Observer O sees a fire-engine leave its station 6363 m due
north from Cape Canaveral, where the super-shuttle Lorentz had been
launched 10^-5 s earlier. A space-cruiser flying north-east sees these two events
also 10^-5 s apart, but with the shuttle launch occurring after the fire-engine
leaves the station.
(a) Show that the speed of the space-cruiser relative to the Earth is 12c/13.
(b) How far apart does the navigator on the space-cruiser measure the two
events to be? [3 marks]
\Delta t ' = \gamma(\Delta t - v\Delta x/c^2)
The Attempt at a Solution
I know this must be quite simple, i'm just not quite sure where to start. I've fiddled with the Lorentz transformation equations to no avail - i think its the time interval concept which confuses me here (what actually is the interval? 10^-5 s?)
I tried putting in the value of the speed given (*cos45 as it's heading north-east) to try obtain the times given or something related to them but also this was pretty useless.
I imagine part b is quite trivial once part a is complete
Helpfull shoves most appreciated,