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Vector addition (?) with acceleration and velocities given in vectors

  1. Jun 12, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    Particle is at (9, 4) on coordinate plane.
    Initial velocity: 5.5 x - 7 y
    Acceleration: 1 x + 8.5 y
    Time: 1.5s

    1: What is the x-component of velocity after that 1.5s?
    2: Y-component?
    3: Magnitude from origin of graph?

    2. Relevant equations

    sqrt(x^2 + y^2) for magnitude, tan^-1(y/x) for angles.

    3. The attempt at a solution

    Dunno if I'm on the right track, but based on the above:

    Initial velocity: 8.9022, -51.8428°
    Acceleration: 8.5586, 83.2902°

    The inclusion of acceleration and velocity as opposed to simple vector addition has got me a bit screwed up. What should I do with this?
     
    Last edited: Jun 12, 2007
  2. jcsd
  3. Jun 12, 2007 #2

    nrqed

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    The question is incomplete. They want the x and y components of what, exactly?? I am *guessing* they mean the x and y components of the *position vector* at 1.5 second, but that's just a guess. You should confirm.
     
  4. Jun 12, 2007 #3

    exi

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    Sorry about that. It's asking for the x- and y- of velocity after 1.5s.
     
  5. Jun 12, 2007 #4

    nrqed

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    Ahh!!!!

    Ok, then you only need to use the following vector formula

    [itex] \vec{v}(t) = \vec{v_i} + \vec {a} t [/itex]

    So just multiply the acceleration vector by the time (the result will be a vector, of course) then add this vector to the initial velocity vector. For this part of the calculation, you should stay in cartesian coordinates, it's the only way to add two vectors together.

    The third question does not make much sense to me (magnitude from the origin of graph???) This is why I thought they were asking about a position or a displacement vector. I don't understand what "magnitude from the origin of graph" means. Magnitude of what vector??
     
  6. Jun 12, 2007 #5

    exi

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    The question is "what is the magnitude of the displacement from the origin after 1.5s?"

    After 1.5s, the velocity is 7 x + 5.75 y (thank you very much, btw), confirmed to be correct answers. I tried to graph this out and see if I couldn't land a number that seemed plausible, which I did - but it was incorrect, at 9.0588.

    What am I doing wrong?
     
  7. Jun 12, 2007 #6

    nrqed

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    Ahh, ok, now it makes a bit more sense. They now want the magnitude of the displacement vector which has nothing to do with th evelocity vector you just calculated. You need another calculation.

    The position vector at anytime is given by

    [itex] \vec{r} (t) = \vec{r_i} + \vec{v_i} t + \frac{1}{2} \vec{a} t^2 [/itex]

    You should find the position at 1.5 second and calculate the magnitude of that. (notice that the initial position vector is given in the question).
     
  8. Jun 12, 2007 #7

    exi

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    Got it. Used that equation twice - once for x- and y- values each - and ended up with 18.375x + 3.0625y, for a magnitude of 18.6285m.

    I dunno why I completely forgot about that equation, especially considering I just used it extensively on a physics exam.

    Thanks for your help. :wink:
     
  9. Jun 12, 2007 #8

    nrqed

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    You are very welcome :smile:
    Notice that the vector equations are the obviosu equivalent of the one-dimensional equations for constant acceleration.
     
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