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Nearest Distance between 2 boats

  1. Jan 20, 2009 #1
    Ship A is 10 km due west of ship B. Ship A is heading directly north at a speed of 30km/hr, whilst ship B is heading in a direction 60 degress west of norht at a speed of 20km/hr.

    a) Deteremine the magnitude and direction of the velocity of ship B relative to ship A.

    b) What will be their distance of closest approach?

    My Attempt for the first part

    B Relative to A, first into compentents, so sin(60)*20=17.3 and cos(60)*20=10
    Then relative to A
    would be 10 km west and 30-17.3= 12.67 km south

    however the actual answer is Velocity is 26.5 km/h in a direction South 41 degrees West.

    How was i supposed to do it, and how do i start part b.
  2. jcsd
  3. Jan 20, 2009 #2
    For part a), you have simply got the components confused. Sin() gives the 'west' component not 'south', and vice versa. Try drawing the triangle to see your mistake (sin = opp/hyp, etc).

    For part b), now that you have the relative velocity vector you can set up t-dependent vector R giving the relative position of the boats. Now the problem is simply a matter of finding the shortest distance between a point and a line. One way to do it is to construct a line which goes through the origin and whose dot-product with R is zero (therefore orthogonal to R). Then evaluate the point of intersection of the two lines and take the magnitude of the separation.
  4. Jan 20, 2009 #3
    Right i see part the mistake, so now i have 20 km/hr south and 17.3km/hr west. Find the hypontuse of this I get 26.4 km/hr, ok i got it. Thank you

    b)i don't see it, by vector r does it mean, r=(10,0)+t(-17.3,-20)
  5. Jan 20, 2009 #4
    Right, R gives the equation of a line in the x-y plane, with direction (-17.3,-20). So a vector perpendicular to the line could be

    L = m(20,-17.3)

    With m a parameter which can take any value (similarly to t).

    We then find the points (x,y) where the two lines cross:

    x = 20m = 10 -17.3t
    y = -17.3m = -20t

    Now solve the simultaneous equations => t=0.25 etc

    A quicker way to do it is to spot that the point we want occurs at

    R (dot) [direction of line] = 0

    i.e. 10*-17.3 + t(-17.3^2 + -20^2) = 0

    => t = 0.25 etc

    A third way to do it is to simply take the modulus of R and minimise it w.r.t. t. Takes a bit more algebra but requires less geometrical thinking.
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