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Homework Help: Constant Acceleration With Two Objects

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A train pulls away from a station with a constant
    acceleration of 0.40 m/s2. A passenger arrives at the
    track 6.0 s after the end of the train has passed the
    very same point. What is the slowest constant speed at
    which she can run and catch the train?

    2. Relevant equations

    X = X0 + V0 * t + 0.5 * a * t2
    V = V0 + a * t

    3. The attempt at a solution

    The passenger arrieves 6 sec. latter, so: tp = tt - 6sec


    X = V0 * ( t - 6 sec)


    X = 0.5 * 0.4 m/s2 * t2

    Both, train and passenger, have to be at the same place at the same time in order to achieve an encounter, so:

    xtrain = x passenger

    V0 * ( t - 6 sec) = 0.2 m/s2 * t2

    So here is the problem. I know I have to find out V0 , but I have 3 variables and 2 equations. Any ideas? This is supose to be easy......... :(
    Last edited: Feb 16, 2009
  2. jcsd
  3. Feb 16, 2009 #2
    For the passenger, X is Vo*(t-6) not +, careful!

    An easy way to check this is plug in t=6s, he should be at x=0.

    Then your strategy of setting X1=X2 is correct.

    You should first get a function of time as a function of Vo. Then try to remember how you can find the minimum value of a function!
  4. Feb 16, 2009 #3


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    Homework Helper

    After the 6 s head start, the train is moving at 2.4 m/s and has gone 7.2 m.
    Easier to let t be the time after this point.

    (edited - I had a mistake earlier)
    Last edited: Feb 16, 2009
  5. Feb 16, 2009 #4
    We know the position of the person is X1 = V0 * ( t - 6 )

    The position of the train is X2 = 0.2 * t^2

    All he needs to do is set X1=X2 and find a function of the time taken to catch up in terms of Vo, he can then use calculus to determine the best choice of Vo :)
  6. Feb 16, 2009 #5


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    Homework Helper

    Okay, I see it - had confused distance and velocity and got a v function with no minimum.
    How interesting, at speed 5 the person catches the train in 9 seconds. At speed 4, she never catches it.
  7. Feb 16, 2009 #6
    So, after some basic algebra, I got this:

    v0 * (t - 6) = 0.2 * t2

    -0.2 t2 + v0 t - v0 6 = 0

    From the quadratic formula, I get:

    (- v0 t +/- √ v02 + 4 * 0.2 * 6 v0) / 0.4

    This should give me the time at which she catches the train, right?
  8. Feb 16, 2009 #7
    Using the quadratic formula would be meaningless here, you don't want the values of Vo such that t=0 (which is what the quadratic formula would give you)!

    Do you know how to take the derivative of a function and find its minimum value? :) If not I can just show you a way to find it =-).
  9. Feb 16, 2009 #8
    Hey, dontdisturbmycircles, first of all let me thank you all your time and dedication.

    Yes, i know how to derivate and find a minimun of a function. I just was not expecting to come to that kind of solution. Having said that, thanks again for the help. I got it now!!
  10. Feb 16, 2009 #9
    You are welcome ;).
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