Average Velocity and Average Speed — Question 4

Question 4

A chord summarizes an interval, not the whole graph

The solid position–time graph is exact and complete. Its vertices are $(0,-4)$, $(6,20)$, $(9,20)$, and $(12,8)$, with time in seconds and position in metres. Each segment is straight. Positive position is eastward.

1. Find $x(2\ \mathrm{s})$ from the first segment. For the interval from 2 to 12 s, calculate displacement, distance traveled, average velocity, and average speed.

2. Calculate the slope of the dashed chord using its endpoint coordinates. Explain its physical meaning and why the chord does not assert that the point followed it.

3. For the full interval from 0 to 12 s, find both averages. Explain why changing the initial event changes the averages even though the final event is unchanged.

4. Does the flat part of the graph add distance? Explain how it affects the elapsed-time denominator for intervals containing it. No derivative or instantaneous-velocity calculation is needed.

Question 4 - Solution

A chord summarizes an interval, not the whole graph

(a) Interpolate on the specified straight segment.

The first segment changes position by 24 m in 6 s, so at 2 s its position is $-4+(2/6)24=\boxed{4\ \mathrm{m}}$. From 2 to 12 s, $$\Delta x=8-4=4\ \mathrm{m},\quad D=(20-4)+0+(20-8)=28\ \mathrm{m},$$ $$T=10\ \mathrm{s},\qquad\boxed{\bar v=+0.40\ \mathrm{m/s}},\quad \boxed{\bar s=2.80\ \mathrm{m/s}}.$$

(b) A chord uses exactly the two selected events.

$$x_{\mathrm{chord}}(t)=4\ \mathrm{m}+(0.40\ \mathrm{m/s})(t-2\ \mathrm{s});\quad x_{\mathrm{chord}}(6\ \mathrm{s})=5.6\ne20\ \mathrm{m}.$$

Its slope is $(8-4)/(12-2)=0.40\ \mathrm{m/s}$, the interval-average velocity. It encodes net position change per elapsed time. The actual solid history rises to 20 m, remains there, and falls to 8 m; the chord suppresses these intervening details.

(c) Averages belong to an interval.

Over 0–12 s, $\Delta x=8-(-4)=12\ \mathrm{m}$ and $D=24+12=36\ \mathrm{m}$. Therefore $$\boxed{\bar v_{0\to12}=+1.00\ \mathrm{m/s}},\qquad \boxed{\bar s_{0\to12}=3.00\ \mathrm{m/s}}.$$ Changing the initial event changes both the numerator and the elapsed time. A final event alone cannot define an average.

(d) Constant position adds time without adding path length.

The 6–9 s plateau is a 3 s stop. It contributes zero distance and displacement, but all 3 s belong in either requested interval’s denominator. Horizontal length in this graph measures time, not physical distance.