Question

If the maximum acceleration that is tolerable for passengers in a subway train is $1.34m/s^2$ and subway stations are located $806m$ apart, what is the maximum average speed of the train, from one start-up to the next? Given the "dead time" at a station is $20s$.

Answer 1

We assume the train accelerates from rest $(v_0=0$ and $x_0=0)$ at $a_1=1.34m/s^2$ until the midway point and them decelerates at $a_2=-1.34m/s^2$ until it comes to a stop $(v_2=0)$ at the next station.

Thus, for the first part of the motion (till midpoint), if the time taken is $t_1$, then:

$s=\frac{806}{2}m=403m$

And, $s=v_{0}t+\frac{1}{2}{a}_1{t}_1^2$

$\Rightarrow t_1^2=\frac{2s}{a_1}=\frac{2\times 403}{1.34}{s}^2$

$\therefore t_1=24.53s$

Since the time interval for the decelerating stage is the same, we double this result and obtain the total time, $t=49.16s$ for the travel time between stations.

Adding a "dead time" of $20s$, the time between one start-up and next:

$t=49.16s+20s=69.16s$

Thus, if average speed is $v_{avg}$:

$v_{avg}=\frac{806m}{69.16s}=11.65m/s$