Question

(a) If the maximum acceleration that is tolerable for passengers in a subway train is$1.34m^2$ and subway stations are located $\mathbf{806}\mathbf{}\mathit{m}$apart, what is maximum speed a subway train can attain between stations? (b)What is the travel time between stations? (c) If a subway train stops for localid="1660818170335" $20s$ at each station, what is the maximum average speed of the train, from one start up to the next? (d) Graph x, v and a versus t for the interval from one start-up to the next?

Short answer

1.The maximum speed of a subway train that can attain between stations is $32.9\frac{m}{s}$.

2.The travel time between stations is $49.1s$

3.The maximum average speed of the train from one start-up to the next$11.7\frac{m}{s}$

Step-by-step solution

Given information

Acceleration of the train $1.34\frac{m}{s^2}$

Distance between the stations $806m$

Understanding the concept

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the formula for third kinematic equation, the maximum speed of a subway train that can attain between the stations can be found. Further, using the formula for the second kinematic equation, the travel time between stations can be calculated.

Formula:

The final velocity is given by,

${V_r}^2={V_0}^2+2ax$

The displacement is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2$

Where $V_0$the initial velocity, x is is the displacement.

(a) Determination of maximum speed attain by train between stations

Let’s consider the train accelerates from $v_0$& $x_0$at $a_1=1.34\frac{m}{s^2}$ up to midway point

$$\begin{gathered} X_1=\frac{806}{2}m \\ =403m \end{gathered}$$

Where velocity is v₁ and then train decelerates at $a_2=-1.34\frac{m}{s^2}$ until it comes to a stop $V_0=0\frac{m}{s}$

So, maximum speed that train can attain is $32.9\frac{m}{s}$.

(b) Determination of travel time between stations

The time $t_1$for the accelerating is

$X_1=V_0{t}_1+\frac{1}{2}{a}_1{t_1}^2$

$t_1=\sqrt{\frac{2\left(403m\right)}{1.34{\displaystyle \frac{m}{s^2}}}}=24.53s$

As time for decelerating is also same, total time $t=2t_1=49.1s$ is for travel between stations.

Therefore, the travel time between stations is $49.1s$
.

(c) Determination of maximum average speed of the train from one start-up to the next

With dead time of $20s$total time from one stop to another stop is

$$\begin{gathered} T=t+20 \\ =69.1s \end{gathered}$$

$\begin{array}{rcl}{V}_{avg}& =& \frac{806}{69.1}m/s\\ & =& 11.7\frac{m}{s}\end{array}$

(c) Graphs of  x, v and a versus t

Graph of distance vs time

Graph of velocity vs time

(iii) Graph of acceleration vs time