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A train pulls away from a station with a constant acceleration of 0.44 m/s2. A passenger arrives at a point next to the track 6.2 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 still catch the train?

  • physics -

    distance train=1/2 a t^2
    distance passenger=V*(t-6)

    set them equal

    1/2 a t^2=Vt-6V



    Well, V^2>.88*6V

    Lets see if it works.

    t=(5.20+-sqrt(0))/.44=12 seconds

    distance train: 1/2 .44*144=you do it
    distance passenger= 5.20(12-6)=you do it

  • physics -


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