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Homework Help: Hi i need some help

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    the parametric equations of the motion are x=(t)^2m and y=(6t-1.5t^2)m i need to find the
    velocity at the time moment of t=2s

    2. Relevant equations



    3. The attempt at a solution
    so this are the four solutions v=o,v=2m/s,v=4m/s,6m/s .
     
  2. jcsd
  3. Feb 19, 2010 #2
    Do you know how to take a derivative?
    dx/dt = 2t m/s
    dy/dt= 6-3t m/s

    At t=2
    dx/dt=4 m/s
    dy/dt=0 m/s

    v= 4 m/s in the x direction.
     
  4. Feb 19, 2010 #3
    No i dont now how if you could post it how to do also the derivate then it would be ok and thanks for this post
     
  5. Feb 21, 2010 #4

    Astronuc

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    Staff Emeritus
    Science Advisor

    See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/vel2.html

    One has two position parameters x(t) and y(t).

    The velocity, or rather speed, in each direction is just the first derivative with respect to time,

    vx(t) = dx(t)/dt, and vy(t) = dy(t)/dt.

    http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html
    http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html#c3

    Then since velocity is a vectors,

    v(t) = vx(t) i + vy(t) j, where i and j are just the unit vectors in x and y directions.

    The magnitude of v(t) is given by the square root of the sum of the squares of the speeds in both direction, i.e.,

    |v(t)| = sqrt (vx2(t) + vy2(t))

    See also - http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html#vsc1
     
  6. Feb 21, 2010 #5
    Thanks very much sir astronuc
     
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