Yes. I follow Thorne, Misler, and Wheeler's derivation of the formula for a uniformly accelerating object, and then my own quick work for the formula for the light catching up. Work is in natural units.
Consider a 1+1 Minkowski space. The four velocity has constant magnitude $-1$, so the 4-acceleration is orthogonal to the velocity vector: $$u^\mu a_\mu=\frac{1}{2}\frac{d}{d\tau}(u^\mu u_\mu)=0$$ Therefore, when the object is at rest initially, the time component of the 4-acceleration is $0$, so $a^\mu=(0,a)=\left(0,\frac{d^2x}{dt^2}\right)$ in the rest frame. Therefore, in all frames, $$a^\mu a_\mu=\left(\frac{d^2x}{dt^2}\right)^2$$ Now, consider a particle undergoing uniform acceleration $g$ in the $x$ direction. We immediately have $$\frac{dt}{d\tau}=u_0,\;\;\frac{dx}{d\tau}=u_1, \;\; \frac{du_0}{d\tau}=a_0,\;\;\frac{du_1}{d\tau}=a_1$$ We also have three algebraic relations: $$ \begin{align} u^\mu u_\mu &= -1 \\ u^\mu a_\mu &= -u^0 a^0 +u^1a^1=0 \\ a^\mu a_\mu &= g^2 \\ \end{align} $$ One can solve these for the acceleration to find: $$a_0=\frac{du_0}{d\tau}=gu^1, \;\; a_1=\frac{du_1}{d\tau}=gu^0$$ This are fairly trivial to solve, and combined with a suitable choice of origin, one finds $$gt=\sinh g\tau, \;\; gx = \cosh g\tau$$ from which we find $$x^2-t^2=\frac{1}{g^2}$$
Now, to solve for the light catching up. We will assume both $x>0$ and $t>0$. One sees that the asymptote is $t= x$. By setting $t=0$, one immediately sees that the x-intercept is $x=\frac{1}{g}$. Since light travels on lines of $t = \pm x - b$, the sign on the $x$ must be positive (or it goes the opposite way), and $0<b\leq \frac{1}{g}$. Solving for $t$, we get $$\bbox[5px,border:2px solid black]{t=\frac{1}{2bg^2}-\frac{b}{2},\;\;0<b\leq\frac{1}{g}}$$
Converting out of natural units, we have $$t=\frac{c^3}{2bg^2}-\frac{b}{2c},\;\;0<b\leq\frac{c^2}{g}$$
Note that the answer is for the rest frame, and not for the accelerating object's frame.