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An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s2. If the acceleration from the wind lasts for 2.70 s, find the magnitude, r, and direction, θ, of the bird\'s displacement during this time period. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.)

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Jordan

  1. Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79
    m/s.

    Vw = a*t = 0.30 * 2.7 = 0.81 m/s.

    Vr. = sqrt(Vb^2+Vw^2) = 1.96 m/s. = Resultant velocity.
    D = Vr*t = 1.96*2.7 = 5.29 m.

    Tan A = Y/X = Vw/Vb = 0.81/1.79=0.45251
    A = 24.3o = Direction.

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    Henry

  2. Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79
    m/s

    To solve the problem, you need to create a formula with a position as a function of time for the bird(x) and the wind(y), and then use R = sqrt(Rx^2 + Ry^2) to find the resultant magnitude and the inverse tangent of y/x

    R = Ri(initial) + Vi*t + 1/2(Ri)t^2

    Rx = 0 + 1.79(2.7) + 1/2(0)(2.7)^2 = 4.833

    Vi for the wind is 0 because it is with respect to the bird for the y coordinate
    Ry = 0 + 0(3.7) + 1/2(0.3)(3.7)^2 = 2.054

    R= sqrt(4.833^2 + 2.054^2) = 5.251 meters

    A = arctan(2.054/4.833) = 23.025 degrees

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    Octavio

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