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Ferry crossing

  1. Feb 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A ferry captain wishes to travel directly across a river. A current of 4km/h is flowing and the ferry can travel at 8km/h.
    a) In what direction should the captain direct the ferry?
    b) What is the resultant velocity of the ferry as seen by someone standing on the riverbank?

    2. Relevant equations

    3. The attempt at a solution
    a) The ferry should be directed to travel 26.6 degrees up stream so that it will travel in a straight line. - same as the answers.

    b) I don’t get. It’s not sqrt(8^2+4^2) because it travels in a straight line now. Would it simply be 8km/h? The answers suggested 6.9m/s.
  2. jcsd
  3. Feb 15, 2007 #2
    to the viewer on the river bank it comes directly at it, so the apparent distance traveled is 8km. But the actual speed toward the river bank is reduced by the angle upstream. I would think that angle would be 30, and
  4. Feb 16, 2007 #3
    I still don't see how you arrived at this answer. Maybe could yo draw a diagram?
  5. Feb 16, 2007 #4
    To overcome the current, is the direction of the ferry, having X ,----I
    a component of 4Km/hr upstream to offset the drift. Since........X----I
    the total velocity is 8Km/hr sin (theta)=4/8. To observer ............X---I
    ferry is bearing directly at them--he doesn't see the angle.............X-,I
    the ferry must make with respect to the shoreline since...................X
    the net velocity upstream is a wash, ie=0. The component
    dircted toward the river is then cos(30)*8. Your approach
    using pythagorans theorum ok, but should be:

    8=sqrt(4^2+Vapperent^2), squaring both sides V^2=64-16
    Last edited: Feb 16, 2007
  6. Feb 16, 2007 #5
    I see, the key idea was that the ship is travelling at an angle of 30 degrees upstream. There is a horizontal component of 4km/h directly upstream and a vertical speed directly crossing the shore. This speed can be calculated by the pythagoras theorem sqrt(64-16)=6.93km/h. Note that the book got a) wrong. The angle upstream should be different to the angle the ship would be tilted had it travelled directly straight. Because in one case, 8km/h was the hypotenus, in the other it was not.
  7. Feb 16, 2007 #6
    Exactly, complain re the text, there is no excuse for these types of errors.
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