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Particle with horizontal velocity and vertical acceleration

  1. Jun 16, 2013 #1
    1. A particle moving at a velocity of 6.5 m/s in the positive x direction is given an acceleration of 0.9 m/s^2 in the positive y direction for 2.9 s. What is the final speed of the particle?



    2. Relevant equations
    v= vi +at


    3. The attempt at a solution

    v= 6.5 + (.9)(2.9)= 9.11 m/s

    I don't know if the velocity is right. the velocity is horizontal, and acceleration is vertical, so it makes me doubt.
     
  2. jcsd
  3. Jun 16, 2013 #2

    gneill

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    Staff: Mentor

    The horizontal and vertical components of the velocity are independent; They should be treated as vector components when you "add" them. The final speed will be the magnitude of the resultant vector.
     
  4. Jun 16, 2013 #3
    so, i have to use the following, vy = viy +ay t and vx= vix + ax t ??
     
  5. Jun 16, 2013 #4

    UVW

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    That's right. Before, you were using v = vy + axt, which doesn't make sense, since those directions are perpendicular. You'll need to take gneill's advice about what to do from there.
     
  6. Jun 16, 2013 #5
    vy = viy +ay t = viy + (.9)(2.9)

    vx= vix + ax t = (6.5) + ax (2.9)

    how do i get the missing values?
     
  7. Jun 16, 2013 #6

    gneill

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    Staff: Mentor

    Initially the particle had only a velocity in the x-direction. So what was the initial y-direction speed? If the acceleration is only in the y-direction, what's the x-direction acceleration?

    Hint: zero is a perfectly good value for a parameter :smile:
     
  8. Jun 16, 2013 #7
    so basically

    vy = viy +ay t = 0 + (.9)(2.9) and because i have already the velocity of the x direction i can do the v=sqrt( x^2 + y^2) right?
     
  9. Jun 16, 2013 #8

    gneill

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    Staff: Mentor

    Yup. That's right.
     
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