Ship $A$ is sailing towards north-east with velocity $\vec {v} = 30\hat {i} + 50\hat {j}\ km/hr$ where $\hat {i}$ points east and $\hat {j}$, north. Ship $B$ is at a distance of $80\ km$ east and $150\ km$ north of Ship $A$ and is sailing towards west at $10\ km/hr$. A will be at minimum distance from $B$ in

The velocity of ship A relative to ship B is found by subtracting B's velocity from A's velocity, giving (30 - (-10))i + (50 - 0)j = 40i + 50j km/hr. The initial relative position vector of B with respect to A is 80i + 150j km. The minimum distance occurs when the relative position vector is perpendicular to the relative velocity vector, meaning their dot product equals zero: (80 - 40t)(40) + (150 - 50t)(50) = 0. Expanding this gives 3200 - 1600t + 7500 - 2500t = 0, which simplifies to 4100t = 10700, yielding t = 2.6 hours.

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