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Help understanding this speed problem

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    2. A train pulls away from a station with a constant acceleration of 0.40 m/s.
    A passenger arrives at a point next to the track 6.0 s after the end of the train has started
    from rest at that very same point. What is the minimum constant speed at which she can
    run and still catch the train? On a single graph, plot the position versus time curves for
    both the train and the passenger.

    2. Relevant equations

    x(t)=x0+v0t+1/2at2

    v(t)=v0+at

    v2-v02=2aΔx

    v (average) = v0+v(t)/2


    3. The attempt at a solution

    I'm having a heck of a time trying to understand the correct way to set this problem up. I know that we need to find the constant speed of the passenger. I also know that the position and velocity of the train need to be solved relative to the position of the passenger and then we set figure out how fast the passenger needs to move relative to the train.

    For the train, I know we solve for its position from the equation:

    x(t)=x0+v0t+1/2at2

    I know we also need to solve for the postion of the passenger. However, this is where I'm stuck. I'm also confused as to how to properly set the train and passenger equal to one another to solve for the final speed of the passenger. Any help is very much appreciated.
     
  2. jcsd
  3. Dec 11, 2012 #2

    lewando

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    Have you tried plotting the position versus time curves for both the (end of the) train and the passenger? This will help, I think.
     
  4. Dec 11, 2012 #3

    haruspex

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    Create an unknown for the passenger's speed and obtain an equation for the passenger's distance from start at time t. Now you have an equation for the motion of each. What variables have to match up for the passenger to board the train?
     
  5. Dec 11, 2012 #4
    I have not. This is from an old quiz for which I still have not figured out how my professor solved it.

    Here is what he did:

    xtrainc = atrain/2 t2c

    xp.c.(tp.c=vp.c(tc-Δt)-Where does this equation come from???

    From here, he set the position of the train and passenger equal to one another and then calculated the average speed of the train relative to the passenger's velocity. I guess I'm not right now where that second equation above comes from and how or why you can make the train and passenger equal to one another. Is it because the passenger comes out and to the same point at the same time the train leaves from rest?

    c=critical conditions
     
  6. Dec 11, 2012 #5

    haruspex

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    Seems unnecessarily complicated notation. Presumably it means, at any time t > Δt:
    xp=vp(t-Δt)
    So what two equations do you have involving xc and tc, the position and time of catching the train?
     
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