= Solution
Write $\Delta x=h(t_B)-f(t_A)$. Substituting $\Delta t=v\Delta x/c^2$ into <4/b/ii> gives $L^2=\Delta x^2(1-v^2/c^2)$. The positive-separation branch therefore has
$$
\boxed{\Delta x=\gamma L,\qquad\Delta t=\frac{\gamma Lv}{c^2}.}
$$
The <Born-rigid front trajectory from a rear worldline> follows by setting $t_A=t$:
$$
\boxed{h\!\left(t+\frac{\gamma(t)Lv(t)}{c^2}\right)=f(t)+\gamma(t)L,
\quad v(t)=f'(t),\quad\gamma(t)=[1-v(t)^2/c^2]^{-1/2}.}
$$
Every $v$ and $\gamma$ in this formula is evaluated at the rear event with laboratory time $t$, not at the shifted argument of $h$.
For completeness, a smooth future-directed front trajectory requires this time mapping to be locally invertible. If the signed laboratory acceleration is $a_S=dv/dt$ and the rear <proper acceleration> is $\alpha_F=\gamma^3a_S$, then
$$
\frac{dt_B}{dt}=1+\frac{L\alpha_F}{c^2},\qquad
\frac{dx_B}{dt}=v\left(1+\frac{L\alpha_F}{c^2}\right).
$$
When $1+L\alpha_F/c^2>0$, their ratio gives $dh/dt_B=v$, so the two simultaneous points indeed share their instantaneous velocity. If that factor reaches zero, an acceleration-horizon obstruction appears; an arbitrary rear trajectory cannot be extended to an arbitrary rigid length while retaining both rockets future-directed. The positive-acceleration example below satisfies the condition.
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