1. The problem involves two velocities: A with speed 10 km/h and B with speed 20 km/h. 2. B moves relative to A in a direction north-east (or upward relative to A). We need to find the relative velocity of B with respect to A and the direction of B's motion. 3. The relative velocity formula is: $$\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$$ where $\vec{v}_{BA}$ is velocity of B relative to A. 4. Given speeds: - $v_A = 10$ km/h - $v_B = 20$ km/h Assuming A moves east and B moves north-east relative to A, we can use vector components. 5. Let’s assume A moves along the x-axis (east) with velocity $\vec{v}_A = 10 \hat{i}$ km/h. 6. B moves at 20 km/h at an angle $\theta$ north of east. The components of $\vec{v}_B$ are: $$v_{Bx} = 20 \cos \theta$$ $$v_{By} = 20 \sin \theta$$ 7. The relative velocity components are: $$v_{BAx} = v_{Bx} - v_A = 20 \cos \theta - 10$$ $$v_{BAy} = v_{By} = 20 \sin \theta$$ 8. The magnitude of relative velocity is: $$v_{BA} = \sqrt{(20 \cos \theta - 10)^2 + (20 \sin \theta)^2}$$ 9. The direction $\phi$ of B relative to A is: $$\phi = \tan^{-1} \left( \frac{20 \sin \theta}{20 \cos \theta - 10} \right)$$ 10. Without a specific $\theta$, this is the general solution for relative velocity and direction. --- For the second problem: 1. Given wind velocity $u$ at angle $\theta$, and velocity $v$ of B relative to A at distance $d$. 2. If $v < u \sin \theta$, then B cannot reach A. 3. If $v > u$, the time $t$ for B to reach A is: $$t = \frac{d}{\sqrt{v^2 - u^2 \sin^2 \theta} - u \cos \theta}$$ This formula comes from relative velocity and component analysis. --- Final answers: - Relative velocity magnitude: $$v_{BA} = \sqrt{(20 \cos \theta - 10)^2 + (20 \sin \theta)^2}$$ - Direction relative to A: $$\phi = \tan^{-1} \left( \frac{20 \sin \theta}{20 \cos \theta - 10} \right)$$ - Time for B to reach A if $v > u$: $$t = \frac{d}{\sqrt{v^2 - u^2 \sin^2 \theta} - u \cos \theta}$$