Irodov 1.8: Crossing Costs More Than Coasting – Amit & Deepali Sinha Foundation

Problem Irodov 1.8

Two boats, $A$ and $B$ , start together from a buoy anchored in the middle of a river and move off along mutually perpendicular straight lines: boat $A$ travels along the river (parallel to the current), boat $B$ travels straight across it (perpendicular to the current). Each boat moves the same distance away from the buoy and then returns to it. Each boat’s own velocity relative to the water is $\eta = 1.2$ times the stream’s velocity. Find the ratio $\tau_A/\tau_B$ of the two boats’ total travel times.

Seen from above: the current runs downstream at v_0. Both boats start from the midstream buoy O, boat A running s along the river and boat B the same s across it. The dashed arrow beside each leg is that boat's return. — Seen from above: the current runs downstream at $v_0$ . Both boats start from the midstream buoy $O$ , boat $A$ running $s$ along the river and boat $B$ the same $s$ across it. The dashed arrow beside each leg is that boat’s return.

Given

$\eta = v/v_0 = 1.2$

Find

The ratio

\tau_A/\tau_B

of the two boats’ round-trip times, for equal distances

s

out and back

Solution

Step 1Boat

A

: straight up and down the current

Boat A's velocity diagram, the triangle collapsed onto a line. Each group of three arrows starts from the same dashed reference. Upstream (left) v and v_0 oppose, leaving the ground speed v-v_0; downstream (right) they add, giving v+v_0. — Boat $A$ ’s velocity diagram, the triangle collapsed onto a line. Each group of three arrows starts from the same dashed reference. Upstream (left) $\vect v$ and $\vect v_0$ oppose, leaving the ground speed $v-v_0$ ; downstream (right) they add, giving $v+v_0$ .

Let $v$ be each boat’s own speed relative to the water and $v_0$ the stream speed. Boat $A$ runs a distance $s$ upstream, ground speed $v-v_0$ , then $s$ downstream, ground speed $v+v_0$ :

\tau_A = \frac{s}{v-v_0}+\frac{s}{v+v_0} = \frac{2vs}{v^2-v_0^2}. \tag{1}

Step 2Boat

B

: angled to cross straight and straight back

Boat B's velocity triangle, the same on both legs. Side XY is its water-relative speed v, angled upstream; side YZ is the stream velocity v_0; side XZ closes as the ground velocity, square across the river, right angle at Z. — Boat $B$ ’s velocity triangle, the same on both legs. Side $XY$ is its water-relative speed $v$ , angled upstream; side $YZ$ is the stream velocity $v_0$ ; side $XZ$ closes as the ground velocity, square across the river, right angle at $Z$ .

To move along the straight line perpendicular to the bank, boat $B$ must angle upstream enough to cancel the current, exactly as in the figure’s right triangle; its straight-across ground speed is the other leg,

v_\perp = \sqrt{v^2-v_0^2},

the same magnitude on the way out and (aimed the mirror way) on the way back, so the round trip covers a distance $2s$ at this speed:

\tau_B = \dfrac{2s}{\sqrt{v^2-v_0^2}}

(2)

Step 3Take the ratio

\frac{\tau_A}{\tau_B} = \frac{2vs/(v^2-v_0^2)}{2s/\sqrt{v^2-v_0^2}} = \frac{v}{\sqrt{v^2-v_0^2}}. \tag{3}

Dividing numerator and denominator by $v_0$ casts this in terms of $\eta$ alone; substituting $\eta=1.2$ ,

\frac{\tau_A}{\tau_B} = \frac{\eta}{\sqrt{\eta^2-1}} = \frac{1.2}{\sqrt{1.2^2-1}}.

\dfrac{\tau_A}{\tau_B} = \dfrac{\eta}{\sqrt{\eta^2-1}} \approx 1.8

(4)

Check

As $\eta\to1$ (the boats barely outrun the current), both round trips get long, but at different rates. Boat $A$ ’s upstream leg speed $v-v_0\to0$ linearly in $(\eta-1)$ , so $\tau_A\sim s/(v_0(\eta-1))$ diverges as $1/(\eta-1)$ ; boat $B$ ’s crossing speed $\sqrt{v^2-v_0^2}\,\propto\sqrt{\eta-1}$ vanishes only as $\sqrt{\eta-1}$ , so $\tau_B$ diverges more slowly, as $1/\sqrt{\eta-1}$ . A quantity blowing up as $1/(\eta-1)$ outpaces one blowing up as $1/\sqrt{\eta-1}$ , so the ratio $\tau_A/\tau_B\to\infty$ : boat $A$ , fighting the current head on, suffers far more than boat $B$ , which only has to crab across it. That matches the formula, which diverges at $\eta=1$ .

Remark

Boat $B$ ’s crossing here is exactly swimmer 1’s triangle from Problem 1.7: the same $\sqrt{v^2-v_0^2}$ leg from angling upstream to cancel drift. What is new in this problem is the direct head-on comparison with boat $A$ ’s along-stream run, which is what exposes how much more a current punishes travel along its own direction than across it.