1. Velocity Analysis

Let the velocity of the wind be $\vec{v}_w$. The direction of the smoke stream is determined by the velocity of the wind relative to the steamer.

• Steamer A: Moves North at $30$ km/h. $\vec{v}_A = 30\hat{j}$.
• Steamer B: Moves South at $10$ km/h. $\vec{v}_B = -10\hat{j}$.

2. Relative Wind Direction

Condition 1 (Steamer A): The smoke spreads towards the West. This means the relative velocity $\vec{v}_{w/A}$ is in the $-\hat{i}$ direction.

$$ \vec{v}_{w/A} = \vec{v}_w – \vec{v}_A = -k\hat{i} \quad (k > 0) $$

$$ \vec{v}_w = 30\hat{j} – k\hat{i} $$

Condition 2 (Steamer B): The smoke spreads towards the North-West. This means $\vec{v}_{w/B}$ is along the vector $\hat{j} – \hat{i}$.

$$ \vec{v}_{w/B} = \vec{v}_w – \vec{v}_B = \vec{v}_w – (-10\hat{j}) = \vec{v}_w + 10\hat{j} $$

Substituting $\vec{v}_w$:

$$ \vec{v}_{w/B} = (30\hat{j} – k\hat{i}) + 10\hat{j} = 40\hat{j} – k\hat{i} $$

For this vector to be North-West (angle $45^\circ$ between North and West), the magnitude of the North component must equal the magnitude of the West component:

$$ |v_y| = |v_x| \implies 40 = k $$

3. Resultant Wind Velocity

Substituting $k=40$ back into the expression for $\vec{v}_w$:

$$ \vec{v}_w = -40\hat{i} + 30\hat{j} $$

Magnitude:

$$ |\vec{v}_w| = \sqrt{(-40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{km/h} $$

Direction:

Let $\theta$ be the angle measured from the North direction towards West.

$$ \tan \theta = \frac{|v_x|}{|v_y|} = \frac{40}{30} = \frac{4}{3} $$

$$ \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ $$

The wind blows at 50 km/h, $53^\circ$ West of North.