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Homework Help: River crossing

  1. Feb 26, 2010 #1
    Problem: a boat has to cross a river that is 1.5 km wide. There is a current of 5 km/h flowing parallel to the banks of the river. If the boat travels at 12 km/h relative to the water, what is the minimum time it will take the boat to cross the river?

    The fastest time will be if all of the boats speed it directed across the river, perpendicular to the current. Since horizontal motion does not affect vertical motion, it will take 1.5/12 hours (.125 hours) or 7.5 minutes to cross the river.

    My teacher says that this approach and the answer it yields is incorrect, but was unable to satisfactorly explain why.

    Her method: the resultant velocity will be 13 m/s at about 22 degrees. 1.5/cos(22) gives the distance that the boat will travel on it's journey across the river. (here is the part thst doesn't make sense to me) since the boat travels at 12 m/s relative to the water (Didnt we just agree that it would travel at 13 m/s?), it will take 1.5/cos(22)/12 hours, which equals .135 hours (8.1 minutes).
     
  2. jcsd
  3. Feb 27, 2010 #2
    You seem right. But you say velocity is w.r.t water- still or moving? Generally the water is assumed still. Nonethless your teacher's method looks wrong to me. To add the velocity of the boat with that of the stream is to consider them as measured from one frame. Then to argue that the velocity of the boat is w.r.t water is inverted reasoning.
     
  4. Feb 27, 2010 #3

    tiny-tim

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    Hi ƒ(x)! :smile:

    Unless there are other parts to this question, your method is quickest and therefore best, but both give the right answer.

    Your teacher correctly finds a speed of 13 at an angle of cos-112/13, so the time taken is the same as if the speed was 12 straight across (ie, your figure).
     
  5. Feb 27, 2010 #4
    Asked someone else and now i know why

    the 12 m/s doesn't belong on one of the legs of the triangle, but on the hypotenuse.

    So its horizontal speed is the square root of 119. Which gets a fastest time of .137 seconds. Apparently when they said relative to the water, they meant the current... :uhh:
     
  6. Feb 27, 2010 #5

    tiny-tim

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    Hi ƒ(x)! :smile:

    (relative to the water, and relative to the current, mean the same thing :confused:)

    No, that's wrong.

    If the boat is pointing at an angle θ to the current, then we draw a vector triangle of the velocities: 5 along the current, followed by 12 at an angle θ, making the third side (length V at an angle φ say) representing the resultant velocity (relative to the ground).

    Obviously, 12sinθ = Vsinφ = 1500.

    The time taken is distance/speed = (1500/sinφ)/V = 1500/12sinθ, which is a minimum at sinθ = 1.
     
  7. Feb 27, 2010 #6
    Can I have a diagram?

    I seem to be getting a lot of problems where I disagree with my teacher, post it here, and then have you guys agree with me...
     
  8. Feb 28, 2010 #7
    Hello everyone!
    First pertaining to tiny tim's first response. Of course f(x)'s teacher was right when she said that velocity is 13m/s at angle acos(12/13). But we should divide distance along this direction by the correct speed along this direction which is 13m/s and not 12m/s.
    Now pertaining to the problem itself. Is it wrong to simply think that time would be minimum if relative velocity of swimmer is always perendicular to the bank,something reinforced by our senses(every velocity is relative)?
     
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