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Irodov 1.8Trickykinematicsrelative-motionriver-crossing

River Boat Time Ratio

Problem 1.8
Two boats, AA and BB, move away from a buoy anchored at a point in the middle of a river. Boat AA moves along the river and boat BB across the river. Having moved off an equal distance from the buoy, the boats returned. Find the ratio of times of motion of boats τA/τB\tau_A / \tau_B if the velocity of each boat with respect to water is η=1.2\eta = 1.2 times greater than the stream velocity.
Answer: τA/τB=1/1−1/η2=1.8\tau_A/\tau_B = 1/\sqrt{1 - 1/\eta^2} = 1.8
Solution Path
Boat A (parallel: downstream then upstream) takes time 2dv/(v^2-u^2). Boat B (perpendicular: across then back) takes time 2d/sqrt(v^2-u^2). Their ratio simplifies to 1/sqrt(1-1/eta^2). With eta=1.2, the ratio is 1.8. Parallel travel is always slower because the upstream penalty exceeds the downstream gain.
01Problem Restatement
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Two boats start from a buoy anchored in the middle of a river. Each boat covers a distance dd away from the buoy and then returns.Boat A travels parallel to the river: it goes downstream a distance dd, then turns around and comes back upstream a distance dd.Boat B travels perpendicular to the river: it crosses a distance dd to one bank, then crosses back a distance dd to the buoy.Given:
- vv = velocity of each boat relative to water
- uu = stream velocity
- η=v/u=1.2\eta = v/u = 1.2 (boat speed is 1.2 times the stream speed)We need to find: the ratio τA/τB\tau_A / \tau_B of the total times taken by the two boats.
Find τA/τB\tau_A / \tau_B
02Physical Picture and Strategy
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Picture a river flowing to the right at speed uu. A buoy sits in the middle, anchored to the riverbed. Both boats have the same speed vv relative to water.Boat A (parallel path): Going downstream, the current helps: effective speed is v+uv + u. Coming back upstream, the current opposes: effective speed is v−uv - u. The two legs have very different speeds.Boat B (perpendicular path): The boat must aim upstream at an angle so that its net motion is straight across the river. By the Pythagorean theorem, the effective crossing speed is v2−u2\sqrt{v^2 - u^2}, and this is the same for both the outward and return trips.Strategy: We compute τA\tau_A and τB\tau_B separately using distance/speed, then take their ratio. The distance dd will cancel. Finally, we express the result in terms of η=v/u\eta = v/u and substitute η=1.2\eta = 1.2.
Parallel boat has asymmetric speeds (fast + slow); perpendicular boat has constant reduced speed
03Step 1: Boat A (Downstream + Upstream)
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Boat A travels along the river. The stream either helps or hinders depending on direction.Using distance = speed ×\times time for each leg:Downstream (current assists): effective speed =v+u= v + u, so time =d/(v+u)= d/(v+u)Upstream (current opposes): effective speed =v−u= v - u, so time =d/(v−u)= d/(v-u)Total time for Boat A:τA=dv+u+dv−u\tau_A = \frac{d}{v+u} + \frac{d}{v-u}Combining the fractions over a common denominator (v+u)(v−u)=v2−u2(v+u)(v-u) = v^2 - u^2:τA=d(v−u)+d(v+u)v2−u2=2dvv2−u2\tau_A = \frac{d(v-u) + d(v+u)}{v^2 - u^2} = \frac{2dv}{v^2 - u^2}This means: Boat A's total time is governed by the harmonic mean of its two speeds. The upstream penalty is always larger than the downstream gain, making τA\tau_A longer than the naive estimate of 2d/v2d/v.
