. D < 8 m, D ≥|∆x| ⇒∆x = −2 m, xf = −5 m. The −3 to +5 m route requires at least 8 m of travel, so it is excluded by the strict bound. The surviving endpoint is xf = −5 m , with ∆x = −2 m . Direct westward travel from −3 to −5 m has length 2 m and satisfies the bound. (d) A side-changing crossing has a different consequence. Starting west and crossing the beacon exactly once leaves the target east of it. With final range 5 m, xf = +5 m and ∆x = +8 m . Touching the beacon and returning to the same side is not such a crossing. This conclusion requires traveled distance at least 8 m, so the crossing log and the strict less-than-8-m bound cannot both be correct under the stated assumptions. Independent data can resolve ambiguity, but incompatible constraints should not be combined into an impossible history.

Question 8

A range is not a signed position

A point target moves continuously along a straight line through a fixed beacon at x=0x=0. East is positive. An ideal range sensor reports only r=|x|r=|x|, in metres, with no side information. Its initial and final readings are ri=3m,rf=5m.r_i=3\ \mathrm{m},\qquad r_f=5\ \mathrm{m}.

Diagram shown in the original SVG worksheet.

  1. List every possible ordered pair (xi,xf)(x_i,x_f) and the corresponding signed displacement. Does subtracting the ranges, 5−35-3, determine the displacement?

  2. Now learn that the target starts west of the beacon. What two endpoint/displacement possibilities remain?

  3. Add an independent bound that the total traveled distance is strictly less than 8 m. Which of those two possibilities is excluded, and why? Give one compatible path for the surviving possibility.

  4. Instead of the path-length bound, suppose a complete crossing log records exactly one passage from one side of the beacon to the other. With the initial side still known, determine the final side and displacement. Could this crossing log and the less-than-8-m bound both hold under the same model?

Question 8 - Solution

A range is not a signed position

Diagram shown in the original SVG worksheet.

(a) Restore the signs that the range sensor omits.

3→5:Δx=+2m,3→−5:Δx=−8m,−3→5:Δx=+8m,−3→−5:Δx=−2m.\begin{aligned}3\to5 &: \Delta x=+2\ \mathrm{m},&3\to-5 &: \Delta x=-8\ \mathrm{m},\\-3\to5 &: \Delta x=+8\ \mathrm{m},&-3\to-5 &: \Delta x=-2\ \mathrm{m}.\end{aligned} Thus all four signed displacements are possible. The range change +2m+2\ \mathrm{m} is not generally the signed position change; it equals that displacement only for the first listed endpoint pair.

(b) Initial side information removes two candidates.

xi=−3m⇒Δx∈{−2,+8}m.x_i=-3\ \mathrm{m}\Rightarrow\Delta x\in\{-2,+8\}\ \mathrm{m}.

Starting west fixes xi=−3mx_i=-3\ \mathrm{m}. The remaining possibilities are xf=+5mx_f=+5\ \mathrm{m} with Δx=+8m\Delta x=+8\ \mathrm{m}, and xf=−5mx_f=-5\ \mathrm{m} with Δx=−2m\Delta x=-2\ \mathrm{m}.

(c) Path length cannot be shorter than endpoint separation.

D<8m,D≥|Δx|⇒Δx=−2m,xf=−5m.D<8\ \mathrm{m},\ D\geq|\Delta x|\Rightarrow\Delta x=-2\ \mathrm{m},\quad x_f=-5\ \mathrm{m}.

The −3-3 to +5m+5\ \mathrm{m} route requires at least 8 m of travel, so it is excluded by the strict bound. The surviving endpoint is xf=−5m\boxed{x_f=-5\ \mathrm{m}}, with Δx=−2m\boxed{\Delta x=-2\ \mathrm{m}}. Direct westward travel from −3-3 to −5m-5\ \mathrm{m} has length 2 m and satisfies the bound.

(d) A side-changing crossing has a different consequence.

Starting west and crossing the beacon exactly once leaves the target east of it. With final range 5 m, xf=+5m\boxed{x_f=+5\ \mathrm{m}} and Δx=+8m\boxed{\Delta x=+8\ \mathrm{m}}. Touching the beacon and returning to the same side is not such a crossing.

This conclusion requires traveled distance at least 8 m, so the crossing log and the strict less-than-8-m bound cannot both be correct under the stated assumptions. Independent data can resolve ambiguity, but incompatible constraints should not be combined into an impossible history.

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