How Does Time Dilation Affect Flight Duration in Special Relativity?

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In the discussion on time dilation and flight duration in special relativity, a plane traveling at 208 m/s is analyzed to determine the difference in flight time as perceived by clocks on the plane versus those on the ground. The formula used for calculations is delta t = delta t(not) / (1-v^2/c^2), and the user applied a binomial expansion due to the low velocity. The initial calculation yielded a result of 0.999, which was questioned for accuracy, indicating a potential math error. Clarifications were provided regarding the meaning of delta(t) and delta(t0), emphasizing that delta(t) represents the difference in time. The discussion highlights the importance of precision in calculations when applying special relativity concepts.
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Homework Statement



A plane flies with a constant velocity of 208 m/s. The clocks on the plane show that it takes exactly 2.00 hours to travel a certain distance. Calculate how much longer or shorter than 2.00 h this flight will last, according to clocks on the ground. ________s

Homework Equations



delta t = delta t(not) / (1-v^2/c^2)



The Attempt at a Solution



Because the velocity of the plane is very low, I used the binomial expansion: delta t = delta t(not)*[1 + 1/2*[208m/s / 3x10^8 m/s]^2]. When I worked it out and solved for delta t, I got .999, which I then added to the 7200s ( 2 hours). I know that the time according to the ground clocks will be longer, but I think there's a math error somewhere which is preventing me from understanding and getting the answer. Any help is appreciated. Thank you!
 
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What are delta(t) and delta(t0) (nought not not) supposed to mean? Get rid of the deltas in that formula and realize delta(t)=t-t0. And you won't get .999 unless your calculator only has three digits of precision.
 
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