91Ó°ÊÓ

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

Distance between the stations $806m$

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.

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 $32.9\frac{m}{s}$.

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 $49.1s$
.

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

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

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

Graph of distance vs time

Graph of velocity vs time

(iii) Graph of acceleration vs time