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The distance between two moving particle...

The distance between two moving particles `P` and `Q` at any time is a.If `v_(r)` be their relative velocity and if `u` and `v` be the components of `v_(r)`, along and perpendicular to `PQ`.The closest distance between `P` and `Q` and time that elapses before they arrive at their nearest distance is

Text Solution

Verified by Experts

The correct Answer is:
C
`t=(AC)/v=(a cos theta)/v because v cos theta=v_(1),t=(32xx2/sqrt5)/(4sqrt5)=2s`
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If the distance between any two moving points at any time be a if V be their relative velocity and u and v be the components of V respectively along and perpendicular to the direction a, then their distance when they are nearest to one another is (av)/(V) and the time that elapses before they arrive at their nearest distance is (au)/( V^(2))

If the distance between any two moving points at any time be a if V be their relative velocity and u and v be the components of V respectively along and perpendicular to the direction a, then their distance when they are nearest to one another is (av)/(V) and the time that elapses before they arrive at their nearest distance is (au)/( V^(2))

Knowledge Check

  • The resultant R of P and Q is perpendicular to P. Also |P| = |R|- The angle between P and Q is [tan 45° = 1]

    A
    `(5pi)/4`
    B
    `(7pi)/4`
    C
    `(pi)/4`
    D
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  • Two particles start moving from the same point along the same straight line. The first moves with constant velocity v and the second with constant acceleration a . During the time that elapses before the sound catches the first, the greatest distance between the particle is.

    A
    `(v^2)/(a)`
    B
    `(v^2)/(2 a)`
    C
    `(2 v^2)/(a)`
    D
    `(v^2)/(4 a)`

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    A
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    B
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    D
    `pi`

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