[Q1] Let $\vec{v}_{w/g}$ be the velocity of the true wind relative to the ground, $\vec{v}_{c/g}$ be the velocity of the cyclist relative to the ground, and $\vec{v}_{w/c}$ be the velocity of the apparent wind relative to the cyclist.

The fundamental relationship for relative velocity is:

$$\vec{v}_{w/c} = \vec{v}_{w/g} - \vec{v}_{c/g}$$

To find the true wind velocity, $\vec{v}_{w/g}$, we rearrange this equation algebraically:

$$\vec{v}_{w/g} = \vec{v}_{w/c} + \vec{v}_{c/g}$$

This shows the true wind velocity is the vector sum of the apparent wind velocity and the cyclist's velocity.

We define a coordinate system where the positive x-axis points East and the positive y-axis points North. The given vectors are:

• Cyclist's velocity: $\vec{v}_{c/g} = 6 \hat{j}$ m/s
• Apparent wind velocity: $\vec{v}_{w/c} = 8 \hat{i}$ m/s

Substituting these into the derived expression for the true wind velocity:

$$\vec{v}_{w/g} = (8 \hat{i}) + (6 \hat{j}) \text{ m/s}$$

The magnitude (speed) of the true wind is given by the Pythagorean theorem:

$$v_{w/g} = |\vec{v}_{w/g}| = \sqrt{(v_{w/g,x})^2 + (v_{w/g,y})^2}$$ $$v_{w/g} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ m/s}$$

The direction of the true wind, $\theta$, measured from the East towards the North, is found using the arctangent:

$$\theta = \arctan\left(\frac{v_{w/g,y}}{v_{w/g,x}}\right)$$ $$\theta = \arctan\left(\frac{6}{8}\right) = \arctan(0.75) \approx 36.9^\circ$$

The true wind is directed $36.9^\circ$ North of East.