A man walks at a constant speed of 4 km h⁻¹ from west to east on level ground - Leaving Cert Applied Maths - Question 2 - 2019

Let the velocity of the wind be $\mathbf{V_w}$. The man's velocity, $\mathbf{V_m}$, is 4 km h⁻¹ to the east, so we can represent it as $\mathbf{V_m} = 4 \hat{i}$.

According to the information given, the wind appears to the man to come from the direction north 50° east, which means it has a component in the $\hat{j}$ direction and a component in the $\hat{i}$ direction. Let’s denote the velocity of wind as follows:

$\mathbf{V_w} = V_x \hat{i} + V_y \hat{j}$.

For the woman, her velocity, $\mathbf{V_w}$, can be split into components: [ V_{wm} = 4 \sin(40) \hat{i} - 4 \cos(40) \hat{j}. ]

Setting these equal gives:

$4 = V_x + 4 \sin(50)$

Using the first equation and solving, we find:

$u = 8.578$ km h⁻¹.

This leads us to find: [ |\mathbf{V_w}| = \sqrt{(\sqrt{7.25})^2 + (\sqrt{5.51})^2} = 6.08 \text{ km h}^{-1}. ]

Finally, finding the angle: $\tan^{-1}(\frac{2.57}{5.51}) \Rightarrow \text{south } 25°$.