• \(u\) = Velocity of river flow (Drift velocity).
• \(v\) = Velocity of wave relative to water (Expansion velocity).
• \(b\) = Width of river.

The wave spreads out as a circle of radius \(R\) growing at speed \(v\). So \(R = vt\).
Simultaneously, the entire circle is carried downstream by the river at speed \(u\). The center drifts by distance \(x = ut\).

For the wave to reach point B (directly opposite A), the circle must be large enough to intersect B. We form a right-angled triangle with:

• Vertical side: River width \(b\).
• Horizontal side: Drift distance \(ut\).
• Hypotenuse: Wave radius \(vt\).

Applying Pythagoras theorem: \[ (vt)^2 = b^2 + (ut)^2 \]

\[ v^2 t^2 – u^2 t^2 = b^2 \] \[ t^2 (v^2 – u^2) = b^2 \] \[ t = \frac{b}{\sqrt{v^2 – u^2}} \]

This solution is valid only if \(v > u\).

• If \(v \le u\), the wave drift is too fast, and the ripple is swept downstream before it can ever cross the width \(b\). It never reaches B.

Answer: \[ t = \begin{cases} \frac{b}{\sqrt{v^2 – u^2}} & \text{if } v > u \\ \infty & \text{if } v \le u \end{cases} \]