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Kinematics of particles

  1. Jan 1, 2006 #1
    it is known from t=2 to t=10 the accelareation of a particle is inversely proportional to the cube of time t.when t=2, v=-15 and when t=10 ,v=0.36.knowing that the particle is twice as far from the origin when t=2 and t=10,determine the position of the particle when t=2 and t=10


    the answer is 35.2 and 17.6.
    pls help....thank you...
     
  2. jcsd
  3. Jan 1, 2006 #2

    arildno

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    And your work so far?
     
  4. Jan 1, 2006 #3
    i have try to find the eqn using a=1/(t^3) and integrate to become s but i found the eqn wrong.so i have no idea where to get the eqn
     
  5. Jan 1, 2006 #4

    arildno

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    Inversely proportional in this case means that there exists a constant c, so that [itex]a*t^{3}=c[/itex], where a denotes the acceleration, and t denotes time.
    See if this helps you.
     
  6. Jan 1, 2006 #5
    then i integrate it and get s=c/2t.is this the eqn??
     
  7. Jan 1, 2006 #6
    no,i still can't find the result
     
  8. Jan 1, 2006 #7
    I think you're ignoring your velocity information.
     
  9. Jan 2, 2006 #8
    may i know which eqn to use??
     
  10. Jan 2, 2006 #9
    I think you were on the right track, but you have to remember that each time you integrate there is an added constant (+ C).
     
  11. Jan 2, 2006 #10
    however,i still can't solve the problem.......
    is there anybody who manage to solve this problem pls.....
     
  12. Jan 2, 2006 #11
    [tex]a = kt^{-3} = \ddot{x} = \frac{dv}{dt}[/tex]

    Integrate this to get

    [tex]v = \int^{t} kT^{-3}dT + C = -k\frac{t^{-2}}{2} + C[/tex]

    Use your two initial conditions to work out the values of k and C. Integrate again to give you an expression for x, and use the condition you're given on the distances to work out what the constant of integration for the distance integral is. Then you have your equation for the position in terms of time.
     
  13. Jan 2, 2006 #12
    i have got the eqn for the x but what is the meaning of (knowing that the particle is twice as far from the origin when t=2)??
     
  14. Jan 2, 2006 #13
    So you can figure out what the constant is.

    What do you have at this point?
     
  15. Jan 3, 2006 #14
    i can't figure out because i don't know what is the meaning of the sentence....can u pls tell me
     
  16. Jan 3, 2006 #15

    Doc Al

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    I assume this means: at time t=2 the particle is twice as far from the origin as it is at time t=10.
     
  17. Jan 3, 2006 #16
    Derive that
    x(t)=64/t+t+6/5
     
  18. Jan 3, 2006 #17
    The sentence means x(2) = 2*x(10). This will lead you to the answer balakrishnan_v has posted.
     
  19. Jan 3, 2006 #18
    twice as far from the origin as it is at time t=10.what does it means??
     
  20. Jan 3, 2006 #19
    then,for part b)the total distance travelled by the particle from t=2s to t=10s??plsssss

    the answer is 18.4
     
  21. Jan 3, 2006 #20
    pls help...........
     
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