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Velocity of Particle

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    The velocity v of a particle moving in the xy plane is given by v= (5.9 t - 4.1 t2)i + 8.7j, with v in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.7 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?


    2. Relevant equations



    3. The attempt at a solution
    Part A is -24.44 i + 0 j m/s2

    Part B is .7195 seconds.

    How do I find part C?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2009 #2

    rock.freak667

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    if velocity is v=xi+yj

    then |v| =speed = √(x2+y2)
     
  4. Sep 17, 2009 #3
    That makes sense, but how do I set that equation up. I am confused with the "t" being in the and the "t^2"?
     
  5. Sep 17, 2009 #4

    rock.freak667

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    the 'y' term is a constant, so you can square both sides of the equation and then make the 'x2' the subject and then take the square root of both sides, then solve. (I hope you understood what I meant)
     
  6. Sep 17, 2009 #5
    No, I'm sorry, It's just not making sense to me. So (5.9t-4.1t^2)^2??
     
  7. Sep 18, 2009 #6

    rock.freak667

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    ok we'd get


    [tex]10=\sqrt{(5.9t-4.1t^2)^2 +(8.7)^2}[/tex]


    so if you square both sides you get rid of the square root sign. Then rearrange and make (5.9t-4.1t^2)2 the subject and take the square root of both sides now.
     
  8. Sep 18, 2009 #7
    would I get sometime like 1.3= t(5.9-4.1t)??
     
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