Ship A is sailing towards north-east with velocity v = 30i + 50j… | JEE Main 2019 (8 Apr, Shift 1)

Q21JEE Main 2019 · 8 APR · SHIFT 1Difficult

Relative Velocity in Two Dimensions

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:

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Answer: (c)

Position of B relative to A: $\vec{r} = 80\hat{i} + 150\hat{j}$ ; $\vec{v}_{BA} = -10\hat{i} - (30\hat{i} + 50\hat{j}) = -40\hat{i} - 50\hat{j}$
Minimum distance at $t = -\frac{\vec{r}\cdot\vec{v}_{BA}}{|\vec{v}_{BA}|^2} = \frac{3200 + 7500}{1600 + 2500}$
$t = \frac{10700}{4100} \approx 2.6\text{ h}$
So the answer is (c).

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Relative Velocity in Two Dimensions question from JEE Main 2019 (8 Apr, Shift 1)

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