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Relativity Problem. I feel so stupid.

  1. Jan 9, 2005 #1
    How fast must a pion be moving in order to travel 15 m before it decays?
    The avg lifetime, at rest, is 2.6 x 10^-8 s.

    The answer is supposed to be 0.89c

    This sounds so easy. Like, just plugging in the formula but I'm not getting it. I haven't done relativity in so long. I think I must've forgotten more than I thought. I can't even find my notes now. Can someone please tell me how to get to the answer?
  2. jcsd
  3. Jan 9, 2005 #2
    We arent even touching on reletivity in my AP Physics BC class so I cant help you
  4. Jan 10, 2005 #3
    Thanks yapper for looking at the problem, at least.

    Anyone else?

  5. Jan 10, 2005 #4
    Try the college boards
  6. Jan 10, 2005 #5
    the answer is not .89c, [tex] \gamma = 1/ \sqrt{1-.89^2} [/tex]~[tex] 2 [/tex] it must travel at much higher speed, at leat .99c up (Sorry, i don't have a calculator handy and cant give you the exact answer), your answer is propabaly right, don't alway trust the textbook answer...
    Last edited: Jan 10, 2005
  7. Jan 10, 2005 #6


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    Find a formula for the time in the stationary frame in terms of the velocity v. Then use that time in the time dilation formula. T'=T(1-v^2/c^2)^(1/2). Then solve for v.

    Remember T' is the time in the moving frame (in the pion's frame).

    I get 2.66*10^-8m/s which is approximately 0.89c
  8. Jan 10, 2005 #7


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    Let the pion's resting lifespan be [itex]\tau[/itex]. The distance it is required to travel from the perspective of a resting observer is [itex]d[/itex], and the pion's velocity relative to the resting observer is [itex]v[/itex]. [itex]c[/itex] is the speed of light.

    From the frame of the pion, it will always "measure" its lifespan as [itex]\tau[/itex]. This is the proper time interval. The resting observer will measure this time interval as [itex]\gamma \tau[/itex] where [itex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex].

    The observer measures the velocity of the pion as [itex]v[/itex], where :

    [tex]\frac{d}{\gamma \tau} = v[/tex]

    [tex]\frac{d}{\tau} = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

    Put the relevant values in and solve for v and the answer is [itex]v = 0.89c[/itex]
  9. Jan 10, 2005 #8
    Awesome. Thanks!
    I got the same answer as vincentchan. And other people did too so I just left it at that. I'm so grateful it's been proved otherwise.

    V.v.v. thankful. :rofl:
  10. Jan 10, 2005 #9
    just wanna point out...
    2.66*10^-8m/s is 0.89*10^-16c
  11. Jan 10, 2005 #10

    y don't you simply use [tex] t = \gamma \tau [/tex]
    and again, v is not .89c
    Last edited: Jan 10, 2005
  12. Jan 10, 2005 #11


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    Show us your reasoning.
  13. Jan 10, 2005 #12
    oh, sorry, I thought its say the particle travel 15 mins.... I didn't realize it is 15 meters, you are right, and so do the answer
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