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Time dilation

by aft_lizard01
Tags: dilation, time
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aft_lizard01
#1
Aug5-09, 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
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.
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ideasrule
#2
Aug5-09, 10:31 PM
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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?
aft_lizard01
#3
Aug5-09, 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.

kuruman
#4
Aug6-09, 08:20 AM
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Time dilation

Quote Quote by aft_lizard01 View Post
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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