River Crossing: Minimum Time, Shortest Path, and a Dock 600 m Downstream
A boatman works a river 800 m wide in which the current is a steady 3.0 m/s; his boat does 5.0 m/s in still water. His jetty is exactly opposite the starting point, and the ferry there leaves 190 s after he casts off. Last week he steered perpendicular to the bank, reasoning that 'the shortest crossing is the straight one', and came ashore 480 m downstream. Today he must reach the jetty or lose a day. A second passenger, however, does not want the jetty at all: she must be landed 600 m downstream on the far bank, and asks whether a single crossing can put her there and how long it will take. The engine setting is fixed, so 5.0 m/s relative to the water is the only speed available.
1. Analyse the perpendicular crossing: crossing time, drift, and whether it is the fastest possible crossing.
Read the worked answer
Time. Steering perpendicular to the bank, the whole 5.0 m/s is directed across the river, so t = 800/5.0 = 160 s. This is the minimum time: any upstream or downstream tilt reduces the crossing component to 5.0 cos θ and increases the time to 160/cos θ.
Drift. The current is unaffected by his heading, so the boat is carried 3.0 × 160 = 480 m downstream - exactly the observed landing point. The path over the ground is a straight line at tan⁻¹(3/5) = 31° to the bank, of length √(800² + 480²) = 933 m.
Judgement. The boatman's reasoning confuses two different optimisations. Steering perpendicular minimises time; the shortest path (a straight line to the jetty) needs a heading partly upstream and takes longer. Neither is 'the right way' in general - the choice depends on which quantity the trip is constrained by, time or landing point.
2. Find the heading required to land at the jetty and decide whether he can catch the 190 s ferry. If not, what minimum still-water speed would be needed?
Read the worked answer
Heading for the jetty. The downstream component of the boat's velocity must exactly cancel the current: 5.0 sin θ = 3.0 → sin θ = 0.6, θ = 37° upstream of the perpendicular.
The crossing component is then 5.0 cos 37° = 4.0 m/s (equivalently √(5² − 3²) = 4.0 m/s), so t = 800/4.0 = 200 s.
Verdict on the ferry. 200 s > 190 s: he misses it by about 10 s.
Could he make it another way? Row perpendicular (160 s), land 480 m downstream and walk to the jetty. A brisk 1.5 m/s walk over 480 m takes 320 s, giving 480 s in all - far worse. Landing part-way and walking at 16 m/s (58 km/h) would be needed to save 30 s, which is not humanly possible.
Required speed for a 190 s straight crossing. The across-river component must be 800/190 = 4.21 m/s while still cancelling 3.0 m/s of current: v = √(4.21² + 3.0²) = √(17.7 + 9.0) ≈ 5.2 m/s. His 5.0 m/s falls short, so the practical answer is to radio the ferry for a 10 s delay.
3. The second passenger wants to be landed 600 m downstream. Find the heading and the crossing time, and explain why the answer is not simply 'aim perpendicular'.
Read the worked answer
Let β be the heading measured from the perpendicular towards downstream. Across-river component = 5.0 cos β, so t = 800/(5.0 cos β) = 160/cos β. Downstream velocity = 3.0 + 5.0 sin β.
Drift: (3.0 + 5.0 sin β)(160/cos β) = 600 → 480 + 800 sin β = 600 cos β → 4 sin β − 3 cos β = −2.4
Write the left side as 5 sin(β − 36.87°): 5 sin(β − 36.87°) = −2.4 → sin(β − 36.87°) = −0.48 → β − 36.87° = −28.7° → β = 8.2°.
So he must aim 8.2° downstream of the perpendicular, and t = 160/cos 8.2° = 161.7 s ≈ 162 s.
Check: downstream velocity = 3.0 + 5.0(0.1426) = 3.71 m/s; drift = 3.71 × 161.7 = 600 m ✓.
The heading looks wrong because 600 m is more drift than the 480 m a perpendicular crossing already produces; the current alone is not enough, so he must add a small downstream component of his own. Tilting downstream costs almost nothing in time (160 → 162 s) but gains 120 m of drift. Had she asked for less than 480 m, the heading would have been upstream.