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Time dilation 
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#1
Aug509, 10:03 PM

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 Dtp= d/v Dtp=Dt/lambda or dt/sqrt(1v^2/c^2) 3. The attempt at a solution Dtp=7.5m/40mph=.1875hr .1875=(5minutes/60minutes per hour)/sqrt(1v^2/40^2) rearranging it becomes: v=sqrt( ((.083hr/.1875hr)^21)*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
Aug509, 10:31 PM

HW Helper
P: 2,323

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?



#3
Aug509, 10:37 PM

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. 


#4
Aug609, 08:20 AM

HW Helper
PF Gold
P: 3,443

Time dilation



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