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Thread: relative velocity angle

  1. #1
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    relative velocity angle

    relative velocity angle-win_20170320_20_58_23_pro.jpgClick image for larger version. 

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    question 1.

    part i distance traveled by man d=s*t .6*20=12 distance traveled by woman =.4*20=8 i cant do part ii. ive tried setting up a triangle with there distances i just worked out and you are given the angle tan^-1 3/4 and i have tried the sine rule to work it out but the right answer is 60 degrees and i am not getting it. previous similar worked examples were solved using the sine rule
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    Re: relative velocity angle

    The distances you have calculated are correct but they are not the sides of a triangle.

    Redraw your diagram, so that the woman ends up due south of the man. Your diagram will consist of 2 triangles. Then try again.

    Come back if you get stuck again.
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  3. #3
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    Re: relative velocity angle

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    Re: relative velocity angle

    Do you know if they are the exact sizes of the sides triangles. 10/sinx = 8/sin(tan^-1 3/4). Does not work
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    Re: relative velocity angle

    No it doesn't work because the triangle(s) you need to work with does not include 8 and 10 as sides.
    relative velocity angle-capture.jpg

    Angle ACB = arctan(3/4)

    CA = 10 (distance apart at start)

    Using basic trig (or recognising a 3:4:5 = 6:8:10 right angled triangle). AB=6 and BC=8

    CP = 12 (distance swum by man) and AR = 8 (distance swum by woman)

    Now concentrate on triangles ABQ and RPQ (which are similar).

    relative velocity angle-capture1.jpg

    See if that helps.
    Last edited by Debsta; Mar 21st 2017 at 12:51 AM.
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    Re: relative velocity angle

    man swimmer starts at (0,0) and swims due W at 0.6 m/s ...

    $x_1(t)=-0.6t$

    $y_1(t)=0$

    woman swimmer starts at (-8,6) and swims at an angle $\alpha$ S of W at 0.4 m/s ...

    $x_2(t) = -8 -0.4\cos{\alpha} \cdot t$

    $y_2(t) = 6 - 0.4\sin{\alpha} \cdot t$

    at $t=20$ sec, the woman swimmer is due S of the man swimmer ...

    $x_2(20)=x_1(20)$

    $-8-8\cos{\alpha}=-12 \implies \cos{\alpha}=\dfrac{1}{2} \implies \alpha = 60^\circ$



    for part (iii) ...

    $d=\sqrt{[x_2(t)-x_1(t)]^2 + [y_2(t)-y_1(t)]^2}$

    expand and simplify the resulting time quadratic under the radical and find the time for minimum distance by either calculus or non-calculus means.
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    Re: relative velocity angle

    Ok thanks. But is there not any way to work it out with the triangle's
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    Re: relative velocity angle

    There is no way to tell distance AQ or PQ. All i know is the size of AB
    Last edited by markosheehan; Mar 21st 2017 at 05:15 AM.
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