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Kinematics / vectors

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle has an initial horizontal velocity of 2.6 m/s and an initial upward velocity of 3.5 m/s. It is then given a horizontal acceleration of 1.3 m/s^2 and a downward acceleration of 1.2 m/s^2.

    What is its speed after 2.2 s?

    What is the direction of its velocity at this time with respect to the horizontal (answer between -180º and +180º)


    2. Relevant equations

    all kinematic equations


    3. The attempt at a solution

    i'm lost on this one
     
  2. jcsd
  3. Jan 10, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    Acceleration is a vector. Velocity is a vector.

    You can treat the horizontal and vertical components separately.

    You know for instance that Final Velocity is equal to Initial velocity + acceleration times time. Vf = Vi + a*t

    Once you determine your components of the resulting Velocity ... combine ... you're done.
     
  4. Jan 10, 2009 #3
    You are going to end up with 2 final velocities, one in the x and one in the y.

    So to find the direction from the positive x axis, take triangles into consideration. You'll have an x component and a y component so you can easily solve for the direction with respect to the horizontal.
     
  5. Jan 10, 2009 #4
    thanks, i solved and got the right answer
     
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