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Velocity and position of the particle as a function of time

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle starts from the origin with velocity 5 m/s at t = 0 and moves in the xy plane with a varying acceleration given by a = (9*squareroot*t)j, where "a" is in meters per second squared and t is in seconds.

    Determine the velocity of the particle as a function of time.
    Determine the position of the particle as a function of time

    2. Relevant equations
    I have no idea


    3. The attempt at a solution
    I have no idea
     

    Attached Files:

  2. jcsd
  3. Oct 14, 2011 #2
    Hi!
    So you first might like to recall how does one determine acceleration, velocity and all that:
    Remember that:
    [itex]
    \large
    \frac{d^2\vec{r}}{dt^2} = \vec{a}
    [/itex]
    Split the components;
    The acceleration is given only in the [itex] \hat{j} [/itex] direction, so that you're left with the simple task of integrating
    [itex]
    \frac{dv_y}{dt} = 9\sqrt{t}
    [/itex]
    Try it,
    Daniel
     
  4. Oct 14, 2011 #3
    i tried that, i didnt get the right answer, can u show me your steps?
     
  5. Oct 15, 2011 #4
    Certainly!
    Look here:
    The velocity vector has two components, i, and j; in the i direction, it was initially constant, and with the abscence of acceleration in that direction it will remain so.
    In in the j direction, after integration, we get:
    [itex]
    \large
    v_j = \displaystyle \int 9\sqrt{t}dt = 9\frac{2}{3}t^{\frac{3}{2}}=6t^{\frac{3}{2}}
    [/itex]
    [itex]
    \large
    v_i = 5
    [/itex]
    [itex]
    \large
    v = \sqrt{{v_i}^2 + {v_j}^2} = \sqrt{25+36t^3}
    [/itex]
    And in order to evaluate the position, all you need is a simple exercise in calculus for [itex] r=\displaystyle \int vdt [/itex], again, by integrating each fraction of the velocity separately, and r = sqrt(r_i^2+r_j^2) as before...
    Daniel
     
    Last edited: Oct 15, 2011
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