g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward t
1 answer:
Answer:
the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
Explanation:
Given that;
speed of the first spacecraft from earth v = 0.80c
speed of the second spacecraft from earth v = -0.60 c
Using the formula for relative motion in relativistic mechanics
u' = ( v - v
) / ( 1 - (v
v
/ c²) )
we substitute
u' = ( 0.80c - ( -0.60c) ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )
u' = ( 0.80c + 0.60c ) / ( 1 - ( -0.48c² / c² ) )
u' = 1.4c / ( 1 - ( -0.48 ) )
u' = 1.4c / ( 1 + 0.48 )
u' = 1.4c / 1.48
u' = 0.9459c ≈ 0.95c { two decimal places }
Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
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