Ship A is sailing towards north-east with velocity $\vec v = 30\hat i + 50\hat j\ \text{km h}^{-1}$ where $\hat i$ points east and $\hat j$, north. The ship B is at a distance of $80\ \text{km}$ east and $150\ \text{km}$ north of Ship A and is sailing towards the west at $10\ \text{km h}^{-1}$. A will be at the minimum distance from B in
Answer: (C) $2.6\ \text{h}$
Work in the frame of A. Position of B relative to A at $t = 0$: $\vec r_0 = 80\hat i + 150\hat j$ km.
Velocity of B relative to A: $\vec v_{BA} = -10\hat i - (30\hat i + 50\hat j) = -40\hat i - 50\hat j$ km/h.
The separation $\vec r_0 + \vec v_{BA}t$ is smallest when it is perpendicular to $\vec v_{BA}$:
$$t = -\frac{\vec r_0\cdot\vec v_{BA}}{|\vec v_{BA}|^2} = \frac{80\times40 + 150\times50}{40^2 + 50^2} = \frac{10700}{4100} \approx 2.6\ \text{h}$$
Solution by Sreeraj P, M.Sc Physics