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OL Exam Question on Relative velocity

  1. Oct 23, 2012 #1
    correction -->it should be a high-level exam question.
    1. The problem statement, all variables and given/known data

    Two ships A and B move with constant speeds 48 km/h and 60 km/h respectively. At a certain instant ship B is 30 km west of A and is travelling due south. Find
    (i) the direction A should steer in order to get as close as possible to ship B
    (ii) the shortest distance between the ships.

    the answer:
    Diagram 1: Vab = Va -Vb

    Vab = (-48cosAi - 48sinAj) - (-60j)

    = (-48cosA)i + (60-48sinA)j

    tanB = j over i = 60-48sinA over 48cosA
    tanB = 5-4sinA over 4cosA

    then differentiate tan B with respect to A and make it equal to 0

    You get 16cos^2 A - 20sinA + 16sin^2 A

    which simplifies to sin A = 4/5 , (and tanB = 3/4)

    Then part 2: (from diagram 2)

    |BX| = 30sinB
    = 30(0.6)
    = 18km

    attachment.php?attachmentid=57908&d=1213476181.gif

    Why I need to differentiate tanB equal to 0? I just dont understand..how can finding the maxium value
    of tanB help me to find the shortest distance?
     
    Last edited: Oct 23, 2012
  2. jcsd
  3. Oct 23, 2012 #2

    haruspex

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    It's the min of tan B, not the max.
    The minimum distance is 30 sin B, so want the value of A that minimises sin B.
    d sin(B)/dA = d sin(B)/d tan(B) * d tan(B) /dA (chain rule), so for d sin(B)/dA = 0 we want either d sin(B)/d tan(B) = 0 or d tan(B) /dA = 0. The first never happens.
     
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