# Confused

• Aug 30th 2009, 08:58 PM
BiGpO6790
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.
• Aug 30th 2009, 09:30 PM
earboth
Quote:

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 ... (Rofl)

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