 Solution

The proper time satisfies $dt/d\tau =\gamma (u)$. The collinear transformation from the Earth frame into the instantaneous rest frame gives proper acceleration $f={\gamma }^{3}du/dt$, soStarting from rest, integration gives $\mathrm{artanh}(u/c)=f\tau /c$. HenceSince $dx/d\tau =u\gamma =c\sinh (f\tau /c)$, a further integration givesUsing $\tau =10$ Julian years and $f=1\mathrm{m}{\mathrm{s}}^{-2}$ gives $f\tau /c\simeq 1.053$, so the star is about $5.77$ light years away in the Earth frame. The Earth-frame travel time is about $11.95$ years and the final speed is about $0.783c$. The model accelerates toward the star throughout; adding a braking leg would be a different itinerary.