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Finding coordinates after finding the acceleration of an object

  1. Sep 15, 2013 #1
    Acceleration and coordinates at time t.

    1. The problem statement, all variables and given/known data
    At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vector v i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 3.60 s, the particle's velocity is vector v = (7.00 i + 3.70 j) m/s. (Use the following as necessary: t. Round your coefficients to two decimal places.)

    a) Find the acceleration of the particle at any time t. Found this a= (4.00i+5.70j)/3.6s
    b) Find its coordinates at any time t. I have no clue how to do this

    2. Relevant equations

    I tried x⃗t=x⃗0+v⃗0t+12a⃗t^2 but webassign says my answer is wrong

    3. The attempt at a solution
    So i used the above equation and plugged in what I am given or know.
    Xo=0
    Vot=(3.00i-2.00j)t
    a=(4.00i+5.70j)/3.60
    soooo
    Xt= 0 + (3.00i-2.00j)t + 1/2[(4.00i+5.70j)/3.60]t^2
    in the end i got
    x=(3.00i)t + (0.56i)t^2
    y=(2.00j)t + (0.79j)t^2

    Can someone help me understand why i got this wrong?
     
    Last edited: Sep 15, 2013
  2. jcsd
  3. Sep 15, 2013 #2
    a = 1.11 i + 1.58 j
    Integrating,
    v = 1.11 t i + 1.58 t j
    Again, integrating,
    s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
     
  4. Sep 15, 2013 #3
    I understand where the a= 1.11i + 1.58j comes from but I dont understand what the other two parts (v and s) have to do with this
    Can you explain?
     
  5. Sep 15, 2013 #4
    s = 1.11 t[itex]^{2}[/itex]/2 i + 1.58 t[itex]^{2}[/itex]/2 j
    so
    x = 1.11 t[itex]^{2}[/itex]/2
    y = 1.58 t[itex]^{2}[/itex]/2
     
  6. Sep 15, 2013 #5
    So your saying I have no need to use Xt= Xo + Vot + 1/2at^2?
    But what is s?
    Cause it looks like your saying that by using the i and j components of "s", the x and y would be constants
    but thats not possible is it? because im looking for the coordinates at any time t
     
  7. Sep 15, 2013 #6
    Unfortunately I never used proper limits in integration. Anyway lets do it with your equation.
    By s I meant displacement vector.
    Now,
    [itex]\overline{s}[/itex] = [itex]\overline{v0}[/itex]t + 1/2 [itex]\overline{a}[/itex]t[itex]^{2}[/itex]
    [itex]\Rightarrow[/itex] [itex]\overline{s}[/itex] = (3i - 2j)t + 1/2 (1.11i + 1.58j)t[itex]^{2}[/itex]
    [itex]\Rightarrow[/itex] [itex]\overline{s}[/itex] = (3t +.56t^2)i + (-2t + .79t^2)j
    Thus finally,
    x = (3t + .56t^2)
    y = (-2t + .79t^2)
     
    Last edited: Sep 15, 2013
  8. Sep 15, 2013 #7
    ohhh ok that makes sense
    I almost had it right but you cleared up my confusion
    thank you so much!!
     
  9. Sep 15, 2013 #8
    Correcting my first approach:

    [itex]\bar{a}[/itex] = 1.11i + 1.58j
    Integrating, [itex]\bar{v}[/itex] - [itex]\bar{v0}[/itex] = 1.11ti + 1.58tj
    Again integrating, [itex]\bar{s}[/itex] - [itex]\bar{s0}[/itex]- [itex]\bar{v0}[/itex]t = 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
    But [itex]\bar{s0}[/itex] = 0 as the particle is at origin at t = 0
    So, [itex]\bar{s}[/itex] = [itex]\bar{v0}[/itex]t + 1.11 (t^2)/2 i + 1.58 (t^2)/2 j
    ...
     
    Last edited: Sep 15, 2013
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