τA=2dv/(v2−u2)\tau_A = 2dv / (v^2 - u^2)
04Step 2: Boat B (Across and Back)
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Boat B travels perpendicular to the river. To move straight across, the boat must aim upstream at an angle. The velocity triangle gives the effective perpendicular speed.Using the Pythagorean theorem on the velocity triangle:veff=v2−u2v_{\text{eff}} = \sqrt{v^2 - u^2}where vv is the hypotenuse (boat speed in water) and uu is the horizontal component (stream). The effective crossing speed v2−u2\sqrt{v^2 - u^2} is the same going out and coming back.Total time for Boat B (two equal legs):τB=dv2−u2+dv2−u2=2dv2−u2\tau_B = \frac{d}{\sqrt{v^2 - u^2}} + \frac{d}{\sqrt{v^2 - u^2}} = \frac{2d}{\sqrt{v^2 - u^2}}This means: Boat B loses some speed to fighting the current (from vv down to v2−u2\sqrt{v^2 - u^2}), but both legs are equally affected, so there is no asymmetry penalty.
τB=2d/v2−u2\tau_B = 2d / \sqrt{v^2 - u^2}
05Step 3: Time Ratio and Why Parallel is SlowerKEY INSIGHT
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Now we take the ratio τA/τB\tau_A / \tau_B. The distance dd cancels completely.τAτB=2dv/(v2−u2)2d/v2−u2=vv2−u2\frac{\tau_A}{\tau_B} = \frac{2dv/(v^2-u^2)}{2d/\sqrt{v^2-u^2}} = \frac{v}{\sqrt{v^2-u^2}}Dividing numerator and denominator by vv:τAτB=11−u2/v2=11−1/η2\frac{\tau_A}{\tau_B} = \frac{1}{\sqrt{1 - u^2/v^2}} = \frac{1}{\sqrt{1 - 1/\eta^2}}Substituting η=1.2\eta = 1.2:τAτB=11−1/(1.2)2=11−1/1.44=11−0.694=10.306≈1.8\frac{\tau_A}{\tau_B} = \frac{1}{\sqrt{1 - 1/(1.2)^2}} = \frac{1}{\sqrt{1 - 1/1.44}} = \frac{1}{\sqrt{1 - 0.694}} = \frac{1}{\sqrt{0.306}} \approx 1.8Why is parallel travel slower? When you travel parallel, the current gives you a bonus on one leg but an equal penalty on the other. But time = distance/speed, and this is a nonlinear function. The time lost going upstream (at the slower speed v−uv - u) always exceeds the time saved going downstream (at the faster speed v+uv + u). This is the harmonic mean being less than the arithmetic mean.
τA/τB=1/1−1/η2≈1.8\tau_A/\tau_B = 1/\sqrt{1 - 1/\eta^2} \approx 1.8
06Verification and Summary
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Verification: Let us check the numerical computation step by step.η=1.2  ⟹  η2=1.44  ⟹  1/η2=0.694\eta = 1.2 \implies \eta^2 = 1.44 \implies 1/\eta^2 = 0.6941−1/η2=1−0.694=0.3061 - 1/\eta^2 = 1 - 0.694 = 0.3060.306=0.553\sqrt{0.306} = 0.553τA/τB=1/0.553=1.81≈1.8  ✓\tau_A/\tau_B = 1/0.553 = 1.81 \approx 1.8 \; \checkmarkDimensional check: the ratio is dimensionless since both τA\tau_A and τB\tau_B have units of time, and dd cancels. ✓\checkmarkLimiting cases: as η→∞\eta \to \infty (boat much faster than stream), the ratio →1\to 1, meaning both paths take the same time. As η→1\eta \to 1 (boat barely faster than stream), the ratio →∞\to \infty because the upstream leg takes forever. Both limits make physical sense. ✓\checkmarkSummary: Boat A (parallel) takes 80% longer than Boat B (perpendicular) when η=1.2\eta = 1.2. The key insight is that traveling parallel to a current creates an asymmetric speed profile, and the slow upstream leg dominates. The perpendicular boat loses some speed to current compensation, but both legs are equally affected. The ratio 1/1−1/η21/\sqrt{1 - 1/\eta^2} is always greater than 1, confirming that parallel travel always takes longer.
τA/τB=1.8\boxed{\tau_A/\tau_B = 1.8}
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