A boatman at a ghat in Varanasi wants to land at the ghat exactly opposite. He does not point his boat at that ghat. He points the bow upstream, at a slant, and rows steadily. The boat moves straight across the Ganga, sideways relative to where it points. A child in a small boat who points straight at the far ghat lands well downstream of it.
Three velocities on one river
Three velocities appear in every river problem. The boat moves through the water with velocity \(\vec v_{br}\), its velocity relative to the river. The water moves relative to the ground with velocity \(\vec v_r\), the river velocity. The boat moves over the ground with velocity \(\vec v_b\). The rule of the last lesson, written for the boat and the river, gives the connection.
\[ \vec v_{br} = \vec v_b - \vec v_r \quad\Longrightarrow\quad \vec v_b = \vec v_{br} + \vec v_r \]
The direction in which the bow points is the heading of the boat, the direction of \(\vec v_{br}\). The line the boat actually follows over the ground is its track, the direction of \(\vec v_b\). In a flowing river the two are different. The distance by which the landing point lies downstream of the point exactly opposite is the drift.
Splitting the motion across and along the river
Take the \(x\) axis along the bank, downstream, and the \(y\) axis straight across. Let the river have width \(d\). Let the boat point at an angle \(\alpha\) upstream of the straight-across line. Then the components of the ground velocity are these.
\[ v_y = v_{br}\cos\alpha, \qquad v_x = v_r - v_{br}\sin\alpha \]
By the independence of motions from the last chapter, the two components act separately. Only \(v_y\) carries the boat across, so the crossing time is \(t = d / (v_{br}\cos\alpha)\). During that time \(v_x\) carries the boat along the bank, so the drift is \(x = v_x t\).
The shortest time: point straight across
The crossing time \(d / (v_{br}\cos\alpha)\) is least when \(\cos\alpha\) is largest. The largest value of \(\cos\alpha\) is 1, at \(\alpha = 0\). So the quickest crossing points the bow straight across. The river still carries the boat downstream for the whole crossing, so the boat lands downstream.
The shortest path: point upstream
The shortest possible path is the straight line across, of length \(d\). The boat follows it only when its drift is zero, that is, when \(v_x = 0\). So the upstream part of the boat's own velocity must cancel the river exactly.
\[ v_{br}\sin\alpha = v_r \quad\Longrightarrow\quad \sin\alpha = \dfrac{v_r}{v_{br}} \]
The boat then moves straight across at the speed that is left, \(v_{br}\cos\alpha = \sqrt{v_{br}^2 - v_r^2}\). This is less than \(v_{br}\), so the shortest path always takes longer than the shortest time.
A river \(120\,\text{m}\) wide flows at \(3\,\text{m/s}\). A boat moves at \(5\,\text{m/s}\) relative to the water. Find (a) the least crossing time and the drift for it, (b) the heading and the time for landing exactly opposite.
\(d = 120\,\text{m}\), \(v_r = 3\,\text{m/s}\), \(v_{br} = 5\,\text{m/s}\). Across is \(y\), downstream is \(x\).
(a) \(t = d / v_{br}\), \(x = v_r t\). (b) \(\sin\alpha = v_r / v_{br}\), \(t = d / (v_{br}\cos\alpha)\).
(a) \(t = 120 / 5 = 24\,\text{s}\), \(x = 3 \times 24 = 72\,\text{m}\). (b) \(\sin\alpha = 0.6\), \(\alpha = 37°\), \(v_{br}\cos\alpha = 5 \times 0.8 = 4\,\text{m/s}\), \(t = 120 / 4 = 30\,\text{s}\).
The quick crossing takes \(24\,\text{s}\) and lands \(72\,\text{m}\) downstream. The straight crossing takes \(30\,\text{s}\), pointing \(37°\) upstream. Six extra seconds buy a landing at the right ghat.
A boat must cross a river in the least possible time. In which direction should it point?
