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Catching a train

  1. Apr 13, 2004 #1
    A train pulls away from a station with a constant acceleration of 0.21 m/s2. A passenger arrives at the track 6.0 s after the end of the train passed the very same point. What is the slowest constant speed at which she can run and catch the train?

    Xtrain=Xo+Vot+1/2(a*t^2)
    Xtrain=3.78m + 1.26t + .5*(.21m/s^2)t^2
    Vtrain=Vo + at
    Vtrain=1.26m/s + (.21m/s^2)t

    Xperson=Vperson*t
    Vperson=X/t

    **at the time the person catches the train the
    Vtrain=Vperson &
    Xtrain=Xperson

    Vtrain=Vperson
    1.26m/s + (.21m/s^2)t=X/t
    since the X's are the same substitute Xtrain for X on right side of equation.
    1.26m/s+(.21m/s^2)t=[3.78m + (1.26m/s)t +(.105m/s^2)t^2]/t
    (1.26m/s)t + (.21m/s^2)t^2=3.78m + (1.26m/s)t +(.105m/s^2)t^2
    (.105m/s^2)t^2=3.78m
    t^2=36s^2
    t=6s

    plug it in
    X=Vot + 1/2*a*t^2
    X=1.26m/s*(6s) + .105m/s^2(36s^2)
    X=7.56m+3.78m=11.34m

    **which in theory would make min velocity needed 11.34m/6s=1.89m/s

    but this isnt correct
    Can anyone see what I have done wrong w/this problem?

    Thanks!
     
  2. jcsd
  3. Apr 13, 2004 #2
    nevermind I got it. I forgot to add the initial 3.78m the train traveled before the person started running.
     
  4. Apr 13, 2004 #3
    Why did you decide that Xo of the train is 3.78m and its Vo is 1.26m/s? :confused: Instead of calculating how much the train has passed in 6 seconds (which is what I suspect you have done, and maybe also where your error lies), I would just give it a 6 seconds fore on the person by changing "t" in its equations with "t + 6".

    So we have:

    [tex]x_{train} = x_{person}[/tex]
    [tex]v_{train} = v_{person}[/tex]

    [tex]\frac{1}{2}a(t + 6s)^2 = vt[/tex]
    [tex]a(t + 6s) = v[/tex]

    Substituting v we get:

    [tex]\frac{1}{2}a(t + 6s)^2 = a(t + 6s)t[/tex]

    And indeed t = 6s. Now just plug that into the second equation above:

    [tex]v = a(t + 6s) = a(6s + 6s) = 2.52 m/s[/tex]
     
  5. Apr 13, 2004 #4
    I find your explanation bewildering. I am restating the problem as I understand it.

    At time = 0 the end of the train is at x = 0 and v = 0. Six seconds later, xend = 0 + 0*t + (0.21)*t^2/2 = 3.78 meters
    vend0 = left to you to calc.

    At this point, xpass = 0, vpass = something, call it vpass

    xpass = vpass*t

    By the time the passenger catches the train,

    xpass = xend
    vpass = vend

    So,
    vpass*t = 3.78 + vend0*t + 0.21*t*t/2
    vend = vpass = 0.21*t

    I believe that's all you need -- solve for t then solve for vpass.
     
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