Question:
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 :
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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Work in the frame of A: B moves with the relative velocity \(\vec v_B-\vec v_A\) from a known relative position. The distance is least when the relative position is perpendicular to the relative velocity.
Updated On: Nov 5, 2026
- 2.2 hrs
- 4.2 hrs
- 2.6 hrs
- 3.2 hrs
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The Correct Option is C
Solution and Explanation
Take A as origin, east = x, north = y. Then \(\vec v_A = (30, 50)\) km/hr, \(\vec v_B = (-10, 0)\) km/hr and B starts at \(\vec r_0 = (80, 150)\) km.
Position of B relative to A: \(\vec r(t) = \vec r_0 + (\vec v_B-\vec v_A)t = (80-40t,\ 150-50t)\).
Relative velocity of B w.r.t. A: \(\vec u = (-40, -50)\), \(|\vec u|^2 = 1600+2500 = 4100\).
The separation is minimum when \(\vec r\cdot\vec u = 0\): \(-40(80-40t) - 50(150-50t) = 0\).
\(-3200 + 1600t - 7500 + 2500t = 0 \Rightarrow 4100t = 10700\).
\(t = 10700/4100 \approx 2.6\) hr.
Answer: (3) 2.6 hrs
Position of B relative to A: \(\vec r(t) = \vec r_0 + (\vec v_B-\vec v_A)t = (80-40t,\ 150-50t)\).
Relative velocity of B w.r.t. A: \(\vec u = (-40, -50)\), \(|\vec u|^2 = 1600+2500 = 4100\).
The separation is minimum when \(\vec r\cdot\vec u = 0\): \(-40(80-40t) - 50(150-50t) = 0\).
\(-3200 + 1600t - 7500 + 2500t = 0 \Rightarrow 4100t = 10700\).
\(t = 10700/4100 \approx 2.6\) hr.
Answer: (3) 2.6 hrs
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