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physics

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A duck has a mass of 2.9 kg. As the duck paddles, a force of 0.07 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.21 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.7 s while the forces are acting. How do you get the direction?

  • physics -

    f = ma,
    0.07 = 2.9a,
    a = 0.07/2.9 = 0.024 m/s^2,
    d1 = Vo*t + 0.5*at^2,
    = 0.12*2.7 + 0.5*0.024*(2.7)^2,
    = 0.3481 + 0.0880 = 0.4361 m., E.

    0.21 = 2.9a,
    a = 0.21 / 2.9 = 0.0724 m/s^2,
    d2 = 0.12*2.7 + 0.5*0.0724*(2.7)^2,
    = 0.3481 + 0.2639 = 0.6120 m.@ 60
    deg S. of E. = 300 CCW.

    RESULTANT(R)

    X = hor = 0.4361 + 0.6120*cos300,
    = 0.4361 + 0.306 = 0.7421 m.

    Y = ver = 0 + 0.6120*sin300 = -0.530m

    tanA = y/x = -0.5300 / 0.7421 = -0.7142
    A = -35.5 deg = 35.5 deg S.of E. = 324.5 deg CCW.

    R = X + iY = 0.7421 - i0.5300

    R = X / cosA = 0.7421 / cos324.5 = 0.9115 m @ 35.5 deg. S.of W.

    DISPLACEMENT(D)

    D = R - Vo*t,
    = (0.7421 - i0.5300) - 0.12*2.7
    = 0.7421 - 0.324 - i5300,
    = 0.4181 - i0.5300,

    tanB = Y / X = -0.5300/0.4181=-1.268,
    B = -51.7 Deg = 308.3 deg CCW.

    Magnitude = x / cosB = 0.4181 / cos308.3 = 0.6751 m.

    Direction = 51.7 deg S. of E.

  • physics -

    there is absolutely no way its that complicated. and at points it looks like you made up some values. be more clear.

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