Two subway stops are separated by $\mathbf{1100}\mathit{m}$. If a subway train accelerates at$\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ from rest through the first half of the distance and decelerates at$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{2}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph x, v and a versus t for the trip.

Consider,

$\begin{array}{c}{x}_1=550\mathrm{m}\\ x_2=550\mathrm{m}\\ a_1=1.2\mathrm{m}/{\mathrm{s}}^2\end{array}$

The problem deals with the kinematic equation of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The displacement in kinematic equation is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2........\left(i\right)$

The velocity is given by,

$v_f=v_0+at..........\left(ii\right)$

To find the time of trip, first find the time taken by the train to reach half the distance that is$550\mathrm{m}$by using the equation (i)

$x_1=v_0{t}_1+\frac{1}{2}a{t_1}^2$

where

${\mathrm{v}}_0= \mathrm{Initial} \mathrm{velocity}$

And take initial velocity = 0 because train start from rest

$550=0\times t_1+\frac{1}{2}\times 1.2\times {t_1}^2$

So,

$t_1=30.3\mathrm{s}$

Now, the time for the second half will get the same as$t_1$

So,
$t_2=30.3\mathrm{s}$

So, the total time of travel is

$30.3+30.3=60.6\mathrm{s}$

$$\begin{gathered} v_f=v_i+a\times t \\ {v}_f=0+1.2\times 30.3 \\ \begin{array}{c}{v}_f=36.36\mathrm{m}/\mathrm{s}\\ \approx 36.3\mathrm{m}/\mathrm{s}\end{array} \end{gathered}$$

So, the maximum speed of the train is$36.3\mathrm{m}/\mathrm{s}$.

Below are the graphs of x, v and a versus t