Average Velocity and Average Speed — Question 7

Question 7

Two averages reveal how much travel opposed the net change

During a 50 s run on a straight east-positive track, a cart has average velocity $\bar v=-0.8\ \mathrm{m/s}$ and average speed $\bar s=2.4\ \mathrm{m/s}$. Both averages refer to the same full interval, including any pauses. Let $R$ and $L$ denote the total eastward and westward traveled lengths.

1. Find displacement and total distance, then recover $R$ and $L$ using $D=R+L$ and $\Delta x=R-L$.

2. What fraction of the total distance was traveled eastward? Does that fraction also give the fraction of elapsed time spent moving east? Explain what is missing.

3. For any finite-length path of positive duration, derive $\bar s\geq|\bar v|$. State the equality condition in terms of $R$ and $L$, allowing pauses.

4. An alternative report for this same 50 s interval lists $\bar v=-2.8\ \mathrm{m/s}$ and $\bar s=2.4\ \mathrm{m/s}$. Diagnose its incompatibility without inventing a path.

Question 7 - Solution

Two averages reveal how much travel opposed the net change

(a) Recover lengths from the two averages.

$$R=\frac{T}{2}(\bar s+\bar v),\qquad L=\frac{T}{2}(\bar s-\bar v).$$

$$\Delta x=\bar vT=-40\ \mathrm{m},\qquad D=\bar sT=120\ \mathrm{m}.$$ Adding and subtracting $R+L=120$ and $R-L=-40$ gives $$\boxed{R=40\ \mathrm{m}},\qquad\boxed{L=80\ \mathrm{m}}.$$ Net westward travel still includes substantial eastward distance.

(b) A length fraction is not generally a time fraction.

The eastward share of total distance is $40/120=\boxed{1/3}$. Direction-specific rates and pause durations are unknown. For example, the same 40 m east and 80 m west could use 10 s east and 40 s west, or 25 s in each direction, with suitable constant rates and no pauses. Both runs last 50 s and have the same two averages, but different eastward time fractions.

(c) Nonnegative lengths impose a universal inequality.

For $R,L\geq0$, $R+L\geq|R-L|$. Dividing by $T>0$ yields $$\boxed{\bar s=\frac{R+L}{T}\geq\frac{|R-L|}{T}=|\bar v|}.$$ Equality holds exactly when $\min(R,L)=0$: all nonzero travel is in only one direction. Pauses are allowed. If both lengths are zero, the stationary path also gives equality. Any positive traveled length in both directions makes the inequality strict.

(d) Test the report against the necessary inequality.

The alternative requires $|\bar v|=2.8>2.4=\bar s$, violating the bound. Equivalently it asks for displacement magnitude 140 m with only 120 m of traveled length. No path in this model can do that. At least one reported value, interval definition, or assumption must be incorrect; a choice of route cannot repair the contradiction.