 Solution

The along-strike rupture time is $L/v_r$, where $v_r$ is rupture velocity. Taking a representative crustal $v_r\simeq 3\mathrm{km}{\mathrm{s}}^{-1}$ gives 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.^[ref]

For distant observers approximately along strike, an elementary seismic travel time construction giveswhere $\beta$ is the shear-wave velocity. With $v_r=3$ and $\beta =3.5\mathrm{km}{\mathrm{s}}^{-1}$, 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.