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Two ships intersecting

  1. Oct 29, 2004 #1
    Hi, I have this question which Im having trouble with

    A ship is steaming parallel to a straight coastline, distance D offshore, at speed [tex]v[/tex]. A coastguard cutter, whose speed is u (u<v) seta out from port to intercept the ship. Show that the cutter must start out before the ship passes a point a distance [tex]\displaystyle{D\frac{\sqrt{v^2 - u^2}}{u}}[/tex] back along the coast.

    Im looking for a hint of how to go about this.

    [​IMG]Heres a picture

    Ive tried breaking the u vector into its horizontal and vertical components but that isnt getting me anywhere. Any ideas?

  2. jcsd
  3. Oct 29, 2004 #2
    Welcome to PF LondonLady!

    If the cutter starts out after the ship passes the mentioned distance, the coastguard will be unable to intercept the ship. Breaking the velocity components is just one of the steps. What have you done?
  4. Oct 30, 2004 #3
    Hi, thankyou! :smile:

    Em, well first i designated the value of the distance that v starts behind u to be 'x'

    So then I broke the u vector into [tex]u\cos \theta i + u\sin \theta j[/tex].

    At the time of intersection 't' the two ships will be in the same place so i evaluated the x - y displacement at that time and got...

    [tex]vt - x = ut\cos \theta[/tex]


    [tex]ut\sin \theta = D[/tex]

    I eliminated t to get

    [tex]\displaystyle{x = \frac{D(u\cos \theta - v)}{u \sin \theta}}[/tex]

    My problem is, I don't know if this right, and if it is, how i would go about getting rid of [tex]\theta[/tex]. Maybe im missing some obvious relation involving [tex]\theta[/tex] :confused:
  5. Oct 31, 2004 #4
    That's good; now you want x to be a minimum to get that particular point so take the derivative - remember at a minimum the derivitave = 0. A little trigonometry and the angles dissappear.
  6. Oct 31, 2004 #5
    Hello, thankyou for your reply :smile:


    I started with [tex]\displaystyle{x = D\frac{u\cos \theta - v}{u \sin \theta}}[/tex]

    and differentiated. I found that at the minimum of x

    [tex]\displaystyle{\cos \theta = \frac{u}{v}}[/tex]

    which would mean (after drawing a triangle) that

    [tex]\displaystyle{\sin \theta = \frac{\sqrt{v^2 - u^2}}{u}}}[/tex]

    When I plug these back in at the top I get

    [tex]\displaystyle{x = D\frac{u^2 - v^2}{u\sqrt{v^2 - u^2}}}[/tex]

    which is wrong! :surprised

    What me doing wrong?
  7. Nov 1, 2004 #6
    I got it!

    Thanks all
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