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Help with question

  1. Jan 30, 2007 #1
    A bird flies in the xy-plane with a velocity vector given by v(vector) = (a-bt^2)(i-vector) + ct (j-vector) , with a= 2.4m/s, b=1.6m/s^3,c=4.0m/s^2. The positive y-direction is vertically upward. At t=0 the bird is at the origin.

    Calculate the position vector of the bird as a function of time.
    Express your answer in terms of a, b, and z. Write the vector r(t) in the form vector(r(t)x,r(t)y , , where the x and y components are separated by a comma.

    Calculate the acceleration vector of the bird as a function of time. also in the form as stated above.

    What is the bird's altitude (y-coordinate) as it flies over x=0 for the first time after t=0?
     
  2. jcsd
  3. Jan 30, 2007 #2

    Pyrrhus

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    I see, good standard questions... :rolleyes:

    Let's see how you are tackling them!
     
  4. Jan 31, 2007 #3
    Well, atiderviative will give me the position vector, but I don't know if I should plug in the values of a, b, etc. prior to me taking the antiderivative. Also I am not sure what form they are talking about.

    For the acceleration the derivative is needed.

    Not sure about the last question
     
  5. Jan 31, 2007 #4

    Pyrrhus

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    a, b, c are constants.

    For the last question look at the position function, what can you notice?
     
  6. Feb 1, 2007 #5
    I notice that x=0, so there is no horizontal motion. What do they mean by birds altitude as it flies over x=0 after t=0, meaning whent t=1. Also so since x=0 I only focus on the y componet of the position vector, which is (ct^2)/2?
     
  7. Feb 4, 2007 #6

    Pyrrhus

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    Look at the acceleration vectorial function, it's x component is negative, so sooner or later the bird will pass the straight line x=0.
     
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