Average Velocity and Average Speed — Question 1

Question 1

The clock reading is not the elapsed time

A walker moves along an east-positive straight path. At clock reading $t=10\ \mathrm{s}$ she is at $x=-12\ \mathrm{m}$. She walks directly to $x=18\ \mathrm{m}$, arriving at $t=20\ \mathrm{s}$, waits there until $t=25\ \mathrm{s}$, then walks directly back to $x=6\ \mathrm{m}$, arriving at $t=31\ \mathrm{s}$.

For an interval of positive duration $T=t_f-t_i$, define $$\bar v=\frac{x_f-x_i}{T},\qquad \bar s=\frac{D}{T},$$ where $D$ is the total distance traveled. Average velocity $\bar v$ is signed; average speed $\bar s$ is nonnegative. Unless explicitly restricted, elapsed time includes stops.

1. Find displacement, distance, and elapsed time for the full recorded trip. Then find $\bar v$ and $\bar s$, with units and direction where appropriate.

2. A student divides the displacement by 31 s. Explain the error and show why resetting the clock to zero at departure cannot alter either average.

3. Another student removes the 5 s wait and reports a larger average speed. Calculate that value and explain which interval convention it describes.

4. Does the positive average velocity prove that the walker always traveled east? Use the record to justify your answer.

Question 1 - Solution

The clock reading is not the elapsed time

(a) Use the whole path for distance and endpoints for displacement.

The two moving legs have signed changes $+30$ and $-12\ \mathrm{m}$. Thus $$\Delta x=6-(-12)=18\ \mathrm{m},\quad D=30+12=42\ \mathrm{m}, \quad T=31-10=21\ \mathrm{s}.$$ Consequently, $$\boxed{\bar v=\frac67\ \mathrm{m/s}\approx+0.857\ \mathrm{m/s}\text{ (east)}}, \qquad \boxed{\bar s=2.00\ \mathrm{m/s}}.$$ The wait contributes time but no distance or displacement.

(b) A time origin must cancel.

$$T'=(t_f-c)-(t_i-c)=t_f-t_i=T,\qquad\bar v'=\bar v,\quad\bar s'=\bar s.$$

The reading 31 s is measured from the clock’s zero, not from departure. Dividing by it would include 10 s outside the selected interval. With $t'=t-10\ \mathrm{s}$ the endpoint readings become 0 and 21 s, but their difference remains 21 s. Both averages therefore remain unchanged.

(c) Moving-time and full-trip averages use different denominators.

Removing the wait leaves $10+6=16\ \mathrm{s}$ of moving time. The corresponding distance per moving time is $$\boxed{42/16=2.625\ \mathrm{m/s}}.$$ This can be useful if explicitly labeled as excluding stops. It is not the average speed over the full departure-to-arrival interval, whose duration is 21 s. A reporting convention should identify both endpoint events and whether stationary periods are included.

(d) An average summarizes the net change.

The walker travels west on the final leg, even though the total displacement is eastward. Positive $\bar v$ establishes a positive net displacement over a positive elapsed time; it does not describe every part of the motion.