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Homework Help: Time Dilation Problem

  1. Nov 24, 2009 #1
    A firecracker goes off in Houston, Texas. A time 0.0133 seconds later as measured by synchronized earth clocks, another firecracker goes off in Great Falls, Montana 2400 km away as measured on earth.

    a. How fast must a rocket ship travel if it is to be present at both events?

    b. What will the rocket ship pilot measure to be the time interval between the two events?

    Relevant equations: t = to / (1- v^2 / c^2)

    to calculate the speed needed by the ship to be in both places I used:

    s = d/t = 2400000 m / .0133 s = 1.8 x 10^8 m/s

    I use this speed for part b to get the time perceived by the pilot

    t = .0133 s / (1 - (180451127.8 ^2 / 299792458 ^2)
    t = .0209s

    I thought time would seem slower to the pilot, what did I do wrong here?
  2. jcsd
  3. Nov 24, 2009 #2
    Any help here? Does this look correct or am I off base with the equation that I am using?
  4. Nov 24, 2009 #3

    Doc Al

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    Staff: Mentor

    You are using the time dilation formula backwards. In that formula, 'to' is the time interval according to a single moving clock and 't' is the time interval as measured in the other frame. In this problem, the single moving clock is in the rocket ship.

    (Note that according to the rocket ship, the two earth clocks are not synchronized. So you can't treat the 0.0133 seconds as a time interval recorded by a single earth clock. According to the rocket ship each earth clock does not record 0.0133 seconds as passing during his flight--in fact they record much less.)
  5. Nov 24, 2009 #4
    Thanks - so the value of t is the earth clock and I am solving for to.

    I get a value of .00848 seconds. This value makes more sense from what I have read.
  6. Nov 24, 2009 #5

    Doc Al

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    Staff: Mentor

    Right. As you know, according to earth observers the rocket ship clock runs slow--thus it must show a smaller time interval.
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