= Solution
The along-strike rupture time is $L/v_r$, where $v_r$ is <rupture velocity>. Taking a representative crustal $v_r\simeq3\,\mathrm{km\,s^{-1}}$ gives \b[about $2$ s] for a $6$ km unilateral <finite-source rupture>. A different assumed <rupture velocity> changes this estimate. The local slip rise time at one patch is not specified and need not equal the whole-fault rupture time.
For comparable distance and ground conditions, the eastern observer lies in the forward <rupture directivity> direction. Radiation from successive patches arrives closer together, producing a shorter, more concentrated and generally larger-amplitude ground-velocity pulse. The western observer receives the contributions spread over a longer interval, generally with less concentrated shaking. This concerns pulse duration and <amplitude>; it does not imply an equally large permanent displacement at the eastern site.<What is Directivity by United States Geological Survey>[Explanation of constructive interference and the relation of rupture speed to S-wave speed]{ref}
For distant observers approximately along strike, an elementary <seismic travel time> construction gives
$$
\Delta t_E\simeq L\left(\frac1{v_r}-\frac1\beta\right),\qquad
\Delta t_W\simeq L\left(\frac1{v_r}+\frac1\beta\right),
$$
where $\beta$ is the <shear-wave velocity>. With $v_r=3$ and $\beta=3.5\,\mathrm{km\,s^{-1}}$, these propagation contributions are about $0.29$ and $3.71$ s, respectively; finite local rise time broadens them. <Double-couple radiation pattern>, receiver geometry and local site response can change the observed <amplitudes>, so location east or west alone is not an unconditional intensity prediction.
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