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SUM1 tell me wat i am doing WRONG

  1. Sep 20, 2006 #1
    PLZ SUM1 tell me wat i am doing WRONG!!

    the follwing Q is really annoying me....because i cant work out where i am goin wrong arrgghhhhh

    At time t= 0 seconds particles A and B are moving with velocities (-10i-2j)m/s and (-2i-6j)m/s respectively, the postion vector of B relative to A being (-16i+13j)m. Assuming A and B maintain these velocities find the least distance of seperation bewteen the 2 particles in the subsequent motion and the value of t for which it occurs.

    (PS...we have to use the scalar product method....and yes i can do it with calculus methods....but im really annoyed i can tget this)

    here is my working:

    A v B = va-vb = (-8i+4j) - (ie consider situation as "B" being staionary and "A" moving)

    B r A (or vector AB)= -16i-13j (ie we are told this)

    Vector PB = Vector AB - Vector AP
    =(-16i-13j)-t(-8i+4j)
    =(-16+8t)i +(-13-4t)j

    Given Vector PB will be perpendicular to (-8i+4j) therfore the scalar product of the 2 will = 0

    such that : -8(-16+8t)+4(-13-4t) = 0

    therfore i get t = 0.95......WHICH IS INCORRECT, the answer is 2.25.....can anyone see my ERROE....i really appreciate this!!!!:smile:
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2006 #2

    HallsofIvy

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    Why "- Vector AP"? If AP is the velocity vector of A relative to B and AB the position vector of A relative to B at time t= 0, then the position vector of A relative to B is AB+ APt.

    I don't follow this. The motion of A relative to B, vector PB, is a straight line. It will not be perpendicular to the initial position vector- that's irrelevant. Instead, minimize the length of vector PB.

     
  4. Sep 20, 2006 #3
    yes i understand that you can simply minimise vector PB....however our topic atm is the scalar product, and it is stipulated that we must use this to answer the question.

    “Vector PB = Vector AB - Vector AP” since if you add them as the way u said, it in no way gives vector PB (ie P to B) I agree yes AB+ APt. = however, do u not agree that AP must be in he opposite direction to give the required vector AB….i just skipped this step and inserted the “-“ stright away….

    I could hav done ur method and said:


    AB+ APt = vector of A relative to B = (-16i+13j)+ -t(-8i+4j)

    If u draw it it makes sense since, since isn’t “A relative to B” = vector BA????? This is why, again if you draw it, vector BA= (Position vector) A – (position vector) B.


    As for the following “I don't follow this. The motion of A relative to B, vector PB, is a straight line. It will not be perpendicular to the initial position vector”

    The point at which these 2 particles are closest, given they don’t collide, is the point at which PB is perpendicular to the position vector of t(AvB), ie relative to time hence the line t(-8i+8j), I am not saying it is perpendicular to the initial position vector, I am trying to say it is perpendicular to the position of A relative to B according to time .

    Do you follow me now, hence the pic I included, do u not agree that at some point it will be perpendicular and this represents the least distance…….
     
  5. Sep 20, 2006 #4

    Gokul43201

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    :bugeye: Are you sure about this?
     
  6. Sep 21, 2006 #5
    LOL....no not now, caus i know how smart u guys r!!! its just that thats the way i always hav done finiding that particular vector....so yes i am incorrect, since if you add them as "hallsofivy" stipulates, u obtain the answer, its just that i dont undertsand why it works, tats all....
     
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