Question

# Ship A is sailing towards north-east with velocity →v=30^i+50^j km/h, where ^i points east and ^j points north. Ship B at a distance of 80 km east and 150 km north of ship A is sailing towards the west at 10 km/h. A will be at minimum distance from B in

A
4.2 h
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B
2.6 h
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C
3.2 h
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D
2.2 h
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Solution

## The correct option is B 2.6 hConsidering the initial position of ship A as origin, the velocity of ship will be →vA=30^i+50^j Position of ship A will be, →rA=0^i+0^j Velocity and position of ship B will be →vB=−10^i and →rB=(80^i+150^j) Hence, the given situation can be represented graphically as After time t, coordinates of ship A and B will be (30t,50t) and (80−10t,150) So, distance between A and B after time t is D=√(x2−x1)2+(y2−y1)2 D=√(80−30t−10t)2+(150−50t)2 D2=(80−40t)2+(150−50t)2 Distance will be minimum when ddt(D2)=0 ⟹2(80−40t)(−40)+2(150−50t)(−50)=0 ⟹−6400+3200t−15000+5000t=0 ∴t=214008200=2.6 h

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