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

Position and Displacement — Question 1

Section 2-2 · Physics pages 152–153

Question 1 A negative position is not a direction of travel A point moves on a straight east–west rail. Its position x is measured from a fixed origin, with east positive. Over any selected interval, define the signed displacement by ∆x = xf −xi. Its magnitude is |∆x|. Displacement describes the net change between endpoints; it does not specify every intervening motion. x (m), east −8 −3 −2 0 4 origin A B C One fixed origin and east-positive axis apply to every run. Three separate runs have these exact endpoint positions: The endpoint pairs (xi, xf), in metres, are (−8, −3) for run A, (−3, −8) for run B, and (+4, −2) for run C. (a) Find ∆x and its magnitude for each run, and state the net direction in words. (b) A student says A must be westward because both positions are negative. Another says C’s displacement is +6 m because the endpoints are 6 m apart. Diagnose both errors. (c) Which runs have equal displacement magnitude but different displacement? Explain why a magnitude alone loses information on this line. (d) For run A, give one path with no reversal and another with a reversal that have the same endpoints. Does positive displacement prove eastward travel at every instant?
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Question 1 - Solution A negative position is not a direction of travel x (m), east −8 −3 −2 0 4 origin A B C (a) Subtract the initial coordinate from the final coordinate. ∆xA = −3 −(−8) = +5 m (5 m east), ∆xB = −8 −(−3) = −5 m (5 m west), ∆xC = −2 −4 = −6 m (6 m west). (b) A location sign and a change sign answer different questions. A negative x locates a point west of the origin. It does not say which way the point changes position. In A the final coordinate is greater than the initial one, so the net change is eastward even though both endpoints are west of the origin. For C, 6 m is the displacement’s magnitude. The signed displacement is −6 m because the final position is west of the initial position. Replacing the signed subtraction by endpoint separation discards direction. (c) Equal magnitudes need not be equal displacements. ∆xA = −∆xB, |∆xA| = |∆xB| = 5 m. Runs A and B both have magnitude 5 m , but their displacements are +5 m and −5 m. On this specified axis, the sign supplies the net direction missing from the magnitude. (d) Endpoints do not determine the whole path. Ddirect = 5 m; D−8→0→−3 = 8 + 3 = 11 m, ∆x = 5 m. A monotone path from −8 to −3 m and a path from −8 to 0 and back to −3 m both have ∆x = +5 m. The latter includes westward travel. Positive displacement therefore describes the net endpoint change, not a guarantee of eastward motion at every instant.

Question 1

A negative position is not a direction of travel

A point moves on a straight east–west rail. Its position xx is measured from a fixed origin, with east positive. Over any selected interval, define the signed displacement by Δx=xf−xi.\Delta x=x_f-x_i. Its magnitude is |Δx||\Delta x|. Displacement describes the net change between endpoints; it does not specify every intervening motion.

Diagram shown in the original SVG worksheet.

Three separate runs have these exact endpoint positions: The endpoint pairs (xi,xf)(x_i,x_f), in metres, are (−8,−3)(-8,-3) for run A, (−3,−8)(-3,-8) for run B, and (+4,−2)(+4,-2) for run C.

  1. Find Δx\Delta x and its magnitude for each run, and state the net direction in words.

  2. A student says A must be westward because both positions are negative. Another says C’s displacement is +6m+6\ \mathrm{m} because the endpoints are 6 m apart. Diagnose both errors.

  3. Which runs have equal displacement magnitude but different displacement? Explain why a magnitude alone loses information on this line.

  4. For run A, give one path with no reversal and another with a reversal that have the same endpoints. Does positive displacement prove eastward travel at every instant?

Question 1 - Solution

A negative position is not a direction of travel

Diagram shown in the original SVG worksheet.

(a) Subtract the initial coordinate from the final coordinate.

ΔxA=−3−(−8)=+5m(5m east),ΔxB=−8−(−3)=−5m(5m west),ΔxC=−2−4=−6m(6m west).\begin{aligned}\Delta x_A&=-3-(-8)=+5\ \mathrm{m}\quad(5\ \mathrm{m}\text{ east}),\\\Delta x_B&=-8-(-3)=-5\ \mathrm{m}\quad(5\ \mathrm{m}\text{ west}),\\\Delta x_C&=-2-4=-6\ \mathrm{m}\quad(6\ \mathrm{m}\text{ west}).\end{aligned}

(b) A location sign and a change sign answer different questions.

A negative xx locates a point west of the origin. It does not say which way the point changes position. In A the final coordinate is greater than the initial one, so the net change is eastward even though both endpoints are west of the origin.

For C, 6 m is the displacement’s magnitude. The signed displacement is −6m-6\ \mathrm{m} because the final position is west of the initial position. Replacing the signed subtraction by endpoint separation discards direction.

(c) Equal magnitudes need not be equal displacements.

ΔxA=−ΔxB,|ΔxA|=|ΔxB|=5m.\Delta x_A=-\Delta x_B,\qquad|\Delta x_A|=|\Delta x_B|=5\ \mathrm{m}.

Runs A and B both have magnitude 5m\boxed{5\ \mathrm{m}}, but their displacements are +5m+5\ \mathrm{m} and −5m-5\ \mathrm{m}. On this specified axis, the sign supplies the net direction missing from the magnitude.

(d) Endpoints do not determine the whole path.

Ddirect=5m;D−8→0→−3=8+3=11m,Δx=5m.D_{\mathrm{direct}}=5\ \mathrm{m};\quad D_{-8\to0\to-3}=8+3=11\ \mathrm{m},\quad\Delta x=5\ \mathrm{m}.

A monotone path from −8-8 to −3m-3\ \mathrm{m} and a path from −8-8 to 00 and back to −3m-3\ \mathrm{m} both have Δx=+5m\Delta x=+5\ \mathrm{m}. The latter includes westward travel. Positive displacement therefore describes the net endpoint change, not a guarantee of eastward motion at every instant.

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