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Relative Velocity and Acceleration

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Heather in her Corvette accelerates at the rate of <3.20, -2.00>m/s^2, while Jill in her Jaguar accelerates at <1.00, 3.20>m/s^2. They both start from rest at the origin of an xy coordinate system. After 5.50 s, (a) what is Heather's speed with respect to Jill, (b) how far apart are they, and (c) what is Heather's acceleration relative to Jill?

    2. Relevant equations



    3. The attempt at a solution
    Vhj = Vh - Vj = 3.77 - 3.35 = .421
    [(3.77cos(327.9))i + (3.77sin(327.9))j] - [(3.35cos(72.6))i + 3.35sin(72.6))j
    = 2.19i - 5.20j
    magnitude is 5.64

    I really don't know if this is correct and how it fits in to the problem of getting the distance at 5.5 seconds and the speed at 5.50 seconds. Can someone help please?
     
  2. jcsd
  3. Mar 1, 2009 #2

    LowlyPion

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    Once you determine the |acceleration| you can use ordinary kinematic means to determine d and v.

    D = 1/2*|a|*t2

    V = Vo + a*t

    (but Vo is 0)
     
  4. Mar 1, 2009 #3
    What about the speed? The answer .42 is incorrect. Did I calculate correctly?
     
  5. Mar 1, 2009 #4

    LowlyPion

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    I didn't check your math.

    I wouldn't do it that way.

    For V just multiply by the scalar 5.5 to get the Vx,y at 5.5 and then do the vector difference to determine the relative |v|.

    Part c is the easiest to do because you just take the vector difference and you get the relative a(x,y). Then calculate |a| and apply the kinematic equation.
     
  6. Mar 1, 2009 #5
    How do I get the acceleration? I would normally differentiate but I don't see how in this problem. Sorry for the bother.
     
  7. Mar 1, 2009 #6

    LowlyPion

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    Acceleration is a vector, just like velocity.

    The relative acceleration then is the vector difference right?

    Just like relative velocity is a vector difference. Speed is the scalar of V, or |V|.

    The distance at any time can be found simply from kinematics by 1/2*|a|*t2 since they both started from rest.
     
  8. Mar 1, 2009 #7
    I'm sorry I just don't get this. I do understand part c, but how to get part a and b in terms of kinematics and vectors is just too confusing for me to understand right now. I don't know why the answer .42 is incorrect. Can you try to explain it?
     
  9. Mar 2, 2009 #8

    LowlyPion

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    You are given the acceleration in x,y.

    For Helen = 3.20 i - 2.0 j

    Using V = a*t at 5.5 sec that makes Helen's velocity = 17.6 i - 11 j

    Now figure Jill. And then figure your relative velocity.
     
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