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I'm trying to solve a particular problem given in Introduction to Special Relativity by R. Resnick, which goes like this:

Two observers in the S frame, $A$ and $B$, are separated by a distance of 60 m. Let $S'$ move at a speed $\dfrac{3}{5}c$ relative to $S$, the origins of the two systems, $O$ and $O'$, being coincident at $t = t'= 3 *10^{-7}$ sec ($90/c$). The S frame has two observers, one at $A'$ and the other at B' such that, according to clocks in the $S$ frame, $A'$ is opposite $A$ at the same time that $B'$ is opposite B. $O$ and $O'$ lie at the centre of $A$ and $B$ and $A'$ and $B'$ respectively. a) What is the reading on the clock of $B'$ when $B'$ is opposite $B$?

My question: I do not really get how is it possible for an observer in the S frame to observe A' and A being coincident at the same time as B and B' are. Isn't the length, and hence the distance between A' and B' contracting in S' w.r.t S?

Before this, I tried the following:

At $t=t'=\dfrac{90}{c}$, $x=x'=0$ where $x$ and $x'$ are the locations of the origins O and O' respectively. We want to find $t'$ at which $b=b'$. Using Lorentz transformations, we know that $t' = \dfrac{t-\dfrac{xv}{c^2}}{\sqrt{1-\left (\dfrac{v}{c} \right )^2}}$. Put $t=90/c, v=\dfrac{3}{5}c$ and $x = +30$ (since that is the location of B' in S frame at the given $t$), and we obtain, $t' = \dfrac{90}{c}$, which does not surprise me, because conditions, in away, imply that A & A', B and B', and O & O' are all coincident at the same instant of time (in S, I think, but then what is this t'?). The answer, however, is $\dfrac{45}{c}$, which is curiously what comes out if we put $x=+60$ instead of $30$. I do not understand where is it that I'm going wrong. I believe that there is a large conceptual hole in my understanding of time dilation, length contraction, Lorentz transforms, and relativity of simultaneity.

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