Question

A ship moves along the equator to the east with velocity $v_0=30$ km/hour. The southeastern wind blows at an angle $\varphi=60^\circ$ to the equator with velocity $v=15$ km/hour. Find the wind velocity $v'$ relative to the ship and the angle $\varphi'$ between the equator and the wind direction in the reference frame fixed to the ship.

Answer 1

39.7 km/hour, $\tan^{-1}\left(\frac{\sqrt{3}}{5}\right)$ North of West

Answer 2

To solve this problem, we need to use the concept of relative velocity. We will define a coordinate system and represent the velocities as vectors.

1. Define the Coordinate System:
   Let the equator be the x-axis, with East as the positive x-direction.
   Let North be the positive y-direction.

2. Velocity of the Ship ($\vec{v}_s$):
   The ship moves along the equator to the east with velocity $v_0 = 30$ km/hour.
   $\vec{v}_s = 30 \hat{i}$ km/hour.

3. Velocity of the Wind ($\vec{v}_w$):
   The wind velocity is $v = 15$ km/hour.
   "Southeastern wind" means the wind originates from the southeast, so its direction of motion is towards the northwest.
   The wind blows at an angle $\varphi=60^\circ$ to the equator. Since it's blowing towards the northwest, this means it makes an angle of $60^\circ$ with the West direction (negative x-axis) towards the North (positive y-axis).
   Therefore, the x-component of the wind velocity is negative, and the y-component is positive.
   $\vec{v}_w = (-v \cos \varphi \hat{i} + v \sin \varphi \hat{j})$
   $\vec{v}_w = (-15 \cos 60^\circ \hat{i} + 15 \sin 60^\circ \hat{j})$
   $\vec{v}_w = \left(-15 \times \frac{1}{2} \hat{i} + 15 \times \frac{\sqrt{3}}{2} \hat{j}\right)$
   $\vec{v}_w = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j})$ km/hour.

4. Wind Velocity Relative to the Ship ($\vec{v}'$):
   The velocity of the wind relative to the ship is given by:
   $\vec{v}' = \vec{v}_w - \vec{v}_s$
   $\vec{v}' = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) - (30 \hat{i})$
   $\vec{v}' = (-7.5 - 30) \hat{i} + 7.5\sqrt{3} \hat{j}$
   $\vec{v}' = (-37.5 \hat{i} + 7.5\sqrt{3} \hat{j})$ km/hour.

5. Magnitude of the Relative Wind Velocity ($|\vec{v}'|$):
   $|\vec{v}'| = \sqrt{(-37.5)^2 + (7.5\sqrt{3})^2}$
   $|\vec{v}'| = \sqrt{1406.25 + (56.25 \times 3)}$
   $|\vec{v}'| = \sqrt{1406.25 + 168.75}$
   $|\vec{v}'| = \sqrt{1575}$
   To simplify $\sqrt{1575}$: $1575 = 25 \times 63 = 25 \times 9 \times 7 = (5 \times 3)^2 \times 7 = 15^2 \times 7$
   $|\vec{v}'| = 15\sqrt{7}$ km/hour.
   Numerically, $15\sqrt{7} \approx 15 \times 2.64575 \approx 39.686$ km/hour.
   Rounding to one decimal place, $|\vec{v}'| \approx 39.7$ km/hour.

6. Angle of the Relative Wind Velocity ($\varphi'$):
   The relative velocity vector $\vec{v}' = (-37.5 \hat{i} + 7.5\sqrt{3} \hat{j})$ lies in the second quadrant (West and North).
   Let $\varphi'$ be the angle this vector makes with the negative x-axis (West direction) towards the positive y-axis (North direction).
   $\tan \varphi' = \frac{\text{y-component}}{\text{|x-component|}}$
   $\tan \varphi' = \frac{7.5\sqrt{3}}{|-37.5|}$
   $\tan \varphi' = \frac{7.5\sqrt{3}}{37.5}$
   $\tan \varphi' = \frac{7.5\sqrt{3}}{7.5 \times 5}$
   $\tan \varphi' = \frac{\sqrt{3}}{5}$
   So, $\varphi' = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)$ North of West.

The final answer is $\boxed{\text{39.7 km/hour, }\tan^{-1}\left(\frac{\sqrt{3}}{5}\right) \text{ North of West}}$.

Solution:
The velocity of the ship is $\vec{v}_s = 30 \hat{i}$ km/h (East).
The wind blows from the southeast at $60^\circ$ to the equator, meaning its velocity vector is towards the northwest, making an angle of $60^\circ$ with the West direction.
The velocity of the wind is $\vec{v}_w = (-15 \cos 60^\circ \hat{i} + 15 \sin 60^\circ \hat{j}) = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j})$ km/h.
The wind velocity relative to the ship is $\vec{v}' = \vec{v}_w - \vec{v}_s = (-7.5 \hat{i} + 7.5\sqrt{3} \hat{j}) - (30 \hat{i}) = (-37.5 \hat{i} + 7.5\sqrt{3} \hat{j})$ km/h.
The magnitude of the relative velocity is $|\vec{v}'| = \sqrt{(-37.5)^2 + (7.5\sqrt{3})^2} = \sqrt{1406.25 + 168.75} = \sqrt{1575} = 15\sqrt{7} \approx 39.7$ km/h.
The angle $\varphi'$ of the relative wind direction with the equator (West) is given by $\tan \varphi' = \frac{7.5\sqrt{3}}{37.5} = \frac{\sqrt{3}}{5}$.
Thus, $\varphi' = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)$ North of West.

Answer: The wind velocity relative to the ship is 39.7 km/hour, and the angle is $\tan^{-1}\left(\frac{\sqrt{3}}{5}\right)$ North of West.