Time — Question 3

Question 3

Two reaction delays can cancel, or reinforce a timing error

An observer starts a perfect stopwatch after event A and stops it after event B. The start reaction delay is $r_s$ and the stop reaction delay is $r_e$, both measured positively after the corresponding event. The stopwatch displays $M=12.480\ \mathrm{s}$.

1. Derive a relation between $M$ and the true event interval $\Delta t=t_B-t_A$. Evaluate $\Delta t$ using nominal delays $r_s=0.180\ \mathrm{s}$ and $r_e=0.260\ \mathrm{s}$.

2. Now replace the nominal delays by hard bounds $r_s\in[0.140,0.220]\ \mathrm{s}$ and $r_e\in[0.200,0.320]\ \mathrm{s}$. Every pair of allowed delays is possible. Find the full allowed interval for $\Delta t$.

3. A student subtracts the sum $r_s+r_e$ from the displayed time. Explain why the two delays enter with opposite signs instead.

4. Under what condition do reaction delays cancel exactly? Does using the same observer guarantee that condition, or justify eliminating the bounds by repeated trials?

Treat stopwatch resolution and timebase error as negligible here; the only errors modeled are the two reaction delays.

Question 3 - Solution

Two reaction delays can cancel, or reinforce a timing error

(a) Write the two click times before subtracting.

$$M=(t_B+r_e)-(t_A+r_s)=\Delta t+r_e-r_s, \qquad \Delta t=M-r_e+r_s.$$ With the nominal delays, $$\Delta t=(12.480-0.260+0.180)\ \mathrm{s} =\boxed{12.400\ \mathrm{s}}.$$

(b) Use the extreme allowed combinations.

The smallest true interval uses the largest stop delay and smallest start delay; the largest uses the opposite pair: $$\Delta t_{\min}=(12.480-0.320+0.140)\ \mathrm{s}=12.300\ \mathrm{s},$$ $$\Delta t_{\max}=(12.480-0.200+0.220)\ \mathrm{s}=12.500\ \mathrm{s}.$$ Thus $\boxed{\Delta t\in[12.300,12.500]\ \mathrm{s}}$. These are hard bounds, not a probabilistic confidence interval.

(c) Starting late omits time; stopping late adds time.

A delayed start shortens the measured interval, whereas a delayed stop lengthens it. Their net contribution to $M$ is $r_e-r_s$, not their sum. Subtracting both would incorrectly treat the delayed start as added running time.

(d) State the cancellation condition.

$$M-\Delta t=r_e-r_s=0\ \Longleftrightarrow\ r_e=r_s.$$

Exact cancellation requires $r_e=r_s$ for the trial. One observer may react differently to different events or cues, so identity of the observer is insufficient. Repetition alone does not remove an unknown systematic difference. Any claim that averaging narrows these bounds would require additional, justified information about the delays or their statistical behavior.