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Kinematics One Dimensional help

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data

    A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s after the end of the train left the very same point. What is the slowest constant speed at which she can run and catch the train. Sketch curves for the motion of the pasenger and the train as functions of time
    2. Relevant equations

    x=x0+v0t+1/2at2



    3. The attempt at a solution
    let origin be when the person reaches the tracks (at t=6).
    let h = the distance the train has already traveled in the first 6 seconds)

    PERSON
    x=vop(initial of person)t

    TRAIN
    x=h+vot(initial of train+att2/2

    let position of both objects equal each other

    vopt=h+vopt+att2/2

    what do i do now? i am stuck. i have to find vop, but i cant seem to find out how to find t.... i know the vlaues of h,vot, and at.
     
  2. jcsd
  3. Apr 9, 2015 #2

    BvU

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    You want to also let the times be equal. That's the seond equation.
    Make a graph of position vs time for both, easier to see what you are doing then.
     
  4. Apr 9, 2015 #3
    i tried doing that. i did make a position vs time graph, but i didnt really know how to find a way to write equations for time without introducing a bunch of new unknown variables (like velocity of train AFTER the time driven). can you give me a hint?
     
  5. Apr 9, 2015 #4
    I don't believe what you have written for the equivalence between the train and the person's position is correct. Did you want to put in the initial velocity of the train on the right side first?
     
  6. Apr 9, 2015 #5
    oh sorry. i copied it wrong. i did it right on my paper. it should be v(ot) on the right side. it wont let me change it now.
     
  7. Apr 9, 2015 #6
    You have the right idea with the setting of the equivalences. You'll want to use that to get the time that the person need to run before he catches the train at the constant velocity. Once you get that, you can solve for how far he must run via how far the train has gone from the original point. Then you can get the original velocity from that. Does that help?
     
  8. Apr 9, 2015 #7

    PeroK

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    It looks simpler to me to start at t=0, so that the person starts running at t = 6. This removes h and the initial velocity of the train from your equations.
     
  9. Apr 9, 2015 #8
    Rearrange the equivalence that you set so that you get a quadratic equation.
     
  10. Apr 9, 2015 #9

    SteamKing

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    Why are you complicating this problem?

    Instead of introducing a new variable for everything, try to make some variables do double duty.

    For instance, there's no need to have multiple time variables. One time variable should be sufficient for this problem.

    Assume t = 0 when the train starts to move. At t = 6 sec. is when the passenger shows up at the station and starts chasing the train.

    Using t, you should be able to use the equations of kinematics to calculate the distance the train travels from the station. The same t, offset by 6 secs., can be used to figure out the distance the late passenger runs trying to catch the train, given a constant velocity vp. These two curves can then be easily plotted.
     
  11. Apr 9, 2015 #10

    SammyS

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    It looks like you have some typos in the above.

    That last equation should be:
    vopt=h+vott+att2/2

    If you did know vop, how would you solve for t ?

    By the way, I think your choice for time, t, is fine.
     
  12. Apr 9, 2015 #11
    You only need to account for the time when the train is originally moving to get the h variable (the distance it is ahead at the time the person gets there). After that, all you need to worry about is the time for which the person is running. Isn't it easier to just break the problem up like that?
     
  13. Apr 9, 2015 #12

    BvU

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    You mean vopt=h+vott+att2/2 , right ?

    Sorry, I confused you earlier on.
    You made the drawing, so you can see how to wiggle the straight line for the passenger to let him/her catch the train. The minimum v0 is when the line is a tangent, right ? I.e. when there is 1 single solution for your equation $$
    v_p t = h + v_{0,{\rm train}} t + {1\over 2} a t^2
    $$ (which is correct). It is a quadratic equation in t. When does that have a single solution ?
     
  14. Apr 9, 2015 #13
    i already know h. h would just be 6 times 0.4 in this case.
     
  15. Apr 9, 2015 #14
    yeah i know i made some mistakes in the notations. t stood for train and time and yeah its confusing. i changed in on my paper. the equation is now



    vopt = h+voct+1/2at^2 . voc is the initial velocity of the train at t=6.


    none of you guys are giving me a clear explanation how to find t. i have 2 unkowns in the above equations (vop AND t). i plotted the xvt graph. ive attached what i drew.
     

    Attached Files:

  16. Apr 9, 2015 #15

    BvU

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    Good. Lousy picture, I have to strain my neck to see it. Never mind.
    vperson is the slope of the straight line. It is easy to see that a lower value of vperson still lets him/her catch the train, right ?
    Up to you to find the lowest slope that still allows him/her to catch the train. Now, in what situation does that happen ?

    Maybe. But in the mean time you have managed to let a whole lot of extremely knowledgeable experts go out of their way to help you further. Not bad at all !
     
  17. Apr 9, 2015 #16

    SammyS

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    Solve for t using the quadratic formula. Yes, you will have vop under the radical. Depending upon the value of vop, you may or may not get a real solution for t. (This is a clue to getting a solution!)

    By the Way: The graphical method others suggest is a more meaningful approach from a physical point of view.

    Added in Edit: " for t "
     
    Last edited: Apr 9, 2015
  18. Apr 9, 2015 #17
    but we dont know the value of Vop. would we able to find t?
     
  19. Apr 9, 2015 #18

    BvU

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    Yes
     
  20. Apr 9, 2015 #19
    h=.4*6 is dimensionally incorrect. h is supposed to have dimensions of length. Why would you multiply the acceleration by the time to get a length? You need to integrate.

    I pose this question now....
    The velocity of the person should be equal to the train at the time that he catches it. True or false (forgive me, but I'm a bit rusty which is why I'm asking)?

    If it is true, you can actually put the velocity of the person at the time that he catches the train in terms of the time at which he catches the train and the acceleration. Once you have that, then you can plug that into the quadratic you get by setting the positions equal to each other. @toesockshoe : Is that a helpful tidbit?
     
  21. Apr 9, 2015 #20
    sorry i was typing that response while i was out. yeah h would be 1/2(0.4)*(6^2)

    and false, the velocity of the person and train are NOT going to be equal when the person catches up.

    i just came home and will solve the quadratic formula.
     
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