 Solution

One image takes $512\mathrm{s}$ if $1\mathrm{Hz}$ means one completed scan line per second. The slow-axis speed is $10/512=0.01953\mu \mathrm{m}{\mathrm{s}}^{-1}$. Recording a wire spanning $1\mu \mathrm{m}$ along that axis therefore takes $51.2\mathrm{s}$. Meanwhile the fast-axis drift is $20/3600=0.005556\mu \mathrm{m}{\mathrm{s}}^{-1}$, giving a transverse displacement of $0.2844\mu \mathrm{m}$. The apparent straight-line length isIts slow-axis projection remains $1\mu \mathrm{m}$; the overestimate refers to length along the slanted trace.

With the wire along the fast axis and drift along the slow axis, its length is acquired within one line, so $L_{\mathrm{app}}\simeq 1.000\mu \mathrm{m}$. Taking a forward sweep of $1\mathrm{s}$ across $10\mu \mathrm{m}$, the wire is crossed in $0.1\mathrm{s}$ and the perpendicular displacement is only $0.556\mathrm{nm}$. The corresponding length change is about $1.5\times 10^{-7}\mu \mathrm{m}$, far below the $19.5\mathrm{nm}$ pixel pitch. For the convention in which one hertz includes a forward and reverse sweep, that already negligible displacement halves. The slow-axis calculation still uses one completed line per second.