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Function of Time

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data

    A bird's flight is given by the following function:
    v = (At^2 - B) i + (Ct) j

    1)If the bird is traveling due north at 3 m/s at t= 6 seconds, find 'C'.
    2) The bird accelerates at (-2 m/s^2)i + (.5 m/s^2)j at t = 4s. Find 'A' and 'B'.
    2. Relevant equations

    ?

    3. The attempt at a solution

    I really don't know what to do. If I just solve part 1 algebraically, I get .5 m/s^2. That can't be all there is to it, because then how do I solve part 2?
     
  2. jcsd
  3. Sep 20, 2008 #2

    olgranpappy

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    what do you mean "solve part 1 algebraically"? what exactly did you do? show the work and it will be easier for us to help you.

     
  4. Sep 20, 2008 #3

    olgranpappy

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    ...for example, you probably used C*6=3... but what else does "due north" tell you?
     
    Last edited: Sep 20, 2008
  5. Sep 20, 2008 #4
    Sorry about that. You were right. All I did was set Ct = 3.00 m/s

    C ( 6 seconds) = 3 m/s

    C = (3 m/s) / (6 s)

    Do I need to break this into vectors? Like (3sin 90) i ? Where does t come in?
     
  6. Sep 20, 2008 #5

    olgranpappy

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    Yes, that's quite the right thing to do... but also there is something else that the statement "due north" tells you.

    It tells you that at t=6 the coefficient of the 'i' term is zero.

    Just like at t=6 the coefficient of the 'j' term is 3.

    So, again, at t=6 the whole coefficient of the 'i' term (the thing in parenthesis next to the 'i') equals zero. Write this equation down. Once you tell us this equation we will move on to the next step.
     
  7. Sep 20, 2008 #6
    A(6^2) - b = 0
     
  8. Sep 20, 2008 #7

    olgranpappy

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    Right, again. This gives us an equation for B in terms of A. B=36A.

    Next we move on to exploit the information given about the acceleration.

    At any time (t) we know that the x-velocity is
    [tex]
    v_x=(At^2-B)
    [/tex]

    The x-acceleration is the derivative with respect to time of the x-velocity
    ...what is the value of this derivative?
     
  9. Sep 20, 2008 #8
  10. Sep 20, 2008 #9

    olgranpappy

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    Yes,
    [tex]
    a_x=2At
    [/tex]

    ...then from "2)" you know that at t=4 the x-acceleration is -2

    Use this in the above equation to find A.
     
  11. Sep 20, 2008 #10
    I understand. What happens for the value of B?
     
  12. Sep 20, 2008 #11

    olgranpappy

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    Well, now you know A... and you have an equation for B in terms of A, remember:

    B=36A
     
  13. Sep 20, 2008 #12
    Oh! Of course. I think I've been sitting at the computer too long. You have been a great help. Thanks a lot!
     
  14. Sep 20, 2008 #13

    olgranpappy

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    You're welcome. Cheers.
     
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