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Math Help - Confused

  1. #1
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    Confused

    (a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 846 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 22.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?


    I solved for the max speed but then i noticed the 22 s part of the problem and i am now very confused and unsure of how it works into the rest of the problem.
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  2. #2
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    Quote Originally Posted by BiGpO6790 View Post
    (a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 846 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 22.0 s at each station, what is the maximum average speed of the train, from one start-up to the next?


    I solved for the max speed but then i noticed the 22 s part of the problem and i am now very confused and unsure of how it works into the rest of the problem.
    We assume that there is a mad subway driver at work ...

    1. The subway is accelerated on the first 423 m and must be decelerated on the following 423 m.
    According to d = \dfrac12 \cdot a \cdot t^2 you'll get

    432\ m = \dfrac12 \cdot 1.74\frac m{s^2} \cdot t^2~\implies~t = 22.05\ s

    According to v = a \cdot t the maximum speed of the train is v_{max} = 38.367\ \frac ms = 138.1\ \frac{km}h

    (Did I mention that the driver is mad?)

    2. According to the results at 1. the train needs 44,1 s from one station to the next.

    3. With a 22 s stop at the station the elapsed time from one stop to the next is 66 s, the total distance is 846 m, thus the average speed is:

    v_{average} = \dfrac{846\ m}{66\ s} = 12.82\ \frac ms = 46.1\ \frac{km}h

    Obviously v_{max} = 3 \cdot v_{average}
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