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Special Relativity, change in distance and time at slow speeds

  1. Sep 10, 2006 #1
    I was able to do pretty much every other problem dealing with this topic except this one which is bothering me. Here is the question:

    The distance from New York to Los Angeles is about 5000 km and should take about 50 h in a car driving at 100 km/h. (a) How much shorter than 5000 km is the distance according to the car travelers? (b) How much less than 50 h do they age during the trip?

    Can someone please give me a hint as to which equation I am looking for to solve this problem? Basically all equations with sqrt(1-(u^2/c^2)) are coming out completely wrong for me.

    I am able to get to sqrt(1-(27.78^2/3x10^8)) but dont really know how to deal with subtracting such a small number. Obviously my calculator is throwing 1 at me which screws up the equation.

    Any help is appreciated.
     
  2. jcsd
  3. Sep 10, 2006 #2

    Doc Al

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    Staff: Mentor

    Use the binomial expansion of (1 + x)^n (where x << 1) to estimate the change in distance: (1 + x)^n = 1 + nx + ...
     
  4. Sep 12, 2006 #3
    I am still at a loss here. I found the expansion to be 1+x/2-(x^2)/8 which really does not help me much. With x = 8.57*10^-15, I am still getting what essentially is 1.
     
  5. Sep 12, 2006 #4

    Janus

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    Do you have a scientific calculator function on your computer? (With Windows, it should be under "accessories".) I know that the one on my computer is capable of handling that many decimal places.
     
  6. Sep 13, 2006 #5

    Doc Al

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    Staff: Mentor

    Realize that you are being asked for the change in length, not the length.

    Try this:
    [tex]L = L_0 \sqrt{1 - v^2/c^2}= L_0 (1 - v^2/c^2)^{1/2}[/tex]

    Thus, using a binomial expansion (for v^2/c^2 << 1) :
    [tex]L \approx L_0 (1 - (1/2)v^2/c^2)[/tex]

    So:
    [tex]\Delta L \approx -(1/2) L_0 v^2/c^2[/tex]
     
    Last edited: Sep 13, 2006
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