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A river 120 m wide has straight, parallel banks…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A river $120\ \text{m}$ wide has straight, parallel banks. The water flows fastest in the middle. At a distance $y$ from the near bank its speed is

\displaystyle u(y) = u_0\,\frac{4y(d - y)}{d^2}

where $d$ is the width of the river and $u_0 = 3.0\ \text{m/s}$ . A motorboat moves at $4.0\ \text{m/s}$ relative to the water. It starts from a point A on the near bank, and B is the point on the far bank directly opposite A, as shown in the figure.

Top view of a river 120 m wide between two straight banks, flowing to the right. Arrows across the river show the water's speed: zero at the banks and largest, u0 = 3.0 m/s, in the middle. Point A is on the near (lower) bank and point B on the far bank directly opposite A.

The boat always points straight across the river, at right angles to the banks. How long does the crossing take, and how far downstream of B does the boat land?
On this crossing, what is the boat's speed over the ground in the middle of the river, and what angle does its path make with the bank there?
Instead, the boat always points at a fixed angle $\alpha$ upstream of the straight-across direction. For what $\alpha$ does it land exactly at B? How long does that crossing take?
On the crossing of (c), the boat does not stay on the line AB. How far upstream of the line AB does it get, at most?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The boat's velocity over the ground is its velocity relative to the water plus the water's velocity $u(y)$ . Its part across the river does not depend on $u$ , so it fixes the time. The downstream drift is the sum of $u(y)$ over the crossing.

Step 1. Crossing straight (a). The boat crosses at $4.0\ \text{m/s}$ , so the time is $\dfrac{120\ \text{m}}{4.0\ \text{m/s}} = 30\ \text{s}$ . With $dy = v\,dt$ , the drift is

\displaystyle x = \int_0^T u\,dt = \frac{1}{v}\int_0^d u_0\,\frac{4y(d - y)}{d^2}\,dy = \frac{u_0}{v} \cdot \frac{4}{d^2}\left(\frac{d^3}{2} - \frac{d^3}{3}\right) = \frac{2u_0 d}{3v}

\displaystyle x = \frac{2 \times 360\ \text{m}^2/\text{s}}{3 \times 4.0\ \text{m/s}} = 60\ \text{m}

The water's average speed over the width is $\tfrac23 u_0 = 2.0\ \text{m/s}$ .

Step 2. In the middle (b). There $u = u_0 = 3.0\ \text{m/s}$ , at right angles to the boat's own $4.0\ \text{m/s}$ . Its speed over the ground is $\sqrt{4.0^2 + 3.0^2}\ \text{m/s} = 5.0\ \text{m/s}$ , and its path makes $\tan^{-1}\dfrac{4.0}{3.0} = 53.1^\circ$ with the bank. Near the banks the water is still, and the path is at right angles to them: the path is a curve.

Step 3. Landing at B (c). Pointing at $\alpha$ upstream, the boat crosses at $v\cos\alpha$ and moves upstream at $v\sin\alpha$ relative to the water. Repeating Step 1, its net drift is

\displaystyle x = \frac{1}{v\cos\alpha}\int_0^d \left[u(y) - v\sin\alpha\right] dy = \frac{d}{v\cos\alpha}\left(\tfrac23 u_0 - v\sin\alpha\right)

This is zero when $\sin\alpha = \dfrac{2u_0}{3v} = \dfrac{2 \times 3.0}{3 \times 4.0} = 0.5$ , so $\alpha = 30^\circ$ . The speed across is $4.0 \cos 30^\circ = 3.464\ \text{m/s}$ , and the time is $\dfrac{120\ \text{m}}{3.464\ \text{m/s}} = 34.6\ \text{s}$ .

Step 4. Off the line (d). The boat moves upstream while $u(y) < v\sin\alpha = 2.0\ \text{m/s}$ , that is near the near bank, until $4\eta(1 - \eta) = \dfrac{2.0}{3.0}$ with $\eta = y/d$ . This gives $\eta = \tfrac12\left(1 - \tfrac{1}{\sqrt3}\right) = 0.2113$ , or $y = 25.4\ \text{m}$ . There it is furthest upstream:

\displaystyle x_{\min} = \frac{d}{v\cos\alpha}\int_0^{\eta}\left[u_0\,4\eta'(1 - \eta') - \tfrac23 u_0\right] d\eta' = \frac{u_0 d}{3v\cos\alpha}\left[6\eta^2 - 4\eta^3 - 2\eta\right] = -\frac{u_0 d}{9\sqrt3\,v\cos\alpha}

using $\eta^2 = \eta - \tfrac16$ , which makes the bracket $\tfrac13(2\eta - 1) = -\tfrac{1}{3\sqrt3}$ . Numerically, $\dfrac{360\ \text{m}^2/\text{s}}{54.0\ \text{m/s}} = 6.67\ \text{m}$ upstream of AB. In the middle it drifts back, past the line, and it is $6.67\ \text{m}$ downstream at $y = d - 25.4\ \text{m}$ , before it comes back to B.

Answer: (a) $30\ \text{s}$ ; $60\ \text{m}$ downstream of B. (b) $5.0\ \text{m/s}$ , at $53.1^\circ$ to the bank. (c) $\alpha = 30^\circ$ ; $34.6\ \text{s}$ . (d) $6.67\ \text{m}$ .

Check. If the river flowed everywhere at its average speed $2.0\ \text{m/s}$ , the drift in (a) would be $2.0\ \text{m/s} \times 30\ \text{s} = 60\ \text{m}$ , the same. The total drift only needs the average speed over the width, because the boat spends equal times in equal strips of the river.

Watch out. You cannot use the speed in the middle, $3.0\ \text{m/s}$ , for the whole crossing. That would give $3.0\ \text{m/s} \times 30\ \text{s}$ , too much.

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