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The acceleration of a particle on a horizontal xy plane

  1. Oct 11, 2006 #1
    The acceleration of a particle on a horizontal xy plane is given by a=3ti + 4tj, where a is in meters per second-squared and t is in sec. At t=0, the particle has the position vector r=(20.0m)i + (40.0m)j and the velocity vector
    v=(5.00m/s)i +(2.00m/s)j. At t=4.00 s, what are its position vector in unit-vector notation and the angle between its direction of travel and the positive direction of the x axis?


    I have no idea what to do at all:surprised https://www.physicsforums.com/images/smilies/surprised.gif
    :surprised
     
  2. jcsd
  3. Oct 11, 2006 #2
    Find the general equations for position and velocity by integration and use the initial conditions.

    "the angle between its direction of travel and the positive direction of the x axis"

    For that you can apply the dot product.
     
  4. Oct 11, 2006 #3
    so would i integrate a to find position and then velocity?
     
  5. Oct 11, 2006 #4
    You integrate accelation to find velovity, and then integrate velocity (after plugging in the initial condition) to find position.
     
  6. Oct 11, 2006 #5
    alright so i got:

    dx = 1.5t^2 + vcos0t + 20
    dy = 2t^2 + vsin0t + 40

    where the 0 is theta
     
  7. Oct 11, 2006 #6
    Bringing in new variables (such as theta) without any need will just make the answer more complicated.

    Here's an example on how to go about the problem:

    [tex]\frac{dv_x}{dt} = 3t[/tex]

    [tex]v_x = \frac{3}{2}t^2 + C[/tex] (C is the constant of integration)

    Now put in the initial condition, the value for v_x when t = 0 and find the value for C. Similaraly find v_y. The process is the same for finding position.
     
    Last edited: Oct 11, 2006
  8. Oct 11, 2006 #7

    radou

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    You can be a bit more formal and use [tex]\vec{v}(t) = \int \vec{a}(t)dt + \vec{v}_{0}[/tex], i.e. [tex]\vec{r}(t) = \int \vec{v}(t)dt + \vec{r}_{0}[/tex], but basically that's the same thing Neutrino wrote.
     
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