Kinematics 3 Question 2

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

(2019 Main, 8 April I)

(a) 4.2h

(b) 2.6h

(c) 3.2h

(d) 2.2h

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Answer:

Correct Answer: 2. (b)

Solution:

  1. Considering the initial position of ship A as origin, so the velocity and position of ship will be

vA=(30i^+50j^) and rA=(0i^+0j^)

Now, as given in the question, velocity and position of ship B will be, vB=10i^ and rB=(80i^+150j^) Hence, the given situation can be represented graphically as

After time t, coordinates of ships A and B are

(8010t,150) and (30t,50t)

So, distance between A and B after time t is

d=(x2x1)2+(y2y1)2d=(8010t30t)2+(15050t)2d2=(8040t)2+(15050t)2

Distance is minimum when ddt(d2)=0

After differentiating, we get

ddt[(8040t)2+(15050t)2]=02(8040t)(40)+2(15050t)(50)=03200+1600t7500+2500t=04100t=10700t=107004100=2.6h



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