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Special relativity problem

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    • Thread Starter

    hi i'm just wondering if anyone could help me with the latter parts of this problem, for the length of the train in its frame i got 80m, if anyone could explain where to go next for the rest of the question i woName:  relativity.png
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    • Thread Starter


    (Original post by Scary)
    You're more likely to get help if you post this in the physics forum.

    The answer to part a) is 100m. Proper length is the length of an object in its own rest frame, which is the length it measures itself as, so the trains length in its own frame is given in the question- it's simpler than it looks! The 80m you calculated is the answer to part b) because the platform does not see the train as at rest so you use the length contraction equation.

    The next two parts are best done with separate diagrams showing the reference frames of the train and platform. From the train's reference frame:
    Name:  Relativity 1.jpg
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    From the train's reference frame it is stationary, so the platform is moving at 0.6c. The rear of the train passes the marker when the platform has moved 100m at 0.6c.
     time=\frac{100}{0.6 \times 3.0 \times 10^8} =5.6 \times 10^-7  s
    See if you can do a similar diagram and calculation for the platforms reference frame.
    • Thread Starter

    hi thanks for responding, i've managed to solve the problem a few days ago, i was unsure where you could calculate the time tried everything from suvat to the others, turns out it was just speed=distance/time, anyway thanks for the response i appreciate it.
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