Average Velocity and Average Speed — Question 2

Question 2

Averaging interval averages requires their durations

A cart has three consecutive motion intervals on an east-positive line. Their durations and interval-average velocities are known exactly: Interval A lasts $4\ \mathrm{s}$ with $\bar v_A=+3\ \mathrm{m/s}$; B lasts $10\ \mathrm{s}$ with $\bar v_B=-1\ \mathrm{m/s}$; C lasts $6\ \mathrm{s}$ with $\bar v_C=+2\ \mathrm{m/s}$. Within each interval the cart never reverses direction, although its rate of motion may vary and it may pause. Use $\bar v=\Delta x/T$ and $\bar s=D/T$.

1. Recover each interval’s displacement. Find the full-run average velocity and average speed.

2. A student takes the simple mean of the three listed velocities. Calculate that mean and explain why it fails for this record.

3. Derive a formula for the average velocity across intervals with durations $T_i$ and averages $\bar v_i$. What duration condition guarantees that the simple arithmetic mean works for arbitrary listed velocities?

4. Remove the no-reversal assumption but keep the interval averages and durations. Which full-run average is still exactly known? What lower bound remains for the other, and why?

Question 2 - Solution

Averaging interval averages requires their durations

(a) Undo each interval’s division before combining intervals.

$$\bar v=\frac{3(4)-1(10)+2(6)}{4+10+6}=\frac{14}{20},\qquad\bar s=\frac{12+10+12}{20}.$$

From $\Delta x_i=\bar v_iT_i$, the displacements are $+12,-10,+12\ \mathrm{m}$. Their sum is $+14\ \mathrm{m}$ over 20 s: $$\boxed{\bar v=14/20=+0.70\ \mathrm{m/s}}.$$ Because each interval has no reversal, its distance equals its displacement magnitude. Hence $D=12+10+12=34\ \mathrm{m}$ and $$\boxed{\bar s=34/20=1.70\ \mathrm{m/s}}.$$ Pauses do not affect these deductions: their time is already included in the interval durations.

(b) Equal weighting treats unequal durations as though they were equal.

The simple mean is $(3-1+2)/3=4/3\ \mathrm{m/s}$. It gives the 4 s interval the same influence as the 10 s interval, so it does not reproduce total displacement divided by total time.

(c) Weight each average by the time it represents.

Displacements and durations both add over consecutive intervals: $$\boxed{\bar v_{\mathrm{all}}=\frac{\sum_i\bar v_iT_i}{\sum_iT_i}}.$$ If all durations are equal, the common duration cancels, yielding the arithmetic mean. Equal durations guarantee the result for arbitrary velocities. Unequal durations can coincidentally give the same mean for special data, but do not justify that rule in general.

(d) Hidden reversals change traveled length, not the recorded net changes.

The interval displacements and full-run $\bar v=+0.70\ \mathrm{m/s}$ remain exact. Each interval must have distance at least $|\Delta x_i|$, so $\boxed{\bar s\geq1.70\ \mathrm{m/s}}$. Reversals can add distance without altering an interval’s endpoints or duration; with no bound on the motion rate, the records give no finite upper bound on average speed.