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BPO 2026, Теория — Задача 3. Istanbul Bosphorus Cross-Continental Swim and current correction

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EIGHTH BALKAN PHYSICS OLYMPIAD (BPO-8) · 29 юни 2026 г. · 25 т.

Условие

The Istanbul Bosphorus Cross-Continental Swimming Race is an international open-water event in which swimmers cross from the Asian side of Istanbul to the European side. In this problem, the Bosphorus current is not neglected.

A simplified race course is modeled as a straight crossing of width

D = 1450\ \mathrm{m}

from point $A$ on the Asian shore to point $B$ on the European shore. The current flows parallel to the shores, from north to south, with constant speed $u = 0{,}65\ \mathrm{m/s}$ . A swimmer swims with the constant speed $\mathrm{v} = 1{,}25\ \mathrm{m/s}$ relative to the water.

Top view of the crossing: the Asian shore (left vertical line) and the European shore (right vertical line) with downward arrows marking the current u; from point A a dashed target line AB goes horizontally to B, labelled 'Target line AB' and 'Corrected ground path'; the aiming direction v points up-right from A at angle θ to AB; a solid line 'Path with no current correction' goes from A down to B′; the segment BB′ is marked 'x drift'; the width is marked D = 1450 m. — Figure 3.1 Current correction during the Istanbul Bosphorus crossing.

The swimmer wants to reach exactly point $B$ , directly opposite to point $A$ . Therefore, the swimmer must aim upstream at an angle $\theta$ with respect to the straight line $AB$ .

A) Find the angle $\theta$ at which the swimmer must aim relative to the line $AB$ in order to arrive exactly at point $B$ . [6 т.]

B) Determine the time required for the swimmer to cross the Bosphorus and reach point $B$ . [6 т.]

C) A second swimmer ignores the current and swims directly toward point $B$ . Determine the time needed to reach the European shore, as well as the downstream displacement from point $B$ . [6 т.]

D) During a training session, three swimmers cross the same route. Their measured crossing times and downstream displacements are given below.

Swimmer	Crossing time $t$ (s)	Downstream displacement $x$ (m)
S1	1320	0
S2	1160	270
S3	1450	-110

Here positive $x$ means downstream from $B$ , and negative $x$ means upstream from $B$ .

For each swimmer, determine the average ground-velocity components

v'_x = \frac{x}{t}, \qquad v'_y = \frac{D}{t},

and decide which swimmer made the best current correction. [7 т.]

Отговори

Решение

Choose the positive $x$ -direction along the straight crossing from the Asian shore toward the European shore. Choose the positive $y$ -direction upstream. Therefore the water current is

\vec{u} = (0, -u), \qquad u = 0.65\ \mathrm{ms^{-1}}.

The swimmer’s speed relative to the water is constant:

v = 1.25\ \mathrm{ms^{-1}}.

The width of the crossing is

D = 1450\ \mathrm{m}.

(a) Angle required to land exactly at point $B$ (6 points)

To land exactly at $B$ , the swimmer must aim upstream so that the upstream component of his velocity relative to the water cancels the downstream current.

If the swimmer aims at angle $\theta$ above the line $AB$ , the velocity components relative to the water are

v_x = v\cos\theta,

(1.5p)

v_y = v\sin\theta.

The ground-velocity component in the $y$ -direction is therefore

v_{y,\mathrm{ground}} = v\sin\theta - u.

For exact arrival at $B$ ,

(1.5p)

v_{y,\mathrm{ground}} = 0.

Thus

v\sin\theta - u = 0

(1.5p)

v\sin\theta = u,

(0.5p)

\sin\theta = \frac{u}{v} = \frac{0.65}{1.25} = 0.52.

Therefore

(1.0p)

\theta = \sin^{-1}(0.52) \approx 31.3^{\circ}.

Hence the swimmer must aim approximately

\boxed{\theta \approx 31.3^{\circ}\ \text{upstream.}}

(b) Time required to cross the Bosphorus (6 points)

The forward speed across the Bosphorus is the $x$ -component of the ground velocity:

(2.0p)

v_{x,\mathrm{ground}} = v\cos\theta.

Since

(1.0p)

\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (0.52)^2} = \sqrt{0.7296} \approx 0.8542,

we get

v_{x,\mathrm{ground}} = 1.25(0.8542) \approx 1.068\ \mathrm{m/s}.

The crossing time is

(2.0p)

t = \frac{D}{v_{x,\mathrm{ground}}} = \frac{1450}{1.068} \approx 1.36 \cdot 10^3\ \mathrm{s}.

Thus

(1.0p)

t \approx 1358\ \mathrm{s} \approx 22.6\ \mathrm{min}.

Now the swimmer does not correct for the current. Therefore the swimmer’s velocity relative to the water is completely in the $x$ -direction:

v_x = v = 1.25\ \mathrm{m/s}.

The current gives the whole $y$ -component of the ground velocity:

v_y = -u = -0.65\ \mathrm{m/s}.

Time to reach the European shore. Since the $x$ -component is 1.25 ms $^{-1}$ , the crossing time is

(2.0p)

t_2 = \frac{D}{v} = \frac{1450}{1.25} = 1160\ \mathrm{s}.

Thus

(1.0p)

\boxed{t_2 = 1160\ \mathrm{s}.}

Downstream displacement. The downstream drift is

(2.0p)

d = ut_2 = 0.65(1160) = 754\mathrm{m}.

Therefore the swimmer lands

(1.0p)

\boxed{754\mathrm{m}}

downstream from point $B$ .

(d) Analysis of the GPS checkpoint data

For each swimmer, the average velocity components are obtained from

v'_x = \frac{x}{t}, v'_y = \frac{D}{t},

where $x$ is the final displacement parallel to the current direction relative to point $B$ .

Swimmer S1

t = 1320\mathrm{s}, \qquad x = 0.

Hence

(1.0p)

v'_x = \frac{0}{1320} = 0,

(1.0p)

v'_y = \frac{1450}{1320} \approx 1.099\,\frac{\mathrm{m}}{\mathrm{s}}.

\boxed{v'_x = 0, \qquad v'_y \approx 1.099\ \mathrm{ms^{-1}}.}

Swimmer S2

t = 1160\ \mathrm{s},\ x = 270\ \mathrm{m}.

Thus

(1.0p)

v'_x = \frac{270}{1160} \approx 0.233\ \mathrm{m/s},

(1.0p)

v'_y = \frac{1450}{1160} \approx 1.25\,\frac{\mathrm{m}}{\mathrm{s}}.

Therefore

\boxed{v_x \approx 0.233\mathrm{ms^{-1}}, \qquad v_y = 1.25\mathrm{ms^{-1}}.}

Swimmer S3

t = 1450\ \mathrm{s}, \qquad x = -110\ \mathrm{m}.

Thus

(1.0p)

v'_x = \frac{-110}{1450} = -0.0759\ \mathrm{m/s},

(1.0p)

v'_y = \frac{1450}{1450} \approx 1.00\,\frac{\mathrm{m}}{\mathrm{s}}.

Therefore

\boxed{v_x \approx -0.0759\ \mathrm{ms^{-1}}, \qquad v_y = 1.00\ \mathrm{ms^{-1}}.}

The best current correction corresponds to the swimmer who lands closest to the target line through $B$ . This means the swimmer with the smallest value of $|x|$ .

The displacements are

|x_{S1}| = 0, \qquad |x_{S2}| = 270\mathrm{m}, \qquad |x_{S3}| = 110\mathrm{m}.

Therefore the best current correction is achieved by

(1.0p)

\boxed{\text{S1.}}

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf