Average Velocity and Average Speed — Question 8

Question 8

A time offset cancels only when it is shared

A marker is observed on a fixed east-positive line. Its initial position is within $[-0.2,0.2]\ \mathrm{m}$ and its final position within $[19.8,20.2]\ \mathrm{m}$. Two readings of the same ideal clock are 100.0 s and 110.0 s. This clock may have an unknown constant offset, but its elapsed intervals are exact. All position bounds are hard bounds, and every combination of endpoint positions within them is allowed.

1. Find the tight interval of possible average velocities. Explain why the clock’s unknown constant offset does not widen this interval.

2. Would these data alone establish an average-speed interval with a finite upper endpoint? State a tight lower bound and explain what additional no-reversal information would change.

3. In a separate measurement setup, the two endpoint readings come from different clocks: $r_i=t_i+b_i$ and $r_f=t_f+b_f$. Each offset independently lies in $[-0.1,0.1]\ \mathrm{s}$, and the readings are still 100.0 and 110.0 s. Derive the possible true elapsed-time interval and the tight average-velocity bounds.

4. Why is subtracting an arbitrary “average clock error” unjustified in the separate-clock setup? No probability distribution is supplied.

Question 8 - Solution

A time offset cancels only when it is shared

(a) Bound displacement and divide by the exact duration.

The smallest displacement is $19.8-0.2=19.6\ \mathrm{m}$ and the largest is $20.2-(-0.2)=20.4\ \mathrm{m}$. A shared offset cancels: $$(r_f-b)-(r_i-b)=r_f-r_i=10.0\ \mathrm{s}.$$ Thus the tight interval is $\boxed{\bar v\in[1.96,2.04]\ \mathrm{m/s}}$. The endpoint extremes are allowed by the independent hard bounds.

(b) Endpoints do not bound hidden traveled length from above.

All allowed displacements are positive, so distance is at least 19.6 m and $\boxed{\bar s\geq1.96\ \mathrm{m/s}}$. Monotone travel between the closest allowed endpoints attains that lower bound. Unrestricted reversals admit arbitrarily large distances in the fixed time when no rate bound is specified. With independent knowledge of no reversal, $D=\Delta x$ and average speed would instead lie in the same finite interval $[1.96,2.04]\ \mathrm{m/s}$.

(c) Independent offsets need not cancel.

$$\Delta x_{\min}=x_{f,\min}-x_{i,\max},\quad\Delta x_{\max}=x_{f,\max}-x_{i,\min};\quad\bar v\in\left[\frac{\Delta x_{\min}}{T_{\max}},\frac{\Delta x_{\max}}{T_{\min}}\right].$$

$$T=(r_f-b_f)-(r_i-b_i)=10.0+b_i-b_f\ \mathrm{s}, \qquad \boxed{T\in[9.8,10.2]\ \mathrm{s}}.$$ With positive numerators and denominators, the smallest ratio uses the smallest displacement and longest time; the largest uses the reverse: $$\boxed{\bar v\in\left[\frac{19.6}{10.2},\frac{20.4}{9.8}\right]\ \mathrm{m/s} =\left[\frac{98}{51},\frac{102}{49}\right]\ \mathrm{m/s}}.$$ These are approximately $[1.9216,2.0816]\ \mathrm{m/s}$; the fractions specify the exact bounds. Independent allowed endpoint positions and offsets attain both extremes.

(d) Hard bounds are not statistical corrections.

Nothing specifies likely offsets or a shared value. Replacing the independent offsets by a chosen average assumes information not provided and can discard allowed cases. Their difference, rather than either absolute offset alone, controls the elapsed-time error.