In the twenty-second century, a hyperfast New York to Los Angeles train roars through St. Louis without stopping at a speed of 0.80c. The front of the train arrives at the St. Louis station at precisely 12:00 noon according to all the clocks on the train which are all synchronized in the train's frame), and the passengers are called to lunch in the dining car. An observer on the station platform manages to catch a glimpse of a clock at the front of the train that reads exactly 12:00 noon. What is the time at this instant at the rear of the train, according to the observer on the platform, if the train is 500 m long in its rest frame?
A) 1.35 μs past noon
B) 1.35 μs before noon
C) noon exactly
D) 1.35 ms past noon
E) 1.35 ms before noon

Question

Quizplus · Accepted Answer

The correct answer is A.The time dilation formula is:$$t' = \frac{t}{\gamma}$$where:* $t'$ is the time interval in the moving frame* $t$ is the time interval in the stationary frame* $\gamma$ is the Lorentz factorIn this problem, the train is moving at a speed of 0.80c, so the Lorentz factor is:$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.80^2}} = 1.67$$The train is 500 m long in its rest frame, so the time it takes for the front of the train to pass the observer on the platform is:$$t = \frac{500 	ext{ m}}{0.80c} = 2.08 \mu	ext{s}$$Therefore, the time interval between when the front of the train passes the observer and when the rear of the train passes the observer is:$$t' = \frac{t}{\gamma} = \frac{2.08 \mu	ext{s}}{1.67} = 1.35 \mu	ext{s}$$Since the front of the train arrives at the station at 12:00 noon, the rear of the train arrives at 12:00 noon + 1.35 μs.

In the Twenty-Second Century, a Hyperfast New York to Los