Thanks! So would it be 9.80cs / .455 = 21.5s?
When I use the Lorenz equation, I get
t' = (1/(sqrt(1-.2^2)) ) * 25 - (1/(sqrt(1-.2^2)) ) .2 * 10, I get 23.47?
Homework Statement
In a given reference frame, event 1 occurs at t1 = 0 s and position x1 = 0 m while event 2 occurs at t2 = 3.6 × 10^−4 s and x2 = 0.60 × 10^5m. Is there a second reference frame in which these events could be at the same place but different times? If so, specify its motion...
Thank you very much!
For the second part, would it be the following --
Vrelative = .455c (as shown above)
L = 10c seconds (sqrt(1-.45^2)) = 8.93
8.93/.455 = 19.6s
I'm not sure if I'm using the correct v for all parts, would you be able to tell me if I did this correct? Thank you...
Homework Statement
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...