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    • Thread Starter

    Please can anyone help me with Ex 1A Q 10 in the Heinemann M4 book? I've tried drawing two velocity triangles but I can't get the direction of the wind to match up.

    In fact does anyone have a general method for solving those sorts of problems where you have to find the velocity of the wind given two different situations? Should you try and combine the two velocity triangles?


    Let the wind have components Wx horizontally to the right and Wy vertically upwards.

    The horizontal and vertical components for the first boat are:
    horiz: Wx - (-12) = Wx + 12 (since this boat is travelling the other direction)
    vert: Wy (no motion here)

    Since the wind appears to be blowing from 160, then:
    (Wx + 12)cos160 = (Wy)sin160
    Wx + 12 = (Wy)tan160

    Similarly, for the second boat:
    horiz: Wx - 15 (since they're travelling in the same direction)
    vert: Wy (no vertical motion)

    Wx - 15 = (Wy)tan120

    Solving these two equations simultaneously should get you:
    Wx = -19.2
    Wy = 19.7

    |W| = sqrt[Wx² + Wy²] = 27.5 m/s, this is the true speed of the wind.

    Now we want to find its direction. Draw the line represing W (it should have -Wx and -Wy as its components, because we want where the wind is from). Then the angle W makes with the positive y-axis is:
    180 - arctan(19.2/19.7) = 136 degrees

    So the wind is blowing with speed 27.5 m/s from 136 deg.
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Updated: June 26, 2005
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