A river \(100\,\text{m}\) wide flows at \(4\,\text{m/s}\). A boat that moves at \(5\,\text{m/s}\) relative to the water lands exactly opposite its starting point. How long does the crossing take, in seconds?
When the river is faster than the boat
Zero drift needs \(\sin\alpha = v_r / v_{br}\). If \(v_r > v_{br}\), this sine would be larger than 1, which no angle has. So a boat slower than the river can never land exactly opposite. It can still choose the heading that gives the least drift, and a drawing finds it without any calculus.
Draw \(\vec v_r\) from a point O. Then draw \(\vec v_{br}\) from the tip of \(\vec v_r\). The boat can point in any direction, so the tip of \(\vec v_{br}\) can lie anywhere on a circle of radius \(v_{br}\). The ground velocity \(\vec v_b\) runs from O to that tip. The track is steepest, and the drift least, when the line from O just touches the circle.
At a tangent, the radius meets the tangent line at a right angle. So at the best heading \(\vec v_{br}\) is perpendicular to \(\vec v_b\). The right-angled triangle has hypotenuse \(v_r\) and one side \(v_{br}\). The angle \(\theta\) of the track with the bank then has \(\sin\theta = v_{br} / v_r\), and the boat points upstream of straight across by the same angle, \(\sin\alpha = v_{br} / v_r\).
A river \(120\,\text{m}\) wide flows at \(5\,\text{m/s}\). A boat moves at \(3\,\text{m/s}\) through the water. Find the heading for the least drift, the least drift, and the time taken.
Heading: \(\sin\alpha = v_{br} / v_r = 3/5\), so \(\alpha = 37°\) upstream of straight across.
Least drift: \(x = d\sqrt{v_r^2 - v_{br}^2} / v_{br} = 120 \times 4 / 3 = 160\,\text{m}\) downstream.
Time: the across component is \(v_{br}\cos\alpha = 3 \times 0.8 = 2.4\,\text{m/s}\), so \(t = 120 / 2.4 = 50\,\text{s}\).
Check with the path: the track is \(\sqrt{120^2 + 160^2} = 200\,\text{m}\) long, and \(d\,v_r / v_{br} = 120 \times 5 / 3 = 200\,\text{m}\). The ground speed is \(200 / 50 = 4\,\text{m/s}\), the third side of the 3, 4, 5 triangle.
The same answer comes from calculus. The drift is \(x(\alpha) = d\,(v_r - v_{br}\sin\alpha) / (v_{br}\cos\alpha)\). Setting \(dx/d\alpha = 0\) gives \(v_r\sin\alpha = v_{br}\), the tangent condition of the drawing. The drawing shows why: the circle of possible velocities has one steepest line from O, and it touches the circle.
A boat crosses a river. Which statements are correct?
A ferry crosses the Hooghly at Kolkata, about \(500\,\text{m}\) wide, where the current is about \(1.5\,\text{m/s}\). The ferry moves at about \(4\,\text{m/s}\) through the water and lands at the jetty exactly opposite. Roughly how many seconds does the crossing take?
A ferry pilot on a tidal river points the bow upstream, so that the ferry slides straight across to the jetty. The angle changes with the tide. A swimmer caught in a river current is told to swim straight across, not against the current. Straight across is the quickest way out of the water, and the drift downstream does not matter to a swimmer who only needs the bank.
- Using \(v_r\) or \(v_b\) in the crossing time.Only the across component of the boat's own velocity carries it across: \(t = d / (v_{br}\cos\alpha)\). The river adds nothing across.
- Writing \(\sin\alpha = v_{br} / v_r\) for the shortest path.Zero drift needs the upstream part to cancel the river, \(v_{br}\sin\alpha = v_r\). The ratio \(v_{br}/v_r\) belongs to the least-drift case of a slow boat.
- Measuring \(\alpha\) from the bank in one formula and from the straight-across line in another.Draw the angle on your triangle before you write a sine or a cosine, and keep one reference line for the whole problem.