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Vector components of velocity

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle starts at t=0 at the origin of a coordinate system with a velocity of 1.32m/s in the negative y direction. There is acceleration on the particle of 2.47m/s^2 in a direction 35 degrees above the x axis.

    a) at what time does the y component of the particle's position equal 3.38
    b)At that time, what is the x component of position
    c)at that time, what is the x component of velocity
    d)at that time, what is the y component of velocity


    For a)........

    I used Rf=Ri+Vit+1/2at^2

    3.38=0+-1.32(t)+1/2(2.47)t^2
    My final answer is t=2.27 Is this correct

    For b)........

    I used Xf=Xi+vxiT+1/2at^2

    Xf=0+-1.32(2.27)+1/2(2.47)(2.27)^2
    My final answer is 3.37 Is this correct
     
    Last edited by a moderator: Oct 19, 2009
  2. jcsd
  3. Oct 19, 2009 #2
    ok, for c)....
    Vx=4.29cos(35)=3.51


    For d)
    Vy=4.29sin(35)=2.46

    Is this the right way to complete the steps for this problem
     
  4. Oct 19, 2009 #3
    or.....

    Do I take the derivative of the position equation to find velocity?
     
  5. Oct 19, 2009 #4

    Redbelly98

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    Oops. I have accidentally edited part of your post #1. My apologies! :blushing:

    No.
    Since this equation is for y, what is the y-component of the acceleration?

    No.
    This equation is for x. So, what are the:
    • x-component of the initial velocity, and
    • x-component of the acceleration?
     
  6. Oct 19, 2009 #5

    Redbelly98

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    I don't understand this; where does the 4.29 come from?

    You could do that, or, equivalently, just use the kinematic equation for velocity vs. time.
     
  7. Oct 20, 2009 #6
    Here are my components.....
    Vyi=-1.32 Vxi=0

    ax=2.47(cos(35))=2.02 ay=2.47(sin(35))=1.42


    a) 3.38=0+-1.32(t)+.5(1.42)t^2= .71t^2-1.32t-3.38 t=3.30


    b) Xf=0+0(3.30)+.5(2.02)(3.30)^2=11.0

    c)v=0+2.02(3.30) =6.67

    d)v=-1.32+1.42(3.30)=3.37

    Does that look better?
     
  8. Oct 21, 2009 #7

    Redbelly98

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    Looks good! :smile:
     
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