Time dilation

[aft_lizard01]

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.

[ideasrule]

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]

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]

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."