# Time dilation

by aft_lizard01
Tags: dilation, time
 P: 3 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.
 HW Helper P: 2,324 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?
 P: 3 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.
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## Time dilation

 Quote by aft_lizard01 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.
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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