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Homework Help: Boat going across River

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A Boat at Dock A wants to go to Dock B Straight across. Docks are 100m Apart. Vh=4m/s
    A) If Vo1m=6m/s find the direction he must aim boat (theta) to go straight to dock B
    B)Find time it takes to get to dock B
    C)If he aims boat directly north where will he land on the other side
    -------------------Dock B----------------------
    Vh=4m/s~~~~~~~~~~~~~~~~~~~~~~~~ |
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 100m
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
    ~~~~~~~~~~~~Boat ~~~~~~~~~~~~~ |
    -------------------Dock A----------------------



    2. Relevant equations
    A) I truly have no idea
    B)Vf=Vi+at maybe?
    C)Once again, if I knew I would probably have more luck


    3. The attempt at a solution
    I wish I knew enough to attempt it..
    Thank you for any help you can offer! I am more interested in methods rather than answers
     
  2. jcsd
  3. Dec 16, 2009 #2

    rl.bhat

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    Use vector method.
    One vector is Vh from west to east. Other vector is velocity of the boat. The resultant of these two vectors should be along north. two vectors are given. Find the resultant.
     
  4. Dec 16, 2009 #3
    Ok so is that something like 24/100? I multiplied the velocities and divided by distance. I'm sorry I am just having a hard time grasping this concept for some reason
     
  5. Dec 16, 2009 #4

    rl.bhat

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    When the boat starts moving, the stream pushes it towards east. To move straight towards north, he must aim towards north-west. Draw the vectors and find the angle between V and Vh.
     
  6. Dec 16, 2009 #5
    Ok, so then is it something like Tan(4/6) ? I have trouble understanding which to use Cos, Sin, or Tan. I would have a triangle like this http://img121.imageshack.us/img121/341/86596525.jpg [Broken] right?
     
    Last edited by a moderator: May 4, 2017
  7. Dec 16, 2009 #6

    rl.bhat

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    Your diagram is correct. In the given problem, what is 6 m/s?
     
  8. Dec 16, 2009 #7
    the initial velocity of the boat aiming north towards dock b. thank you for your patience
     
  9. Dec 16, 2009 #8

    rl.bhat

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    No. It can't be. If it so, why they ask the angle?
    It must be the velocity of the boat in still water.
    So in the diagram, diagonal should be 6 m/s.
    Find the resultant velocity of the boat towards north.
     
  10. Dec 16, 2009 #9
    Pythag. Theor. 6^2+4^2=C^2 = 7.2 north? and then use arcsin6/4 to find the angle? Like i said I am so confused on how to tell what to use and when (as far as sin cos and tan goes) to use them so any quick tip would be much appreciated. I know SohCahToa, but in this case I just don't understand. I know we are looking for the North vector which is y, which would be sin i believe, but I'm not too sure.
     
  11. Dec 16, 2009 #10

    rl.bhat

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    According to Pythag. Theor. it should be 4^2 + Vn^2 = 6^2.
    Your angle is correct.
     
  12. Dec 16, 2009 #11
    so then it would be 4.5, and the angle is 41.81. I know I said it but could you give me a little tip or something on figuring out when to use sin and all that, because I sorta just guessed because we were looking for Y and sin is usually going with Y. So then to find time i would just
    divide 100/4.5 ?
     
  13. Dec 16, 2009 #12

    rl.bhat

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    Depending on the given quantities, you have to select either sin or cos. In the given problem, opposite side and hypotenuse with respect to the required angle is given. So you have to use sin to find the angle.
    Your time is correct.
     
  14. Dec 16, 2009 #13
    Thank you, you have been a huge help! I understand it now, it really isn't a difficult question at all.
     
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