Irodov 1.7: Two Routes Across, One Arrival Time – Amit & Deepali Sinha Foundation

Problem Irodov 1.7

Two swimmers start together from a point $A$ on one bank of a river, both aiming for a point $B$ directly opposite $A$ on the other bank. The first swimmer swims along the straight line $AB$ . The second swims perpendicular to the current (straight across) and then walks along the bank, at a velocity $u$ , to cover the distance the current has carried him past $B$ . The stream’s velocity is $v_0 = 2.0$ km/h, and each swimmer’s velocity relative to the water is $v' = 2.5$ km/h. Given that both swimmers reach $B$ at the same moment, find the walking velocity $u$ .

Seen from above: the current runs downstream at v_0 between banks d apart. Both swimmers go from A to B opposite. Route 1 is the line AB; route 2 crosses square to C, then walks to B at speed u. — Seen from above: the current runs downstream at $v_0$ between banks $d$ apart. Both swimmers go from $A$ to $B$ opposite. Route 1 is the line $AB$ ; route 2 crosses square to $C$ , then walks to $B$ at speed $u$ .

Given

$v_0 = 2.0$ km/h
$v' = 2.5$ km/h

Find

The walking speed

u

of the second swimmer

Solution

Step 1Swimmer 1: aim upstream to travel straight across

Swimmer 1's velocity triangle. Side XY is his water-relative speed v', aimed partly upstream; side YZ is the stream velocity v_0; the closing side XZ is his ground velocity, running straight across the river. The right angle sits at Z. — Swimmer 1’s velocity triangle. Side $XY$ is his water-relative speed $v'$ , aimed partly upstream; side $YZ$ is the stream velocity $v_0$ ; the closing side $XZ$ is his ground velocity, running straight across the river. The right angle sits at $Z$ .

To end up moving along the straight line $AB$ (straight across, no net drift), swimmer 1 must aim his fixed water-relative speed $v'$ partly upstream, enough that its upstream component exactly cancels the current $v_0$ . The figure’s right triangle then gives his straight-across ground speed as the other leg,

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

(1)

so he crosses a river of width $d$ in a time

t_1 = \frac{d}{\sqrt{v'^2-v_0^2}}. \tag{2}

Step 2Swimmer 2: cross square, then walk back the drift

Swimmer 2's velocity triangle. Side XY is his water-relative speed v', square across the stream; side YZ is the stream velocity v_0; the closing side XZ is his ground velocity, slanted downstream to C. The right angle sits at Y. — Swimmer 2’s velocity triangle. Side $XY$ is his water-relative speed $v'$ , square across the stream; side $YZ$ is the stream velocity $v_0$ ; the closing side $XZ$ is his ground velocity, slanted downstream to $C$ . The right angle sits at $Y$ .

Swimmer 2 instead aims his water-relative speed $v'$ straight across, so the whole of it goes into the crossing and, as the figure shows, his crossing takes

t_2 = \frac{d}{v'} \tag{3}

and during that time the current carries him a distance $v_0 t_2 = v_0 d/v'$ downstream of $B$ . He then walks that distance back at his own speed $u$ , taking

t_3 = \frac{v_0 d}{v' u}. \tag{4}

Step 3Equate the two total times

Simultaneous arrival means $t_1=t_2+t_3$ :

\frac{d}{\sqrt{v'^2-v_0^2}} = \frac{d}{v'}+\frac{v_0 d}{v' u}. \tag{5}

Dividing through by $d$ and then by $1/v'$ ,

\frac{v'}{\sqrt{v'^2-v_0^2}} = 1+\frac{v_0}{u}. \tag{6}

Step 4Solve for the walking speed

\frac{v_0}{u} = \frac{v'}{\sqrt{v'^2-v_0^2}}-1 = \frac{1}{\sqrt{1-(v_0/v')^2}}-1. \tag{7}

Substituting $v_0=2.0$ km/h and $v'=2.5$ km/h,

u = \frac{v_0}{\left(1-v_0^2/v'^2\right)^{-1/2}-1} = \frac{2.0}{\left(1-(2.0/2.5)^2\right)^{-1/2}-1}\ \text{km/h}.

u = \dfrac{v_0}{\left(1-v_0^2/v'^2\right)^{-1/2}-1} = 3.0\text{ km/h}

(8)

Check

As $v_0\to v'$ (the current as fast as the swimmer), swimmer 1’s crossing speed $\sqrt{v'^2-v_0^2}\to0$ , so $t_1$ diverges and swimmer 1 never gets across, while swimmer 2’s own swim time $t_2=d/v'$ stays finite: he still crosses, and only has to walk back a finite drift. To keep $t_1=t_2+t_3$ with a diverging left side and a bounded first term on the right, the walking time $t_3$ must itself diverge, which means the walk gets easier, not harder, per unit of that diverging time: $u\to0$ . Numerically, with $v'=2.5$ km/h, $u(v_0=2.4)\approx0.93$ km/h, $u(v_0=2.49)\approx0.24$ km/h, $u(v_0=2.499)\approx0.07$ km/h, confirming the trend. The runaway the OTHER way, $u\to\infty$ , sits at the opposite end, $v_0\to0$ (no current to fight at all), where $u\approx 2v'^2/v_0$ .

Remark

Swimmer 1 pays for canceling the drift by losing across-river speed itself, $\sqrt{v'^2-v_0^2}<v'$ (Step 1), so aiming upstream is never free. Swimmer 2 keeps the full $v'$ across instead and pays the drift back afterward on land; which strategy arrives first depends entirely on how fast $u$ is compared to swimming, and this problem is built around tuning $u$ to the one value that makes the two strategies a dead heat.