Actually, the wording of the problem is ambiguous. The distance traveled depends on the way acceleration changes.
EDIT:
I will give a formula for the distance traveled if the proper-acceleration vs. proper-time is given as a_0(\tau).
<br />
\frac{d v}{d t} = \frac{ \frac{ d v' \, \left( 1+ \frac{v' \, u}{c^2} \right) - (v' + u) \, \frac{u \, d v'}{c^2} }{\left( 1+ \frac{v' \, u}{c^2} \right)^2} }{ \frac{d t' + \frac{u \, d x'}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}} } = \frac{d v'}{d t'} \, \frac{ \left( 1 - \frac{u^2}{c^2} \right)^{3/2} }{ \left( 1 + \frac{u \, v'}{c^2} \right)^3 }<br />
If v' = 0, and u = v (instantaneous proper frame), then d v'/d t' = a_0. Also, since we want to use proper time instead of LAB time, we have:
<br />
\frac{d t}{d \tau} = \left( 1 - \frac{v^2}{c^2} \right)^{-1/2}<br />
Combining these two equations, we have:
<br />
\frac{d v}{d \tau} = \frac{d v}{d t} \, \frac{d t}{d \tau} = a_0 \, \left( 1 - \frac{v^2}{c^2} \right)^{3/2} \, \left( 1 - \frac{v^2}{c^2} \right)^{-1/2} = a_0 \, \left( 1 - \frac{v^2}{c^2} \right)<br />
The variables in this ODE may be separated:
<br />
\frac{d v}{1 - v^2 / c^2} = a_0(\tau) \, d\tau<br />
and then integrated:
<br />
\int_{v_i}^{v}{ \frac{d \tilde{v}}{1 - \tilde{v}^2 / c^2} } = \int_{0}^{\tau}{ a_0(\tilde{\tau}) \, d\tilde{\tau} }<br />
For shorthand, we denote A(\tau) \equiv \int_{0}^{\tau}{ a_0(\tilde{\tau}) \, d\tilde{\tau} }. The integral over velocity is performed by introducing the hyperbolic trigonometric substitution (the parameter \eta is called rapidity):
<br />
\frac{\tilde{v}}{c} = \tanh(\eta) \Rightarrow d\tilde{v} = c \, \mathrm{sech}^2(\eta) \, d\eta, \ \frac{1}{1 - \tilde{v}^2 / c^2} = \cosh^2(\eta)<br />
and we have:
<br />
\eta(\tau) - \eta_i = A(\tau)<br />
Thus, we have the following implicity dependence of velocity on proper time:
<br />
v(\tau) = c \, \tanh^{-1} \left(\eta_i + A(\tau) \right), \eta_i \equiv \tanh \left(\frac{v_i}{c} \right)<br />
Once we know \eta(\tau), we can find the dependence of LAB time t on proper time τ:
<br />
\frac{d t}{d \tau} = \left( 1 - \frac{v^2}{c^2} \right)^{-1/2} = \cosh \left( \eta(\tau) \right)<br />
which may be integrated:
<br />
t(\tau) = \int_{0}^{\tau} {\cosh \left( \eta (\tilde{\tau}) \right) \, d\tilde{\tau} }<br />
Finally, the displacement is:
<br />
dx = v \, dt = c \, \tanh \left( \eta(\tau) \right) \, \cosh \left( \eta(\tau) \right) \, d\tau = c \, \sinh \left( \eta(\tau) \right) \, d\tau<br />
<br />
\Delta x = c \, \int_{0}^{\tau} { \sinh \left( \eta(\tilde{\tau} ) \right) \, d\tilde{\tau} }<br />
