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Distance between ships.

  1. Oct 26, 2012 #1
    Hello,

    I'd like to verify whether my solution to the problem as described below is indeed correct.

    Ship A is traveling east at a constant velocity of 30km/h. It notices ship B 15 km away and 30° from the north. Ship B is traveling at a constant velocity. 20 minutes later ship A notices ship B 45° from the north, and 20 more minutes after that ship B is observed 60° from the north. The distance between ship A and B at that point in time ought to be calculated, including the velocity (magnitude and direction) of ship B.

    Proposed solution:

    Supposing ship A first spots ship B from O, (0,0):

    1) (1/3Va)(OB + 1/3Vb) / |(1/3Va)||(OB + 1/3Vb)| = 1/SQRT(2)

    2) (2/3Va)(OB + 2/3Vb) / |(2/3Va)||(OB + 2/3Vb)| = SQRT(3)/2

    [Va, Vb, OB are vectors; 1/SQRT(2) = cos 45°; SQRT(3)/2 = cos 30°]

    I got Vb = (27.56,15.91), |Vb| = 31.82 km/h, distance between ship A and B 40 minutes later = 12.14 km

    Are these equations correct? Is their solution?
     
  2. jcsd
  3. Oct 26, 2012 #2

    berkeman

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    Staff: Mentor

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