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A particle moving in parabola (Kinematics)

  1. Jul 2, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle is moving along a parabola y = x2 so that at any time vx = 3 ms-1. Calculate the magnitude and direction of velocity and acceleration of the particle at the point x = 2/3 m.

    2. Relevant equations

    Kinematics

    3. The attempt at a solution

    vx = 3 (given)
    y = x2 so vy = 2x

    ax = 0
    ay = 2vx

    When x = 2/3, vx = 3 and vy = 4/3
    So v = √{32 + (4/3)2} = 3.3
    Direction tan-1{(4/3)/3} = tan-1(4/6) = 33.7°

    When x = 2/3, ax = 0 and ay = 2 x 3 = 6
    So a = √{02 + 62} = 6
    Direction in +y direction

    Above is how I tried to solve the problem. But then when I checked the answer, I found
    v = 5, direction of velocity tan-1(4/3) or 53°, a = 18 and direction of acceleration +y

    Would someone please help me pointing out my mistakes?
     
  2. jcsd
  3. Jul 2, 2014 #2

    BiGyElLoWhAt

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    when x = 2/3, vx =3 and vy/vx = 4/3
     
  4. Jul 2, 2014 #3

    BiGyElLoWhAt

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    derivative is the slope, not the y value.
     
  5. Jul 2, 2014 #4
    Oh I should have written vy = 2xvx
    That was a super silly mistake. :shy:
    Thanks!
     
  6. Jul 2, 2014 #5

    BiGyElLoWhAt

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    No problemo XD
     
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