Time Dilation: 1.25yr Trip at 2.40x10^8m/s

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An astronaut is traveling at a constant speed of 2.40x10^8 m/s relative to Earth through space. according to timing devices aboard the space vehicle the trip took 1.25 years. how long did the trip last if measured relative to earth
HELP PLEASE!
 
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I guess the answer is 2.08333 years.
1.25=x{1-(v/c)^2}^.5
x=2.08333
 
Last edited:
hemantc007 said:
I guess the answer is 2.08333 years.
1.25=x{1-(c/v)^2}^.5
x=2.08333

thats right!
but how did you get that?
do you need to convert the velocity?
 
pmor@hotmail. said:
thats right!
but how did you get that?
do you need to convert the velocity?

no you don't because , when we apply the formula (v/c)^2 both v and c are in same units i.e m/s...
 
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hmm I am still not getting the right answer when i plug it into my calculater..
 
Set the calculator aside, write it out on paper first.
 
see the formula is t1 = t2 * { 1 - (v/c)^2 }^(1/2)
where t1 is time in space shuttle (at 2.4...velocity)
t2 is time measured on Earth
c is velocity of light
v is velocity of space ship
The above formula is time dilation formula ...
now try it again

ya you could do it on paper too, its easy calculation...
 
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DaveC426913 said:
Set the calculator aside, write it out on paper first.

okay so it would be something like
1.25/ (1-(2.4x10^8)2)^(.5)
which gives me a negative square root
 
pmor@hotmail. said:
hmm I am still not getting the right answer when i plug it into my calculater..

The equation should be x= 1.25/{1-(v/c)^2}^(1/2) with (v/c) and not (c/v).

Use v= 2.40x10^8 m/s and c= 2.99792458x10^8 m/s and you should get the right answer.
 
  • #10
pmor@hotmail. said:
okay so it would be something like
1.25/ (1-(2.4x10^8)2)^(.5)
which gives me a negative square root

dude it is not v^2 ...but...it is v^2/c^2...
where c=299792458 m/s and
v is 2.4*10^8 m/s
 
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  • #11
okay i got ittt finally
thanks guyyys
 
  • #12
The lesson here, I hope you see, is that, if you rush through the work just to get an answer, you're not paying attention. Setting aside the calculator forces you to ensure that the process leading up to the number at the end is the important part.
 
  • #13
pmor@hotmail.
I am sorry for that mistake i.e replacing c by v...
 

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