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Homework Help: Question about average acceleration

  1. Jul 1, 2006 #1
    A bird takes 8.5s to fly from position A to Position B. Determine the birds average acceleration.

    Velocity A = 4.4 m/s [31 Degrees S of E]
    Velocity B = 7.8 m/s [25 Degrees N of E]
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  3. Jul 1, 2006 #2


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    ?? Do you know the definition of "average acceleration"??
  4. Jul 1, 2006 #3
    Change in velocity/change in time = 0.4 m/s^2. I'm having a hard time believing that this is all that is expected as an answer, seems too simple. (This if for a grade 12 physics course). The question is accompanied by a diagram showing the birds curved flight path.
  5. Jul 1, 2006 #4


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    That's the definition except that you are forgetting that velocity is a vector .So the average acceleration is given by
    [tex] { {\vec v_f} - {\vec v_i} \over \Delta t} [/tex].
    You need to do a vector subtraction and divide by delta t. Your answer will be a vector! (the average acceleration is a vector, as is the instantaneous acceleration)

  6. Jul 1, 2006 #5
    I am totally lost on this one. I don't know how to even start to do a vector subtraction with this one. I think my confusion stems from the fact that the bird is flying slightly upward on a curved path. The other examples in the book are nothing like this one. I just started this course, getting started is always the hardest part. Hoping someone can help.
  7. Jul 1, 2006 #6


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    Your first error was thinking of the velocities as "positions".

    a vector has 3 components for example: (East, North, Up).
    Each component is added or subtracted *separately* from the others
    using regular algebra (with signs ... South is NEGATIVE North!).
    Just keep each resulting component separate, in a list.

    Do you know enough "triangle trig" to figure out how much of A is East?

    Each component is divided by a scalar (like time) SEPARATELY.
    Just keep the resulting components separate, in a list.
  8. Jul 1, 2006 #7


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    lightgrav already helped out a lot..

    Just to add my two cents: you do NOT need to worry about the path of the bird. You just need to know the initial velocity and the final velocity and these you gave in your first post. Just find the x and y components of both velocities and subtract them separately.

    To be more precise
    [tex] A_{average,x} = { {v_{f,x} - v_{i,x} \over \Delta t} [/tex]
    and the same for the y component.
  9. Jul 9, 2006 #8
    Thank for the help. If anybody is ambitious enough to work this one out, can you confirm if the correct answer is:

    0.41 m/s^2 [17 Degrees N of E]

    (I arrived at this using for B, v(X) = 7.8cos25, v(y) = 7.8sin 25. For A, v(x) = 4.4cos31, v(y) = 4.4sin31. Then, a(x) = (7.1-3.8)/8.5 = 0.39, and a(y) = (3.3-2.3)/8.5 = 0.12. Then I found the tangent, to arrive at 17 Degrees)
  10. Jul 10, 2006 #9
  11. Jul 10, 2006 #10


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    Watch out. For A the angle is given as being *south of east*. That affects how you calculate the x and y components. Do you see how?
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