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Physics / Chapter 3: Vector / Section 3-2

Adding Vectors Geometrically — Question 3

Section 3-2 · Physics pages 336–337

Question 3 Reordering a polygon changes the route but not the sum Four straight displacement legs are 5 m east, 3 m north, 2 m west and 3 m south. Let the start be O. Route I uses that order. Route II uses 3 m north, 3 m south, 5 m east and 2 m west. Each leg preserves its original direction and magnitude when reordered. resultant O Route I The resultant is drawn slightly below the route for readability; its direction and length are preserved. (a) Construct both head-to-tail chains and find their common resultant. Use geometric cancellation rather than a component table. (b) Find both route lengths. Does equality of resultant and distance imply the same set of intermediate positions? (c) Group the north and south arrows first. Explain why replacing this adjacent pair by its resultant is valid for finding the final point but loses route information. (d) A student reverses the order of Route I and also reverses every arrow. Compare this new route’s resultant with simply reversing the order while leaving arrows unchanged.
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Question 3 - Solution Reordering a polygon changes the route but not the sum resultant O Route I (a) Cancel opposite equal arrows and combine collinear arrows.⃗ R = (5 −2)ˆi + (3 −3)ˆj = 3ˆi m The north and south arrows cancel, leaving the uncanceled eastward part. Reordering translates the arrows but preserves their directions and lengths. (b) The list of leg lengths fixes total route length. s = 5 + 3 + 2 + 3 = 13 m for both orders Route I visits (5, 3) m; Route II never does. Equal endpoints and total distance do not specify intermediate positions or timing. (c) Regrouping preserves net change, not traveled length. 3ˆj −3ˆj =⃗0, sexcursion = 3 + 3 = 6 m Replacing the closed excursion by its zero resultant preserves the final point but removes 6 m of actual travel from the represented route. (d) Reversing the arrows changes the physical displacements. ∑ k⃗ d5−k =⃗R, ∑ k (−⃗d5−k) = −⃗R = −3ˆi m Reversing order alone preserves the eastward sum. Reversing every arrow also reverses the resultant; starting at the original finish then returns to O. Total distance remains 13 m.

Question 3

Reordering a polygon changes the route but not the sum

Four straight displacement legs are 5 m east, 3 m north, 2 m west and 3 m south. Let the start be O. Route I uses that order. Route II uses 3 m north, 3 m south, 5 m east and 2 m west. Each leg preserves its original direction and magnitude when reordered.

Diagram shown in the original SVG worksheet.

  1. Construct both head-to-tail chains and find their common resultant. Use geometric cancellation rather than a component table.

  2. Find both route lengths. Does equality of resultant and distance imply the same set of intermediate positions?

  3. Group the north and south arrows first. Explain why replacing this adjacent pair by its resultant is valid for finding the final point but loses route information.

  4. A student reverses the order of Route I and also reverses every arrow. Compare this new route’s resultant with simply reversing the order while leaving arrows unchanged.

Question 3 - Solution

Reordering a polygon changes the route but not the sum

Diagram shown in the original SVG worksheet.

(a) Cancel opposite equal arrows and combine collinear arrows.

R→=(5−2)î+(3−3)ĵ=3îm\boxed{\vec R=(5-2)\hat i+(3-3)\hat j=3\hat i\,\mathrm m} The north and south arrows cancel, leaving the uncanceled eastward part. Reordering translates the arrows but preserves their directions and lengths.

(b) The list of leg lengths fixes total route length.

s=5+3+2+3=13mfor both orderss=5+3+2+3=13\,\mathrm m\quad\text{for both orders} Route I visits (5,3)(5,3) m; Route II never does. Equal endpoints and total distance do not specify intermediate positions or timing.

(c) Regrouping preserves net change, not traveled length.

3ĵ−3ĵ=0→,sexcursion=3+3=6m3\hat j-3\hat j=\vec0,\qquad s_{\mathrm{excursion}}=3+3=6\,\mathrm m Replacing the closed excursion by its zero resultant preserves the final point but removes 6 m of actual travel from the represented route.

(d) Reversing the arrows changes the physical displacements.

∑kd→5−k=R→,∑k(−d→5−k)=−R→=−3îm\sum_k\vec d_{5-k}=\vec R,\qquad\sum_k(-\vec d_{5-k})=-\vec R=-3\hat i\,\mathrm m Reversing order alone preserves the eastward sum. Reversing every arrow also reverses the resultant; starting at the original finish then returns to O. Total distance remains 13 m.

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