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A ship A is moving westwards with a speed of 10 km $h^{-1}$ and a ship B 100 km south of A is moving northwards with a speed of 10 km $h^{-1}$. The time after which the distance between them becomes shortest is
A
0 h
B
5 h
C
$5\sqrt{2}$ h
D
$10\sqrt{2}$ h
Detailed Solution
$\vec v_A = 10(-\hat i)$ km/h and $\vec v_B = 10\hat j$ km/h.
Velocity of B relative to A: $\vec v_{BA} = 10\hat j + 10\hat i$, so $|\vec v_{BA}| = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ km/h, directed at $45^\circ$.
Relative to A, B moves along BC; the distance is least when B reaches the foot C of the perpendicular from A.
$BC = 100\cos 45^\circ = 50\sqrt{2}$ km
$t = \frac{50\sqrt{2}}{10\sqrt{2}} = 5$ h
Velocity of B relative to A: $\vec v_{BA} = 10\hat j + 10\hat i$, so $|\vec v_{BA}| = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ km/h, directed at $45^\circ$.
Relative to A, B moves along BC; the distance is least when B reaches the foot C of the perpendicular from A.
$BC = 100\cos 45^\circ = 50\sqrt{2}$ km
$t = \frac{50\sqrt{2}}{10\sqrt{2}} = 5$ h