Let the velocity of the wind be $\vec{v}_w$, the velocity of the ship be $\vec{v}_s$, and the observed velocity of the smoke relative to the ship be $\vec{v}_{obs}$. The fundamental relationship for relative velocity is:

$$ \vec{v}_{obs} = \vec{v}_{smoke/ground} - \vec{v}_s $$

Assuming the smoke travels with the wind, its velocity relative to the ground is $\vec{v}_{smoke/ground} = \vec{v}_w$. Therefore,

$$ \vec{v}_{obs} = \vec{v}_w - \vec{v}_s $$

We establish a coordinate system where the x-axis points East and the y-axis points North. The wind velocity vector is $\vec{v}_w = (v_{wx}, v_{wy})$.

Case 1: Ship sailing due North The ship's velocity is $\vec{v}_{s1} = (0, v_{s1})$, where $v_{s1} = 15$ km/h. The smoke is observed drifting due East, so its observed velocity has only an x-component: $\vec{v}_{obs1} = (u_1, 0)$ for some speed $u_1 > 0$. Substituting into the relative velocity equation:

$$ (u_1, 0) = (v_{wx}, v_{wy}) - (0, v_{s1}) = (v_{wx}, v_{wy} - v_{s1}) $$

By equating the y-components, we find the northerly component of the wind's velocity:

$$ 0 = v_{wy} - v_{s1} \implies v_{wy} = v_{s1} $$

Case 2: Ship sailing due East The ship's velocity is $\vec{v}_{s2} = (v_{s2}, 0)$, where $v_{s2} = 24$ km/h. The smoke is observed drifting due Northwest. This means the x-component of its observed velocity is negative, the y-component is positive, and their magnitudes are equal: $\vec{v}_{obs2} = (-u_2, u_2)$ for some speed $u_2 > 0$. Substituting into the relative velocity equation:

$$ (-u_2, u_2) = (v_{wx}, v_{wy}) - (v_{s2}, 0) = (v_{wx} - v_{s2}, v_{wy}) $$

From the y-components, $u_2 = v_{wy}$. From the x-components, $-u_2 = v_{wx} - v_{s2}$. Substituting $u_2 = v_{wy}$ into the x-component equation gives:

$$ -v_{wy} = v_{wx} - v_{s2} \implies v_{wx} = v_{s2} - v_{wy} $$

Solving for Wind Velocity We now solve for the components of the wind velocity using the results from both cases:

$$ v_{wy} = v_{s1} $$ $$ v_{wx} = v_{s2} - v_{s1} $$

Substituting the given values, $v_{s1} = 15$ km/h and $v_{s2} = 24$ km/h:

$$ v_{wy} = 15 \text{ km/h} $$ $$ v_{wx} = 24 - 15 = 9 \text{ km/h} $$

The wind velocity is $\vec{v}_w = (9, 15)$ km/h.

[Q1] Find the speed of the wind. The speed is the magnitude of the wind velocity vector:

$$ v_w = |\vec{v}_w| = \sqrt{v_{wx}^2 + v_{wy}^2} = \sqrt{(v_{s2} - v_{s1})^2 + v_{s1}^2} $$ $$ v_w = \sqrt{9^2 + 15^2} = \sqrt{81 + 225} = \sqrt{306} = 3\sqrt{34} \text{ km/h} $$

[Q2] Find the direction of the wind. The direction $\theta$ is the angle the wind vector makes with the positive x-axis (East).

$$ \tan\theta = \frac{v_{wy}}{v_{wx}} = \frac{v_{s1}}{v_{s2} - v_{s1}} $$ $$ \tan\theta = \frac{15}{9} = \frac{5}{3} $$

The direction is $\theta = \arctan(5/3)$ North of East.