= 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 $f=\gamma^3du/dt$, so
$$
\frac{du}{d\tau}=\frac{f}{\gamma^2}=f\left(1-\frac{u^2}{c^2}\right).
$$
Starting from rest, integration gives $\operatorname{artanh}(u/c)=f\tau/c$. Hence
$$
\boxed{u=c\tanh(f\tau/c),\qquad t=\frac cf\sinh(f\tau/c).}
$$
Since $dx/d\tau=u\gamma=c\sinh(f\tau/c)$, a further integration gives
$$
\boxed{x(\tau)=\frac{c^2}{f}\left[\cosh(f\tau/c)-1\right].}
$$
Using $\tau=10$ Julian years and $f=1\,\mathrm{m\,s^{-2}}$ gives $f\tau/c\simeq1.053$, so the star is \b[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.
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