= Solution
Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2)^{1/2}$ and $\vartheta=\arg(x+iy)$, <superposition> of uniform <potential flow> and a two-dimensional <point source> gives
$$
\boxed{\Phi=Ux+\frac{Q}{2\pi}\log r,\qquad
\psi=Uy+\frac{Q}{2\pi}\vartheta,}
$$
where $(u,v)=\nabla\Phi=(\psi_y,-\psi_x)$. The angular branch cut records the <volume flux> from the source; the logarithm's arbitrary reference length only changes an additive constant in the <velocity potential>.
The <stagnation point> lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid is
$$
\boxed{x_s=-\frac{Q}{2\pi U}=-\frac b\pi,\qquad b=\frac{Q}{2U}.}
$$
The stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing <streamlines> through it have $\psi=\pm Q/2$ when the angle is taken in $(-\pi,\pi)$. Far downstream $\vartheta\to0$, so these <streamlines> approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the <Rankine half-body> between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta/\pi)$, $x=y\cot\vartheta$, $0<\vartheta<\pi$.
For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source is
$$
T_s(x)=\int_0^x\frac{dX}{U+Q/(2\pi X)}
=\frac xU-\frac{Q}{2\pi U^2}\log\left(1+\frac{2\pi Ux}{Q}\right).
$$
The first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi(x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is therefore
$$
\boxed{\tau=\frac xU-T_s(x)=\frac b{\pi U}\log\left(1+\frac{\pi x}{b}\right).}
$$
This is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau-t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote <Darcy velocity> and bulk-area source flux in a medium of uniform <porosity> $\phi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\phi$. No <porosity> is specified here, so the displayed time formula fixes the kinematic convention.
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