Two observers viewed from different sources

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HakemHa
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Homework Statement
Here is a question from "Introduction to Special Relativity Ch2, Robert Resnick" that got me confused:
 "Two observers in the S frame, A and B are separated by a distance of 60m. Let S' move at a speed of 3/5c, relative to S, the origins of the two systems O' and O, being coincident at t′=t=90/c.
 The S' frame has two observers, one at A' and one at a point B' such that, according to clocks in the S frame, A' is opposite A at the same time that B' is opposite B:

a) What is the reading on the clock of B' when B' is opposite B?

  b)The system S' continues moving until A' is opposite B. What is the reading on the clock of B   when he is opposite A'?

  c)What is the reading on the clock of A' when he is opposite B'?
Relevant Equations
Lorentz transform
I builded the translated lorentz transform, at t=0 t'=-22.5 and x'(x=0)=67.5 after that I just didn't the question
 
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PeroK said:
:welcome:

I think you need a diagram for the S frame. Does it say where A and B are relative to the common origin?

Also, are you sure that ##t' = t = 90/c## is corrrect?
A is in the origin I think
 
PeroK said:
Okay, do you have a diagram? I'm still not sure what ##t' = t = 90/c## means. What's your plan for dealing with that? Normally we have ##t = t' = 0## at the common origin.
I mean I just wrote the regular lorentz transform ## \vec X' = \Lambda \vec X + (67.5, -22.5) for \vec X = (x, t)## such that the origins intersect at t=t'=90/c
 
HakemHa said:
I mean I just wrote the regular lorentz transform ## \vec X' = \Lambda \vec X + (67.5, -22.5) for \vec X = (x, t)## such that the origins intersect at t=t'=90/c
The Lorentz transformation demands that ##t= t' = 0## at the origin.

I'm offline for a bit. I suggest you post your answers.