Average Velocity and Average Speed — Question 10

Question 10

Audit a trip summary against its interval records

A cart follows a continuous path on one east-positive line. A logger divides a 30 s run into three consecutive intervals. Each entry is an interval average, including all stops within that interval: Interval A lasts $8\ \mathrm{s}$ with $(\bar v_A,\bar s_A)=(+2,2)\ \mathrm{m/s}$. Interval B lasts $12\ \mathrm{s}$ with $(\bar v_B,\bar s_B)=(-1,3)\ \mathrm{m/s}$. Interval C lasts $10\ \mathrm{s}$; both of its averages are unknown. The full-run summary reports $\bar v=+0.8\ \mathrm{m/s}$ and $\bar s=2.4\ \mathrm{m/s}$. Initial position is $x=0$. Treat all data as exact. No maximum speed is imposed; piecewise-linear position histories are permitted.

1. Recover the full-run displacement and distance, then infer both averages for C. Check the necessary inequality $\bar s_C\geq|\bar v_C|$.

2. Find the positions at the three interval boundaries. Find the eastward and westward lengths required within B. Explain what C’s inferred pair says about reversals during C.

3. Construct one timed piecewise-linear position history matching every entry and the full-run summary. Give all vertices as $(t,x)$ pairs and verify lengths and durations.

4. Does the record uniquely determine the history or the presence of a stop? Exhibit a second compatible history with a stop.

5. Suppose only the full-run average-speed entry is changed to $2.0\ \mathrm{m/s}$, leaving every other supplied value fixed. Is that revised record feasible? Explain why testing only the full-run inequality would miss the problem.

Question 10 - Solution

Audit a trip summary against its interval records

(a) Subtract known contributions from the full totals.

The full displacement is $0.8(30)=24\ \mathrm{m}$ and distance is $2.4(30)=72\ \mathrm{m}$. A contributes displacement/distance $16/16\ \mathrm{m}$; B contributes $-12/36\ \mathrm{m}$. Thus C requires $$\Delta x_C=24-(16-12)=20\ \mathrm{m},\quad D_C=72-(16+36)=20\ \mathrm{m},$$ $$\boxed{\bar v_C=+2\ \mathrm{m/s},\qquad\bar s_C=2\ \mathrm{m/s}}.$$ The inequality holds with equality.

(b) Recover the boundaries and directional lengths.

At 8, 20, and 30 s, the positions are $\boxed{16,4,24\ \mathrm{m}}$. For B, $R_B+L_B=36$ and $R_B-L_B=-12$, giving $R_B=12$ and $L_B=24\ \mathrm{m}$. Both directions occur. C has distance equal to positive displacement, so it has no westward traveled length; eastward motion and pauses are allowed.

(c) Give a constructive consistency check.

Connect the following timed vertices by straight segments: $$\boxed{(0,0)\to(8,16)\to(12,28)\to(20,4)\to(30,24)}.$$ A covers 16 m in 8 s. B covers 12 m east in 4 s and 24 m west in 8 s: distance 36 m, displacement $-12\ \mathrm{m}$, duration 12 s. C covers 20 m east in 10 s. The total length is 72 m, displacement 24 m, and duration 30 s, reproducing both summary averages.

(d) Averages do not specify every interior event.

Keep A and B as above, but replace C by $(20,4)\to(25,4)\to(30,24)$. The cart waits 5 s and then covers 20 m east in 5 s. C still lasts 10 s with the same distance and displacement. Thus the records allow both a stop-free construction and one containing a stop.

(e) Test the residual interval, not only the full summary.

$$\bar s_C^{\mathrm{revised}}=\frac{60-16-36}{10}=0.8\ \mathrm{m/s}<|\bar v_C|=2\ \mathrm{m/s}.$$

The revised average speed permits only $2.0(30)=60\ \mathrm{m}$ overall, leaving $D_C=60-52=8\ \mathrm{m}$. But C still needs displacement $+20\ \mathrm{m}$, which requires at least 20 m of distance. The record is impossible, although its full-run inequality $2.0\geq|0.8|$ passes. Necessary conditions must also hold for the constrained subintervals.