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

Position and Displacement — Question 7

Section 2-2 · Physics pages 164–165

Question 7 A moving origin contributes to a ground displacement A crate moves along a straight truck deck parallel to a ground-fixed east-positive axis X. A local deck coordinate x′ is also east positive, with its origin at a marked point O on the truck. At each instant the positions satisfy Xcrate = XO + x′. The initial and final data are exact: XO,i = 100 m, XO,f = 105 m, x′ i = +2 m, x′ f = −1 m. initial XO = 100 m crate: x′ = +2 m final XO = 105 m crate: x′ = −1 m Both axes point east. The ground reference is fixed; the origin O of the crate’s local coordinate moves. (a) Find the crate’s initial and final ground positions. Find its displacement relative to the deck and relative to the ground. (b) Derive the relation between these two displacements and the ground displacement of O. Why does the moving origin not cancel as a single constant would? (c) A student says “the crate moved 3 m west, so its ground displacement must be negative.” Diagnose the reference error. Do the endpoint data determine the crate’s total ground distance traveled? (d) Keep the same local endpoint coordinates, but replace the known origin motion with ∆XO ∈[3, 5] m. Find the allowed ground displacement. Which net directions, including zero change, remain possible?
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Question 7 - Solution A moving origin contributes to a ground displacement initial XO = 100 m crate: x′ = +2 m final XO = 105 m crate: x′ = −1 m (a) Combine coordinates at the same instant. (Xi, Xf) = (100 + 2, 105 −1) = (102, 104) m. Xi = 100 + 2 = 102 m , Xf = 105 −1 = 104 m . Hence the local displacement is ∆x′ = −1 −2 = −3 m (net west relative to the deck), while ∆X = 104 −102 = +2 m (net east relative to the ground). (b) Subtract the full endpoint relations. ∆X = (XO,f + x′ f) −(XO,i + x′ i) = ∆XO + ∆x′ . Here ∆XO = +5 m, so ∆X = 5 −3 = +2 m. The two origin coordinates differ. Only a shift that has the same value at both instants cancels as a constant; the moving origin contributes its own displacement. (c) A displacement belongs to its declared reference. The 3 m westward net change is relative to the deck. During the same interval the deck origin shifts east relative to the ground. Combining the two gives a ground net change to the east, with no contradiction. Neither endpoint pair specifies all intervening ground motion. The ground traveled distance is at least 2 m, but may be larger if the ground path reverses. No unique distance follows without more path information. (d) Propagate the origin-motion interval through the relation. ∆X = ∆O −3 m, ∆O ∈[3, 5] m ⇒∆X ∈[0, 2] m. With ∆x′ = −3 m fixed, ∆X ∈[3 −3, 5 −3] m = [0, 2] m . Zero net ground change and positive/eastward net change are possible; negative/westward net change is excluded. Zero displacement would still not prove the crate was stationary throughout the interval.

Question 7

A moving origin contributes to a ground displacement

A crate moves along a straight truck deck parallel to a ground-fixed east-positive axis XX. A local deck coordinate x′x' is also east positive, with its origin at a marked point O on the truck. At each instant the positions satisfy Xcrate=XO+x′.X_{\mathrm{crate}}=X_O+x'. The initial and final data are exact: XO,i=100m,XO,f=105m,xi′=+2m,xf′=−1m.X_{O,i}=100\ \mathrm{m},\quad X_{O,f}=105\ \mathrm{m},\qquad x'_i=+2\ \mathrm{m},\quad x'_f=-1\ \mathrm{m}.

Diagram shown in the original SVG worksheet.

  1. Find the crate’s initial and final ground positions. Find its displacement relative to the deck and relative to the ground.

  2. Derive the relation between these two displacements and the ground displacement of O. Why does the moving origin not cancel as a single constant would?

  3. A student says “the crate moved 3 m west, so its ground displacement must be negative.” Diagnose the reference error. Do the endpoint data determine the crate’s total ground distance traveled?

  4. Keep the same local endpoint coordinates, but replace the known origin motion with ΔXO∈[3,5]m\Delta X_O\in[3,5]\ \mathrm{m}. Find the allowed ground displacement. Which net directions, including zero change, remain possible?

Question 7 - Solution

A moving origin contributes to a ground displacement

Diagram shown in the original SVG worksheet.

(a) Combine coordinates at the same instant.

(Xi,Xf)=(100+2,105−1)=(102,104)m.(X_i,X_f)=(100+2,105-1)=(102,104)\ \mathrm{m}.

Xi=100+2=102m,Xf=105−1=104m.X_i=100+2=\boxed{102\ \mathrm{m}},\qquad X_f=105-1=\boxed{104\ \mathrm{m}}. Hence the local displacement is Δx′=−1−2=−3m\Delta x'=-1-2=\boxed{-3\ \mathrm{m}} (net west relative to the deck), while ΔX=104−102=+2m\Delta X=104-102=\boxed{+2\ \mathrm{m}} (net east relative to the ground).

(b) Subtract the full endpoint relations.

ΔX=(XO,f+xf′)−(XO,i+xi′)=ΔXO+Δx′.\Delta X=(X_{O,f}+x'_f)-(X_{O,i}+x'_i) =\boxed{\Delta X_O+\Delta x'}. Here ΔXO=+5m\Delta X_O=+5\ \mathrm{m}, so ΔX=5−3=+2m\Delta X=5-3=+2\ \mathrm{m}. The two origin coordinates differ. Only a shift that has the same value at both instants cancels as a constant; the moving origin contributes its own displacement.

(c) A displacement belongs to its declared reference.

The 3 m westward net change is relative to the deck. During the same interval the deck origin shifts east relative to the ground. Combining the two gives a ground net change to the east, with no contradiction.

Neither endpoint pair specifies all intervening ground motion. The ground traveled distance is at least 2 m, but may be larger if the ground path reverses. No unique distance follows without more path information.

(d) Propagate the origin-motion interval through the relation.

ΔX=ΔO−3m,ΔO∈[3,5]m⇒ΔX∈[0,2]m.\Delta X=\Delta O-3\ \mathrm{m},\qquad\Delta O\in[3,5]\ \mathrm{m}\Rightarrow\Delta X\in[0,2]\ \mathrm{m}.

With Δx′=−3m\Delta x'=-3\ \mathrm{m} fixed, ΔX∈[3−3,5−3]m=[0,2]m.\Delta X\in[3-3,5-3]\ \mathrm{m}=\boxed{[0,2]\ \mathrm{m}}. Zero net ground change and positive/eastward net change are possible; negative/westward net change is excluded. Zero displacement would still not prove the crate was stationary throughout the interval.

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