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Find the time of a certain velocity

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A time-dependent force, F = 8i – 4tj N, is exerted on a 2.00 kg object initially at rest. a) At what time will the object be moving with a speed of 15 m/s? b) How far is the object from its initial position when its speed is 15 m/s? c) Through what total displacement has the object traveled at this moment?

    2. Relevant equations
    F = ma
    vf = v0 + at


    3. The attempt at a solution

    I took the mag of the F which is sq(80) = ma so: 8.9 = 2a, therefore a should be 4.47.
    Then I used vf = vo + at. It is at rest so vo = 0 so: 15 = 4.47t, t = 3.35 and it is incorrect. Can someone please tell me what I did wrong? thanks.
     
  2. jcsd
  3. Mar 7, 2009 #2

    Doc Al

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    Since the force, and thus the acceleration, is time-dependent, you can't use formulas that only apply for constant acceleration. Hint: Use a bit of calculus.
     
  4. Mar 7, 2009 #3
    What I did was integrate the force vector to get <8t, -2t^2> = 2a but I still have two unknowns.
     
  5. Mar 7, 2009 #4

    Doc Al

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    (You mean = 2v.) The only unknown is the time. (Find the magnitude of that velocity vector.)
     
  6. Mar 7, 2009 #5
    Ok tell me if this will work: F = ma so a = F/m so a = <4, -2t>. v = <4t, -t^2> but speed is the mag of velocity so 15 = sq(16t^2 + t^4). So 225 = 16t^2 + t^4 which gives me four roots one of them being 3. Is this good?
     
  7. Mar 7, 2009 #6

    Doc Al

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    Perfecto!
     
  8. Mar 7, 2009 #7
    Great thanks. But now I am having a bit of trouble with the position. It is asking me for the vector here so I split it up into components.
    For x I get xf = xi + vxit + 1/2(ax)t^2 which is xf = 0 + 4t(t) + (1/2)4t^2 at t = 3 gives 54
    For y I get yf = yi +vyit + 1/2(ay)t^2 which is yf = 0 + (-t^2)t + (1/2)(-2t)t^2 at t = 3 is -54. They give a number for the answer so I should mag the vector and I get 76.4 but it is incorrect. Please?
     
  9. Mar 7, 2009 #8

    Doc Al

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    Same issue as before: Don't try to apply constant-acceleration formulas when the acceleration is not constant. Same advice as before: Use calculus.
     
  10. Mar 7, 2009 #9
    But I have already done the calculus. I am just using the v and a that I had before and just plugging them in. I really don't understand why this doesnt work.
     
  11. Mar 7, 2009 #10

    Doc Al

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    You used calculus to find the velocity; now use it to find the displacement.
    Because the acceleration is not a constant over the interval from t = 0 to t = 3 seconds. So those kinematic formulas do not apply.
     
  12. Mar 7, 2009 #11
    I do not know any other formulas that give me a position. Could you give me a little bigger hint?
     
  13. Mar 7, 2009 #12

    Doc Al

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    Sure. Just like a = dv/dt, v = dx/dt.
     
  14. Mar 7, 2009 #13
    Thanks. Just curious. Since the kinematics do not apply, what would we do if we were given an initial velocity and an initial position?
     
  15. Mar 7, 2009 #14

    Doc Al

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    Whenever you integrate, you'll have constants of integration. The initial conditions will determine those constants.

    For example, since the object is initially at rest, you know that at t = 0, v = 0. And since all we care about are displacements from the initial position, we can say that at t = 0, the object is at x = y = 0.
     
  16. Oct 26, 2011 #15
    For the first part how do you get 225 = 16t^2 + t^4?
     
  17. Oct 27, 2011 #16

    Doc Al

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    See post #5. To go from a to v, integrate.
     
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