Travelling to a Star 40 Light Years Away: Time and Distance Calculations

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usahockey7
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A star is 40 light years from Earth.
(a) How far would you measure this distance to be if you traveled it in a spaceship moving at 1.00X10^8 m/s

(b) How long would the trip last (for you)?

Can someone please help me out with this question and what equations I should use to solve this, thanks in advance
 
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Look up length contraction, this should be a fairly simple question.
 
Pengwuino said:
Look up length contraction, this should be a fairly simple question.

okay so for part a, convert light years into meters and that will be the rest length or the distance observed by someone from Earth and then I need to solve for the relativistic length of the distance observed from someone in the spaceship?

And then for part b, use the rest length I calculated in part a, and use the velocity equation v = d/t and solve for the time, which in this case would be the time it would take the space shuttle to reach the star observed by someone on Earth and use that to find the time that the person in the space shuttle would experience?
 
usahockey7 said:
okay so for part a, convert light years into meters and that will be the rest length or the distance observed by someone from Earth and then I need to solve for the relativistic length of the distance observed from someone in the spaceship?

Yes. It would be shorter, of course.

And then for part b, use the rest length I calculated in part a, and use the velocity equation v = d/t and solve for the time, which in this case would be the time it would take the space shuttle to reach the star observed by someone on Earth and use that to find the time that the person in the space shuttle would experience?

Sure, but the more direct approach is using the contracted length from part a) and the fact that you know what speed the ship is traveling at, and immediately solve for the time you see as someone on the ship.
 
Pengwuino said:
Yes. It would be shorter, of course.



Sure, but the more direct approach is using the contracted length from part a) and the fact that you know what speed the ship is traveling at, and immediately solve for the time you see as someone on the ship.

Perfect, thanks for helping and explaining this rather than just giving me an answer.