p38_026

# p38_026 - t = L v = 1 . 0 ly . 625 c = 1 . 6 y . (c)...

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26. (a) In the messenger’s rest system (called S m ), the velocity of the armada is v 0 = v v m 1 vv m /c 2 = 0 . 80 c 0 . 95 c 1 (0 . 80 c )(0 . 95 c ) /c 2 = 0 . 625 c. The length of the armada as measured in S m is L 1 = L 0 γ v 0 =(1 . 0ly) p 1 ( 0 . 625) 2 =0 . 781 ly . Thus, the length of the trip is t 0 = L 0 | v 0 | = 0 . 781 ly 0 . 625 c =1 . 25 y . (b) In the armada’s rest frame (called S a ), the velocity of the messenger is v 0 = v v a 1 vv a /c 2 = 0 . 95 c 0 . 80 c 1 (0 . 95 c )(0 . 80 c ) /c 2 =0 . 625 c. Now, the length of the trip is
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Unformatted text preview: t = L v = 1 . 0 ly . 625 c = 1 . 6 y . (c) Measured in system S , the length of the armada is L = L = 1 . 0 ly p 1 (0 . 80) 2 = 0 . 60 ly , so the length of the trip is t = L v m v a = . 60 ly . 95 c . 80 c = 4 . 0 y ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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