The following problem is part of a homework from last year. I never did understand the solution, and it's been nagging at me ever since. I've discussed the problem with my tutor and with my friends, and they all come to the same answer. They have tried (patiently) to explain this answer to me over and over, but I fail to understand their reasoning.(adsbygoogle = window.adsbygoogle || []).push({});

Below, I have described as clearly as I can my attempt at the problem. Could somebody explain to me in very simple terms where I go wrong?

Ok, so part A: 7.5 miles. That's fine. You can ignore the part about speeding tickets. Suppose that the speed of light in vacuum is c=40.0 mph only and Einstein postulates holds. Assume further that times for acceleration and deceleration are negligible, and one can get to c easily.

A) A hungry student likes to go for lunch in his one hour break. He knows that it takes usually 30 min to get served and to eat. What is the farthest restaurant the student can go to without being late for class? The local speed limit is 30 mph and he would not like to get a speeding ticket which would cost him 30 GBP for each mph he is over the limit.

B) According to the student’s watch, how much time does he actually have for being in the restaurant (in min) if he likes to go to farthest restaurant taking into account the travel time? Will it be more, less or equal 30 min?

Part B is the problem. The official solution is, "The student has longer than 30 minutes to spend in the restaurant, according to his watch". I can't get my head around that. My answer is that the student has exactly 30 minutes in the restaurant. My reasoning is below.

——————————————————

1. Student departs the classroom at [itex]{t_0} = {t'_0} = [/itex] 12:00.

- Let the classroom-restaurant frame be [itex]S[/itex].
- While the student is travelling between the classroom and restaurant at 30 mph, let his frame be [itex]S'[/itex].
- Assume that the classroom clock is synchronised with the restaurant clock. Call the time on these clocks [itex]t[/itex].
- Call the time on the student's wristwatch [itex]t'[/itex].
- Let [itex]{t_0} = {t'_0} = [/itex] 12:00.

2. Student travels 7.5 mi from the classroom to the restaurant at 30 mph (w.r.t. the [itex]S[/itex] frame). [itex]\gamma=[/itex] 1.51.

2. Student arrives at the restaurant at [itex]t=[/itex] 12:15, [itex]t'=[/itex] 12:10 (approx.). Upon reaching the restaurant, the student stops his car, thus exiting the [itex]S'[/itex] frame and entering the [itex]S[/itex] frame.

3. The student's watch is now currently ticking with [itex]S[/itex]-frame minutes. Suppose the student stays in the restaurant for 30 of these minutes.

4. Student departs the restaurant at [itex]t=[/itex] 12:45, [itex]t'=[/itex] 12:40 (approx.).

5. Student travels 7.5 mi from the restaurant back to the classroom at 30 mph.

6. Student arrives at the classroom at [itex]t=[/itex] 13:00, [itex]t'=[/itex] 12:50 (approx.).

It seems to me that if the student spends longer than 30 minutes in the restaurant then he arrivesafter[itex]t=[/itex] 13:00 and is therefore late for class. However, my friends and tutor disagree (adamantly!). What is the real solution?

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