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Relativity in X & Y Axis'-a, V, r

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Liza is driving her Lamborghini with an acceleration of a=(3.00i - 2.00j) m/s^2, while Jill is driving her Jaguar with an acceleration of (1.00i + 3.00j) m/s^2. Both begin from a state of rest, beginning at the start of the XY axis'. After 5.00s:

    a) What's the magnitude of the Liza's Velocity relative to Jill/from Jill's perspective?
    b) What's their distance?
    c) What's Liza's acceleration relative to Jill/from Jill's perspective?

    2. Relevant equations

    X: Vf = Vi + a*t || Xf = Xi +Vi*t + 1/2*a*t^2

    Y: Vf = Vi + a*t || Yf = Yi +Vi*t + 1/2*a*t^2

    3. The attempt at a solution

    I did all the necessary actions to find the Velocities and all that, but I took each Axis differently, found the relative quantities in each axis, and then I did the Pythagorean Theorem to find the combined one.


    Liza: X: Vf = 3m/s^2 * 5s = +15 m/s^2 || Y: Vf = ... = -10 m/s^2

    Jill: X: Vf = ... = +5 m/s^2 || Y: Vf = ... = +15 m/s^2

    Now, the problem is that I haven't really grasped how relativity works when I'm dealing with two Axis'. My book has only two examples, which are fairly basic, and don't really help me with such problems. The first one is the classic "man on sliding treadmill", with one woman on the ground and another on the treadmil. The other example is the boat and the river's stream.

    So, I don't really have anything to look at that'll help me understand how this really works, as the pages devoted to it are just about two. So, I tackled it a bit with logic, and tried to match the numbers in order to get the correct results from the book, but I'd really appreciate some help in understanding why this happens, and how I'll use it in other problems.

    Anyway, back to the question, I followed with this:

    X: Vlj = Vlo - Vjo = (15 - 5) m/s = +10 m/s

    Y: Vlj = Vlo - Vjo = (-10 -15) m/s = -25 m/s

    I thought about it like this: Say I'm in Jill's car (the j) and Liza's in front of us (the l). Liza's velocity relative to mine's should be: Her initial velocity, relative to the starting point/Earth (the o), minus my velocity relative to the starting point/earth.

    And I went ahead and did the same with the other quantities. I found the correct results, but I just want to see if I'm tackling this the right way, or I just happened too find the same results, but with a flawed way of thinking.
  2. jcsd
  3. Dec 11, 2016 #2

    Simon Bridge

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    ... works the same way as one axis, only you are doing a vector subtraction.

    ie. If ##\vec v_x|_y## is the velocity of X wrt Y, while ##v_x## and ##v_y## are the velocities of X and Y wrt a common reference frame (ie the ground), then
    ##\vec v_x|_y = \vec v_x-\vec v_y## You can see this is right because ##\vec v_y|_y=0## ... that is to say that Y is stationary wrt itself.

    Looks good to me. Check - if Jill was going faster that Lisa, then Liza would be going backwards wrt to Jill. Or - your own velocity with repect to yourself would be zero.
  4. Dec 11, 2016 #3
    Thanks for the tip!
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