= Solution
\b[Boost and <velocity> transformation.] For the moving frame,
$$
t'=\gamma_v\left(t-\frac{vx}{c^2}\right),\quad
x'=\gamma_v(x-vt),\quad y'=y,\quad z'=z.
$$
The inverse has $v\mapsto-v$. Taking differentials of that inverse and dividing by $dt$ gives
$$
\boxed{u_x=\frac{u_x'+v}{1+vu_x'/c^2},\quad
u_y=\frac{u_y'}{\gamma_v(1+vu_x'/c^2)},\quad
u_z=\frac{u_z'}{\gamma_v(1+vu_x'/c^2)}}.
$$
\b[Constant <proper acceleration>.] For collinear motion, the instantaneous rest-frame acceleration is $a=\gamma^3dv/dt$. Since $dt/d\tau=\gamma$,
$$
\boxed{\frac{dv}{d\tau}=a\gamma^{-2}
=a\left(1-\frac{v^2}{c^2}\right)}.
$$
Write $v=c\tanh\eta$; then $d\eta/d\tau=a/c$. Starting from rest gives $\eta=a\tau/c$, $dt/d\tau=\cosh\eta$ and $dx/d\tau=c\sinh\eta$. <integration> with the stated initial origin yields
$$
\boxed{v=c\tanh(a\tau/c),\quad
t=\frac ca\sinh(a\tau/c),\quad
x=\frac{c^2}a[\cosh(a\tau/c)-1]}.
$$
\b[<photon rocket> mass ratio.] In the instantaneous <rest frame>, ejecting backward <photon> <energy> $dE_\gamma$ gives, to first order, $c^2dM=-dE_\gamma$ and $Mc\,d\eta=dE_\gamma/c$. Hence $dM/M=-d\eta$, and
$$
\boxed{\frac M{M_i}=e^{-\eta}
=\sqrt{\frac{1-\beta}{1+\beta}}}.
$$
Here $M$ includes all stored fuel and <energy>. This assumes complete conversion into a perfectly collimated <photon> exhaust without other losses.
\b[<photon-rocket fuel for acceleration and braking>.] With effectively instantaneous changes of speed, the cruise <proper time> is $D/(\gamma v)=10\ \mathrm{yr}$, giving
$$
\gamma\beta=10,\qquad
\gamma=\sqrt{101},\quad\beta=\frac{10}{\sqrt{101}},\quad
\eta=\operatorname{arsinh}10=2.99822.
$$
Braking to rest costs another factor $e^{-\eta}$, so $M_f/M_i=e^{-2\eta}=(\sqrt{101}-10)^2=0.00248758$. If equal rest <masses> of matter and <antimatter> annihilate, the consumed <antimatter> is half the total consumed fuel:
$$
\boxed{\frac{M_{\bar m}}{M_i}=\frac12(1-e^{-2\eta})=0.498756
\quad\text{(about $49.88\%$)}}.
$$
This is the fraction of initial total mass, not the <antimatter> fraction of the consumed matter–<antimatter> fuel, which is one half.
\b[<finite-acceleration correction to a relativistic cruise>.] Each constant-proper-acceleration leg has proper duration $\tau_a=c\eta/a$ and Earth-frame length $x_a=(c^2/a)(\gamma-1)$. Thus
$$
\tau_{\rm new}=2\frac{c\eta}{a}+
\frac{D-2x_a}{\gamma v}.
$$
Subtracting the original $D/(\gamma v)$ gives
$$
\boxed{\Delta\tau=\frac{2c}{a}
\left[\eta-\frac{\gamma-1}{\gamma\beta}\right]
\simeq3.98\ \mathrm{yr}}.
$$
For $a=10\ \mathrm{m\,s^{-2}}$, each accelerating leg takes about $2.85$ proper years and covers $8.60$ light years in the Earth frame, leaving positive cruise distance. The new total is about \b[13.98 proper years]. Adding $2\tau_a$ to ten years would overcount the cruise segment replaced by acceleration and braking.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2014/paper-1-photon-journey.png]
{title=Original photon-drive journey with instantaneous and finite acceleration, separating Earth time from traveller time}
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