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Math Help - kinematics in 2d problem

  1. #1
    Junior Member
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    kinematics in 2d problem

    Hi, Im having problems with this question:

    A boat heads north and 20kph
    A wind is blowing SE with a constant speed of 2ms^1

    a) Find the magnitude of the speed of the boat
    b) Find the bearing it travels at

    **************

    Im having problems with the triangle, drawing it out makes the 2 ms^1 the hypotenuse when, to find the speed shouldnt be the hypotenuse i.e the hypotenuse has to be found?

    The using pythagoras I get the resultant being 5.2ms^1 which is complete rubbish.

    Please could anyone give me some helpful pointers please?

    Thankyou very much in advance!
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  2. #2
    Super Member Random Variable's Avatar
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    The angle between the velocity vector of the boat (20 km/hr) and the velocity vector of the wind (7.2 km/hr) is 45 degrees.

    The speed of the boat can be found by the law of cosines.

     c^{2} = 20^{2}+7.2^{2} - 2(20)(7.2) \cos (45)

    c = 15.75 km/hr

    The direction of the boat can be found by the law of sines.

     \frac{15.7542}{\sin(45)} = \frac{7.2}{\sin B}

     \sin B = \frac{(7.2)(\frac{\sqrt{2}}{2})}{15.7542}

    B = 18.9 east of north
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  3. #3
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    thankyou

    How did you work out what the angle was though?

    thanks!
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by fishkeeper View Post
    thankyou

    How did you work out what the angle was though?

    thanks!
     B = \arcsin\Big(\frac{(7.2)(\frac{\sqrt{2}}{2})}{15.75  42} \Big) and then I used a calulator (making sure I was in degrees mode)

    and according to the triangle, the vector is to the east of the velocity vector of the boat

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