As in the figure below, reference frame S \' passesreference frame S with a cert

Question

As in the figure below, reference frame S' passesreference frame S with a certain velocity. Events 1 and 2are to have a certain temporal separation t'according to the S' observer. However, their spatialseparation x' according to that observer has notbeen set yet. Figure 37N-4 gives their temporal separationt according to the S observer as a functionof x' for a range of x' values. Whatis t'? (The vertical axis is marked in increments of500 µs and the horizontal axis ismarked in increments of 500 m.)

Explanation / Answer

Given : The spatial separation in S ' frame is                x' = 2 (50 m)                       =100 m          ...............1 The velocity of the S ' frame with respect to S frame be v   then temporal separation is S' frameis               t ' = x ' / v                      =100 / v          ..............2 we know t ' = t / [ 1 - ( v / c) 2]            Where temporal separation in S frame t = 50 s + ( 500 s / 3 )                                                                  = ( 650 / 3 ) s      ..........3 so that he velocity of frame S' is                             v= c [ 1 - ( t / t' ) 2]                                 = 3 x 108m/s [ 1 - ( 650s *v / 3*100 ) 2 ]   Solving for v we get                             v = ------ m /s           ...............4                                  Substituting 4 in 2 we get        t ' = 100/ v             =----- s Solve it I hope it helps you The spatial separation in S ' frame is                x' = 2 (50 m)                       =100 m          ...............1 The velocity of the S ' frame with respect to S frame be v   then temporal separation is S' frameis               t ' = x ' / v                      =100 / v          ..............2 we know t ' = t / [ 1 - ( v / c) 2]            Where temporal separation in S frame t = 50 s + ( 500 s / 3 )                                                                  = ( 650 / 3 ) s      ..........3 so that he velocity of frame S' is                             v= c [ 1 - ( t / t' ) 2]                                 = 3 x 108m/s [ 1 - ( 650s *v / 3*100 ) 2 ]   Solving for v we get                                                                  = ( 650 / 3 ) s      ..........3 so that he velocity of frame S' is                             v= c [ 1 - ( t / t' ) 2]                                 = 3 x 108m/s [ 1 - ( 650s *v / 3*100 ) 2 ]   Solving for v we get                             v = ------ m /s           ...............4                                  Substituting 4 in 2 we get        t ' = 100/ v             =----- s Solve it I hope it helps you

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