physics

two ends of a train moving with constant acceleration pass a certain point with velocities u and v. what is the velocity with which the middle point of the train passes the same point?
I have asked this question before
you have answered u+v/2. but my mam says it is (u^2+v^2/2)^1/2
please tell me how it comes.

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  1. I agree with mam. What you are figuring is the rms velocity.

    Here is why:

    velocity u is front, then velocity rear, u is
    u^2=v^2+2aL where L is the length of the train. so
    l= (u^2-v^2)/2a

    now at midpoint,
    m^2=v^2+2a L/2
    m^2=v^2+a (u^2-v^2)/2a
    =(u^2+v^2)/2

    m= your answer.

    Sorry for the earlier post.

  2. we know that v2-u2=2as
    NOTE :[2 behind u or v is power of them]
    u= head of train
    3u= tail of train [v=3u]
    so (3u)2-u2=2as
    9u2-u2=2as
    8u2=2as
    a=4u2/s
    Now at middle of train s'=s/2
    v'2-u2=2as'
    v'2=u2+2as/2
    put the value of a
    v'2= u2+2*4u2/s*s/2
    v'2=5u2
    v'= root5u

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  3. siddharth sharma why are you taking taking tail of train as 3u

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    posted by vishal
  4. Let velocity at mod point is V .



    <---------L------->.<---------L ---------->
    +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
    v. V. u

    (V^2-v^2)/2a=(u^2-V^2)/2a


    So at midpoint velocity = √{(u^2+v^2)/2}

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    posted by GOHAN
  5. Let velocity at mod point is V .



    <---------L------->.<---------L ---------->
    +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
    v. V. u

    (V^2-v^2)/2a=(u^2-V^2)/2a

    So at midpoint velocity V = √{(u^2+v^2)/2}

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    posted by GOHAN

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