1. ## 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.

2. Originally Posted by BiGpO6790
(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 $v_{max} = 3 \cdot v_{average}$