Spaceship #1 moves with a velocity of .2c in the positive x direction of reference frame S. Spaceship #2, moving in the same direction with a speed of .6c is 3 x 10^9 m behind. At what times in reference frames S, and in the reference frame of ship #1, will 2 catch up with 1.
The Attempt at a Solution
vrelevative = .6c-.2c / (1-.6(.2)) = .455c
Then, 10 light seconds / .455c = 22.0s
Or...is it done classically .6-.2 = .4, making it 25s?
For the second part, is it done like so?
beta = 1/(sqrt (1-.2)) = 1.02
Then t = 22/1.02 = 21.6s or if it was 25s, 25/1.02 = 24.5s
Can someone explain to me which of these two (if either) are correct? I am not sure if it is done classically or with relativity, so can you explain why it is either.
Thank you very much.