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) 5√(2) h (b) 10√(2) h (c) 0 h (d) 5 h

Answer: (d). Take A at origin. Relative to A, B starts at (0, -100) with velocity vB - vA = (10, 10) km/h. Closest approach when rrel = 0: 10(10t) + 10(-100 + 10t) = 0 200t = 1000 ⇒ t = 5 h So the answer is (d).

A ship A is moving Westwards with a speed of 10 km h−1 and a ship B… | NEET 2015 (AIPMT, Cancelled Paper)

Q23NEET 2015 · AIPMT · CANCELLED PAPERMedium

Relative Velocity in Two Dimensions

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

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Answer: (d)

Take A at origin. Relative to A, B starts at $(0, -100)$ with velocity $\vec{v}_B - \vec{v}_A = (10, 10)\text{ km/h}$ .
Closest approach when $\vec{r}\cdot\vec{v}_{rel} = 0$ : $10(10t) + 10(-100 + 10t) = 0$
$200t = 1000 \Rightarrow t = 5\text{ h}$
So the answer is (d).

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Relative Velocity in Two Dimensions question from NEET 2015 (AIPMT, Cancelled Paper)

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