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Solving in multiple frames

  1. Nov 1, 2006 #1
    A train of length 15cs moves at speed 3c/5. How much time does it take to a pass a person standing on the ground? [That is, what time elapses on the person's watch between the time when he is next to the front of the train, and the time when he is next to the back of the train?] Solve this by working in the frame of the person, and then again by working in the frame of the train.
  2. jcsd
  3. Nov 1, 2006 #2
    Do you know the lorentz transformations? Time dilation and length contraction? I'm not familar with the unit cs (usually would mean a centisecond) in this context.
  4. Nov 1, 2006 #3
    Oh, sorry.

    1 cs equals (3 x 10^8 m/s)(1s) = 3 x 10^8 m

    Also, I know a little bit about time dilation, though I didn't really understand my professor's lecture very well.
  5. Nov 1, 2006 #4
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