Position and Displacement — Question 5

Question 5

Read a position graph as position, not accumulated travel

A point travels on a fixed line. The graph specifies its complete position $x$ during $0\leq t\leq8\ \mathrm{s}$; each drawn segment is exact. The vertices are $(0,-2)$, $(2,4)$, $(4,4)$, $(6,-2)$, and $(8,2)$, with time in seconds and position in metres.

Displacement over an interval is its final position minus its initial position. For a segment with no reversal, traveled distance equals the magnitude of its position change. No calculation of velocity, slope, or graph area is required.

1. Find the displacement and total distance traveled from $t=0$ to $t=8\ \mathrm{s}$.

2. Interpret the horizontal segment from 2 to 4 s. Why does its visible horizontal length on the page not add physical distance traveled?

3. Compare the displacement and traveled distance from 0 to 6 s. Does returning to the initial coordinate mean the point was at rest during that interval?

4. A student reports a traveled distance of 2 m because the final graph height is 2 m. Explain which quantity that height represents and which features of the graph must instead be used to find distance.

Question 5 - Solution

Read a position graph as position, not accumulated travel

(a) Use endpoints for displacement and all legs for distance.

$$\Delta x=2-(-2)=4\ \mathrm{m},\qquad D=|4+2|+|4-4|+|-2-4|+|2+2|=16\ \mathrm{m}.$$

The initial and final positions are $-2$ and $+2\ \mathrm{m}$, so $$\Delta x=2-(-2)=\boxed{+4\ \mathrm{m}}.$$ The four segments contribute lengths $6$, $0$, $6$, and $4\ \mathrm{m}$: $$D=|4-(-2)|+|4-4|+|-2-4|+|2-(-2)|=\boxed{16\ \mathrm{m}}.$$ The graph is specified throughout, so there are no hidden additional reversals to include.

(b) Horizontal on this graph means unchanged position.

From 2 to 4 s, $x=4\ \mathrm{m}$ throughout. The point is at rest relative to this axis during that interval and contributes zero traveled distance. The graph’s horizontal coordinate measures time, not another spatial direction. The length of the inked segment on the page is therefore not a physical path length.

(c) A return cancels displacement but not traveled length.

$$\Delta x_{0\to6}=0,\qquad D_{0\to6}=6+0+6=12\ \mathrm{m}.$$

At 6 s the position is again $-2\ \mathrm{m}$. Thus $$\Delta x_{0\to6}=\boxed{0\ \mathrm{m}},\qquad D_{0\to6}=6+0+6=\boxed{12\ \mathrm{m}}.$$ The point moved to 4 m and returned. Zero displacement means equal endpoint positions, not rest throughout the interval.

(d) Position is a location relative to the chosen origin.

The final height $+2\ \mathrm{m}$ locates the point at the final instant. It is neither the net change of 4 m nor the accumulated travel of 16 m. Distance requires adding the magnitudes of the successive position changes, including both directions, using the supplied complete path and its turning points.