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A particle leaves the origin (motion in 2 dimensions)

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle leaves the origin with an initial velocity v = 3.92i , in m/s. It experiences a constant acceleration
    a = -1.00i -0.80j , in m/s2. What is the velocity of the particle when it reaches its maximum x coordinate?

    i-component of velocity?

    j-component of the velocity?

    When does it reach its maximum x coordinate?

    Where is the particle at this time? i-component of position?

    j-component of position?


    2. Relevant equations
    Trying to figure this out...


    3. The attempt at a solution

    I've been pondering this problem for a while now and I'm stumped. I realize that the object is given an initial velocity in the x direction. Due to its acceleration, it will eventually stop moving altogether. However, I have no clue how the "maximum x" value comes into play, or how I can find it. I'm looking over the kinematics formulas and I can't seem to get them to where I do not have two unknown variables. If anyone could give me a pointer as to how I should approach this problem, I would greatly appreciate it.
     
  2. jcsd
  3. Sep 9, 2008 #2

    Dick

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    Split the motion into x and y components. It reaches maximum x when x'(t)=0. Use equations like s(t)=s0+v0*t+(1/2)*a*t^2.
     
  4. Sep 9, 2008 #3
    Well, no not in this case. the object starts with a velocity to the right, and is under constant acceleration down and to the left, so it will always be moving even when it is furthest to the left.


    Forget the formulae. There's only two you need to know for these kinds of problems: [tex]\vec{a}=\frac{d}{dt}\vec{v}[/tex] and [tex]\vec{v}=\frac{d}{dt}\vec{r}[/tex], where [tex]\vec{r}[/tex] is the object's position vector, [tex]\vec{v}[/tex] is the object's velocity vector, and [tex]\vec{a}[/tex] is the object's acceleration vector. Your object has a constant acceleration vector, an initial velocity vector, and an initial position vector. From there, you can set [tex]\vec{a}[/tex] to a constant and integrate from t = 0 to t to get your velocity vector. Then to find the maximum x value, set the velocity in the x direction to 0 and solve for t.
     
  5. Sep 9, 2008 #4
    Thank you both, I will try and use your advice to try and solve the problem. I don't understand the integration though...what exactly would I integrate to get the velocity? Velocity is the derivative of position with respect to time, so would the velocity be the integral of acceleration with respect to time? If so, would I integrate the expression a = -1.00i -0.80j?

    Thanks so much for the help.
     
  6. Sep 9, 2008 #5

    Dick

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    I don't think you need to integrate anything. The kinematics equations are already integrated. They are for courses that don't use differential equations. Just use them. Just split the motion into components.
     
  7. Sep 9, 2008 #6
    Thanks...I ended up getting the correct answers. Here's what I did:

    I set the first derivative of the equation you gave (which is essentially the basic formula [tex]v=v_0+at[/tex] to zero to find at which time the maximum x would occur. Then I found the x and y velocity using that same formula but working in the dimensions as you suggested. Then I used the position formula in each dimension [tex]r=v_0t+\frac{1}{2}at^2[/tex] to find the position components.

    Thanks again for the help!
     
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