1. Mar 19, 2013

### Sixty3

1. The problem statement, all variables and given/known data
Two spaceships A and B are launched from a point X, in opposite directions.
At time t=15 minutes, spaceship A crashes.
The velocity of the spaceships relative to X is 1.3x10⁸m/s.

How far did the collision happen from B, as observed by astronauts on the spaceship?

2. Relevant equations
x'=ɣ(x-vt)
x=vt

3. The attempt at a solution
I've calculated ɣ to be 1.11

The distance from spaceship A to the point X is 1.17x10¹¹m, so the distance from point B to x is also 1.17x10¹¹m.

x'=ɣ(x-vt)
x'=1.11(2*1.17x10¹¹-1.3x10⁸(60*15))
x'=1.30x10¹¹m

So the distance is 2(1.17x10¹¹)+1.30x10¹¹=3.64x10¹¹m.

I am probably using the equations incorrectly, so if anyone could help me out, it will be appreciated.

2. Mar 19, 2013

### TSny

The Lorentz transformations relate the coordinates of an event as measured in the unprimed frame (call it the "earth frame") to the coordinates of the same event in the primed frame (the B rocket frame). Think of the crash of A as the event. What are the earth-frame values of x and t for this event? Note that the x coordinate of the event in the earth frame is not 2*1.17 x 1011m. After finding the corresponding x' value for the event as measured in the B rocket frame, interpret the meaning of that x' coordinate to decide on the answer to the question.

Last edited: Mar 19, 2013