Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 months ago

A pedestrian walks east at a rate of 6 km/hr. The wind pushes him northwest at a rate of 13 km/hr. Find the direction and magnitude?

of resultant vector.

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  • 8 months ago
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    It's unlikely that the wind would have much effect, due to friction between the pedestrian and the surface (e.g. sidewalk), and that friction is generally necessary for the pedestrian to walk in the first place. However, if we imagine that this is an airplane, not a pedestrian:

    c^2 = a^2 + b^2 - 2ab*cos(C)

    c^2 = 6^2 + 13^2 - 2*6*13*cos(45 degrees)

    c^2 = 36 + 169 - 156*sqrt(0.5)

    c = sqrt(205 - 156*sqrt(0.5)), note that a distance (e.g. c) can't be negative

    c =~ 9.3982647921954007091260564482919

    sin(B)/b = sin(C)/c

    sin(B)/13 = sin(45 degrees)/sqrt(205 - 156*sqrt(0.5))

    sin(B) = 13*sqrt(0.5)/sqrt(205 - 156*sqrt(0.5))

    B = arcsin(13*sqrt(0.5)/sqrt(205 - 156*sqrt(0.5)))

    Since 90 degrees <= B <= 180 degrees, as the "airplane" will end up travelling approximately northwest:

    B =~ 109.151 degrees

    This means that we start pointing east and rotate approximately 109.151 degrees to the left, so the final direction is about 19.151 degrees west of north.

    Note: Since the wind speed is greater than the "airplane's" speed, the "airplane" shouldn't have "taken off" in the first place! It's too dangerous!

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  • Bryce
    Lv 7
    8 months ago

    6cos0°i + 6sin0°j + 13cos135°i + 13sin135°j

    ≈ -3.19i + 9.19j

    magnitude= √(-3.19² + 9.19²)≈ 9.73 km/hr

    Θ= arctan(9.19/-3.19)≈ -70.9°

    180 - 70.9= 109.1° or N19.1°W (19.1° west of north)

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  • 8 months ago

    The problem is: you can't combine a wind speed with a walking rate as you don't know how that wind speed changes his walking rate and direction. If he is determined, it does not change it at all.

    But I'll do a simple vector addition.

    Walking, Ax = 6, Ay = 0

    Wind, Bx = –13 cos 45 = –13/√2 = –9.19

    By = +9.19

    add the x and y components

    Tx = 6–9.19 = –3.19

    Ty = 9.19

    T = √(3.19² + 9.19²) = 9.74 km/hr

    θ = arctan (9.19/3.19) = 70.9º

    from a drawing, that is angle N of E

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