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+### Statement
+
+$1.4.6.$ [Insert the problem statement]
+
+### Solution
+
+### Problem
+
+$1.4.6^*.$ An iceboat is a sailing sled. It can only move along the line in which its runners are pointed. The wind blows with a speed $v$ perpendicular to the direction of the iceboat's motion. The sail makes an angle of $30^\circ$ with the direction of motion. What maximum speed can the iceboat reach under this wind?
+
+### Solution
+
+At the initial moment, when the iceboat is stationary, the air moves relative to it with a velocity $\vec{v}$.
+
+As the iceboat starts moving and acquires its own velocity $\vec{v}_b$, the relative velocity of the wind changes. According to the velocity addition formula, the relative velocity of the wind is:
+$$ \vec{v}_{rel} = \vec{v} - \vec{v}_b $$
+
+Since the absolute wind velocity $\vec{v}$ is perpendicular to the iceboat's velocity $\vec{v}_b$, these vectors form a right-angled triangle.
+
+The wind will continue to exert pressure on the sail and accelerate the iceboat as long as the relative wind velocity $\vec{v}_{rel}$ pushes against it. The iceboat reaches its maximum speed when the wind stops exerting force on the sail. This condition is met when the relative wind velocity vector $\vec{v}_{rel}$ becomes parallel to the sail.
+
+According to the problem, the sail is set at an angle of $30^\circ$ to the direction of the iceboat's motion. Therefore, in the limiting case, the angle between the vectors $\vec{v}_b$ and $\vec{v}_{rel}$ in our right-angled triangle is exactly $30^\circ$.
+
+From this velocity vector triangle, we can express the relationship:
+$$ \tan 30^\circ = \frac{v}{v_b} $$
+
+Solving for the maximum speed of the iceboat $v_b$:
+$$ v_b = \frac{v}{\tan 30^\circ} = v \cot 30^\circ $$
+
+Since $\cot 30^\circ = \sqrt{3}$, we obtain:
+$$ v_b = v\sqrt{3} $$
+
+#### Answer
+
+$$ v_{max} = v\sqrt{3} $$
+
+#### Answer
+
+[Insert a concise answer or boxed result]