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Thread: relative velocity

  1. #1
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    relative velocity

    A ship, A is travelling due east at 24 km/h. A second ship, B, travelling at 33 km/h is first sighted 20 km due south of A. From A the ship B appears to be moving north east.
    Find
    (i) the direction in which B is actually moving


    there are a few ways to do this. the easiest is to set up a triangle.

    relative velocity-win_20170317_11_26_08_pro.jpg


    so using the sin rule. 33/sin45 = 24/sinx when i solve for x i get 30.1 degrees. the answer at the back of the book is 14.1 degrees. i thought i was doing everything right
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    Re: relative velocity

    $v_B = v_A + r$


    $\dfrac{\sin(45^\circ-\theta)}{24} = \dfrac{\sin(135^\circ)}{33}$

    $\sin(45^\circ-\theta) = \dfrac{24 \sin(135^\circ)}{33}$

    $45^\circ-\theta = \arcsin\bigg[\dfrac{24 \sin(135^\circ)}{33}\bigg]$

    $\theta = 45^\circ - \arcsin\bigg[\dfrac{24 \sin(135^\circ)}{33}\bigg] \approx 14.1^\circ$

    Thanks from markosheehan
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    Re: relative velocity

    i dont understand your diagram. is ship B not sighted south of ship A, so it would be heading towards it.
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  4. #4
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    Re: relative velocity

    the sketch shows velocity vectors, not position vectors.
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  5. #5
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    Re: relative velocity

    ok and i solved for the wrong angle in the OP thats why i got it wrong
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