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Kinematics in 2 dimensions; position vector problem.

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data

    "A particle moves in the horizontal plane that contains the perpendicular unit vectors i and j. Initially it is at the origin and has velocity 18ims^-1. After accelerating for 10 seconds its velocity is (30i + 8j)ms^-1. Assume that the acceleration of the particle is constant.

    a) Find the acceleration of the particle
    b) Find the position vector of the particle when the velocity is (36i + 12j)ms^-1.

    2. The attempt at a solution

    For (a) i managed to find out the acceleration being (1.2i + 0.8j)ms^-2 using v=at+u and rearranging it to a=(v-u)/t.

    But for (b) it says the answer is 405i + 90j

    I've tried it over and over again but I don't think the question was specific enough (probably just me not understanding it).

    I've tried it this way using:

    u=18i + 0j
    v=36i + 12j
    t=10
    R (position vector; i used R=(t/2)(u+v))=5(u+v) which gives the vector R= 240i + 40j

    I've tried it by replacing the initial velocity with 36i + 12j but that didn't work either. I can probably work out the question if I understood what it basically is asking for.

    Someone Please run me through this, I'm not good at mechanics :'(
     
  2. jcsd
  3. Aug 19, 2012 #2

    eumyang

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    Homework Helper

    You have the wrong value of t. At t = 10 the velocity vector is 30i + 8j m/s, not 36i + 12j m/s. What would t have to be in order for the velocity vector to be 36i + 12j m/s?
     
  4. Aug 19, 2012 #3
    I only said t=10 because that was the only time value in the question and i couldn't use acceleration from my previous answer because I'm using a different velocity value now.

    I'm so confused...
     
  5. Aug 19, 2012 #4

    Ray Vickson

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    Homework Helper

    If [itex]\vec{v}_0[/itex] is the initial velocity vector and [itex] \vec{a}[/itex] is the (constant) acceleration (both of which you know from part (a)), the velocity vector v(t) and position vector r(t) at time t, starting from the origin, are
    [tex] \vec{v}(t) = \vec{v}_0 + \vec{a} t,\\
    \vec{r}(t) = \vec{v}_0 t + \frac{1}{2} \vec{a} t^2.
    [/tex]

    RGV
     
  6. Aug 19, 2012 #5
    It is a constant acceleration.
    Use equation v2=u2+2as
     
  7. Aug 20, 2012 #6

    HallsofIvy

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    Yes, you are. You can use acceleration from your previous answer- you have to- you are told that the acceleration is constant. Having a different velocity is due to the acceleration not being zero, not that the acceleration changes.
     
  8. Aug 20, 2012 #7
    Oh i see :) I got confused because when i did pure I was taught that you couldn't divide matrices (i was looking at r=(v^2-u^2)/2a) and i was working all these out in matrix form (as a column vector) and it didn't occur to me to dividing it through because i was deliberately avoiding that. I tried dividing the i and j units and it came to the right answer.

    Thank you :)
     
  9. Aug 20, 2012 #8
    Shouldn't stumble into this problem again :D Thank you!
     
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