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Homework Help: Acceleration from unit vectors

  1. Sep 17, 2006 #1
    I have a problem asking me to find the acceleration of a particle when its
    [tex]v_i = (3.00 \hat{i} -2.00 \hat{j} ) m/s [/tex]

    and then 3 seconds later,

    [tex]v = (9.00 \hat{i} + 7.00 \hat{j} ) m/s [/tex]

    The big problem here is that my book doesn't say anything whatsoever about getting an acceleration when dealing with unit vectors. The only thing it says about acceleration at all is that a=dv/dt, but taking the derivatives of those velocities doesn't really get me anywhere.

    Help?! This is a major bottle-neck in my homework and my book is totally worthless.:mad:
    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2


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    Staff: Mentor

    Hi Bob, welcome to the PF. When you are given components in an orthogonal basis system (like the rectangular coordinates you have shown), you can just take the derivatives in each axis, and combine them with the unit vectors in that same system. So just differentiate the x-axis numbers and differentiate the y-axis numbers, and show them as an (x,y) acceleration with the same unit vectors. Make sense?

    EDIT -- Oops, sorry. Your unit vectors are i and j. My comments still hold, as long as i and j are orthogonal. Maybe even if they aren't orthogonal, but I'm not sure about that. In your problem statement, is it apparent that i and j are at right angles in 3-space?
  4. Sep 18, 2006 #3
    yeah, they're pretty much the same as x and y (i being x and j being y [k is z if it's 3d...i have no clue why they can't just use x,y,z]).

    What I don't understand though is how exactly I'm supposed to take the derivative of a constant. If I derive (3.00i-2.00j) I'd get 3.00-2.00dj/di, as far as I can tell (or zero if i don't include the i and j).
    Last edited: Sep 18, 2006
  5. Sep 18, 2006 #4


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    Homework Helper

    First, note that something expressed by unit vectors i and j is a vector too, so you have to write [tex]\vec{v}[/tex] for your velocities. All you have to do is apply [tex]a=\frac{\Delta v}{\Delta t}[/tex] to your vectors, and get the acceleration vector.
  6. Sep 18, 2006 #5
    Well, I tried to do that. I figured it should just be final minus initial, but I tried [tex]([9.00-3.00] \hat{i} + [7.00 + 2.00] \hat{j} ) = (6.00 \hat{i} + 9.00 \hat{j} )[/tex]...oohhhh...but I forgot to divide by three, which would give me 2i and 3j, which is the answer I've been trying to get (lol, yes, I figured this out as I was typing my response).

    Alright, problem solved, thanks a lot!
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