1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Passenger

    X = V0 * ( t - 6 sec)

    Train

    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

    Delphi51

    User Avatar
    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

    Delphi51

    User Avatar
    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 ;).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook