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Relative Veloctiy Of Approach

  1. Oct 25, 2013 #1
    Given in my book-
    1."Velocity Of approach - It is the rate at which distance between two moving particle decreases.
    V(Approach) =V1cosα + V2cosβ "
    2."When the distance between two particles is minimum, the relative veloctiy of approcah is 0"
    attachment.php?attachmentid=63276&d=1382679140.jpg
    3."V1cosα + V2cosβ = 0 .....when distance is minimum(diagram in attachment below)"
    Please Explain Why?
    And, the above equation will become 0 only if cosα = 0 and cosβ = 0 at same time, but if α and β are not equal, then cosα and cosβ cannot be 0 at same time. Thus the Equation is worng but giving correct answers.
    Please Explain Why?
    paint.jpg
     
  2. jcsd
  3. Oct 25, 2013 #2

    Simon Bridge

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    Welcome to PF;
    What about when the situation: v1=v2=v, α = 0, and β=180deg.

    For the two particles to be at the point f closest approach, they must be moving neither (1) towards each other nor (2) away from each other.

    if (1) then they are getting closer,
    if (2) then they have previously been closer,
    thus, their velocity vectors must have no component pointing towards each other.

    In your diagram, the two objects are not at closest approach - they are pictured getting closer together.
     
  4. Oct 25, 2013 #3

    jbriggs444

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    Your claim is that the equation is solved only when α and β are both right angles. Yes, that is one solution.

    To make Simon's example more general (and less explicit), what if α is acute and β is obtuse?

    What about cases where β > 180 degrees?
     
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