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A question on kinematics

  1. Aug 17, 2005 #1
    A train having some lenght.its front part passes through a point 'N' with velocity 'v' while its end part passes through same point with velocity 'u'.Prove that the mid point passes through the same point with velocity √v2+u2 /2.all the parts has acceleration a.
  2. jcsd
  3. Aug 17, 2005 #2
    I'm sorry, is this the equation you're trying to prove?

    [tex]\frac {\sqrt{v^2+u^2}}{2}[/tex]

    Or this?

    [tex]\frac {\sqrt{2v+2u}}{2}[/tex]
    Last edited: Aug 17, 2005
  4. Aug 17, 2005 #3


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    Let the length of the train be s, then midpoint travelles a distance s/2 after the front passes the point. Try to set equations for s and s/2
  5. Aug 18, 2005 #4


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    It's been a couple of days since this was posted, and I started obsessing on it! Here's how I did it. (Warning! I do not get the solution shown and the way I did it seems much to difficult for k-12 problem!)

    Let L be the length of the train and T the time interval between the front of the train passing point "N" until the end of the train passes point "N". Assuming that acceleration a is a constant, [itex]a= \frac{u- v}{T}[/itex].

    The basic kinematic equation are now [itex]v(t)= v+ \frac{u-v}{T}t[/itex]
    and [itex]L(t)= vt+ \frac{u-v}{2T}t^2[/itex] where v(t) is the velocity of the train at time t after the front passes point "N" and L(t) is the distance the front of the train has gone in time t.

    Since, by definition of L and T, the front of the train will have gone distance L in time T, we have
    [tex]vT+ \frac{u-v}{2T}T^2= vT+ \frac{u-v}{2}T= L[/tex]
    [tex]/frac{u+v}{2}T= l[/tex] so
    [tex]T= \frac{2L}{u+v}[/tex]
    Putting that value for T in the two equations
    [tex]v(t)= v+ \frac{u^2- v^2}{2L}t[/tex] and
    [tex]L(t)= vt+ \frac{u^2- v^2}{4L}t^2[/tex].
    We can use that L(t) equation to determine the time when the middle of the train passes point "N":
    [tex]L(t)= vt+ \frac{u^2- v^2}{4L}t^2= L/2[/tex] or
    [tex]t^2+ \frac{4Lv}{u^2- v^2}t= \frac{2L^2}{u^2- v^2}[/tex]
    Completing the square:
    [tex]t^2+ \frac{4Lv}{u^2- v^2}t+ \frac{4L^2v^2}{(u^2-v^2)^2}= \frac{2L^2(u^2-v^2)+ 4L^2v^2}{(u^2-v^2)^2}[/tex]
    [tex](t+ \frac{2Lv}{u^2-v^2})^2=\pm\frac{L\sqrt{2(u^2+v^2}}{u^2-v^2}[/tex]
    [tex] t= \frac{L\sqrt{2(u^2+v^2)}-2v}{u^2-v^2}[/tex]

    Now plug that into [tex]v(t)= v+ \frac{u^2-v^2}{2L}t[/tex] (noting that both the "L" and "u2-v2" terms cancel) we get, for the speed at the time the middle of the train passes point "N":
    [tex]\frac{\sqrt{2(u^2+v^2)}}{2}- v[/tex].
    Last edited by a moderator: Aug 18, 2005
  6. Aug 18, 2005 #5


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    Alternative solution

    All points on the train are connected, so travel at the same velocity.
    i.e. the front of the train is travelling at the same velocity as the end of the train, so when the end of the train reaches velocity 'u', so also is 'u' the velocity of the front of the train. And by this tine the train has travelled a distance L
    When the mid-point of the train reaches 'N', the front of the train will have travelled a distance ½L.

    Let a be the (constant) accln of the train.

    u² = v² + 2as
    u² = v² + 2aL
    a = (u² - v²)/(2L)

    To find velocity, w say, of mid-point of train when reaching 'N', i.e. after having travelled ½L

    w² = v² + 2a½L
    w² = v² + aL
    w² = v² + (u² - v²)/(2L)*L
    w² = v² + (u² - v²)/2
    w² = (u² + v²)/2
    Last edited: Aug 18, 2005
  7. Aug 18, 2005 #6


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    Let the length of the train be L, and the acceleration be a.
    The rear end of the train passes the point after the train has travelled distance L, so that
    v^2 - u^2 = 2*a*L
    The mid point of the train passes the point after the train has teavelled distance L/2, with velocity v' hence
    v'^2 - u^2 = 2*a*(L/2)
    Solve the two equation to get v' and your answer.
  8. Aug 18, 2005 #7


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    Am I onto a graphical solution?

    On the XY plane, let the horizontal axis be the train's speed when its front passes a mark, at time 0. Let the vertical axis be the speed when its rear passes the same mark at time T. Connect point v on the hor. axis to point u on the ver. axis with a straight line. I have a triangle with hypotenuse length = [itex]\sqrt{u^2+v^2}[/itex].

    One can visualize the train's speed moving with uniform acceleration from the horizontal intercept (at t = 0) to the vertical intercept (at t = T) on the hypotenuse. At the midpoint, the speed is the length of the vector that connects the midpoint of the hypothenuse with the origin = [itex]\sqrt{u^2+v^2}\left/2\right.[/itex].
    Last edited: Aug 19, 2005
  9. Aug 21, 2005 #8
    for some reason I got HallsofIvy's answer, minus the minus v
    let [tex]s =[/tex] total length of train
    let [tex]t =[/tex]total time from front at point N to back at point N
    let [tex]t_2 =[/tex] time when midpoint reaches M
    [tex]\frac{u-v}{t} = a[/tex]
    [tex]s = vt + \frac{1}{2}at^2[/tex]
    [tex]\frac{s}{2} = vt_2+ \frac{1}{2}at^2_2[/tex]
    [tex]s = 2\frac{s}{2}[/tex]
    [tex]vt + \frac{1}{2}at^2 = 2vt_2 + at^2_2[/tex]
    [tex]at^2_2 + 2vt_2 - (vt + \frac{1}{2}at^2 )= 0[/tex]
    [tex]t_2 = \frac{-2v \pm\sqrt{4v^2+4at(\frac{1}{2}at+v)}}{2a}[/tex]
    I am too lazy to type out the steps where I simplify the radicand
    the [tex]u[/tex] comes from substituting the acceleration for the formula
    [tex]t_2 = \frac{-2v + \sqrt{2u^2+2v^2}}{2a}[/tex]
    let [tex]v_m =[/tex] velocity at midpoint when passing M
    [tex]v_m = v + at_2[/tex]
    [tex]v_m = v + a \frac{-2v + \sqrt{2u^2+2v^2}}{2a}[/tex]
    [tex]v_m = v + \frac{-2v + \sqrt{2u^2+2v^2}}{2}[/tex]
    [tex]v_m = \frac{\sqrt{2u^2+2v^2}}{2}[/tex]

    how is my logic flawed?
    Last edited: Aug 21, 2005
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