$1.4.6^*.$ 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: $$v_b = v\sqrt{3}$$
