Time dilation

  1. 1. The problem statement, all variables and given/known data

    Now, suppose the student wishes to bring back some ice cream from the restaurant for her friends at school, but since it is such a hot day, the ice cream will melt away in the car in only 5 minutes. How fast will the student have to drive back to get the ice cream to her friends before it completely melts?

    Knowns:
    C=40mph(for these problems)
    Time=5 minutes
    Distance=7.5 miles



    2. Relevant equations
    Dt= change in time
    p= proper interval
    Dt-p= d/v

    Dt-p=Dt/lambda or dt/sqrt(1-v^2/c^2)

    3. The attempt at a solution

    Dt-p=7.5m/40mph=.1875hr

    .1875=(5minutes/60minutes per hour)/sqrt(1-v^2/40^2)

    rearranging it becomes:

    v=sqrt( -((.083hr/.1875hr)^2-1)*40^2)

    answer I get is 35.8mph

    The online homework tells me I am wrong either by sig figs or by bad rounding.
     
  2. jcsd
  3. ideasrule

    ideasrule 2,322
    Homework Helper

    Well then, don't round until you get to the final answer, and try out different numbers of sig figs. Have you tried 36 mph?
     
  4. I have tried 36, 35.8,35.83

    Edit:
    I went back and put 37 and it said I was correct, the correct answer was actually 36.6. I haven't the faintest clue as to how it is 36.6 mph.
     
    Last edited: Aug 5, 2009
  5. kuruman

    kuruman 3,448
    Homework Helper
    Gold Member

    Yes, the correct answer is 36.55 mph. The problem is not with the number crunching but with the set-up of the equations. The correct time interval for fetching the ice-cream in the Earth's frame is 7.5/v not 7.5/40. The student does not travel at the speed of "light."
     
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