In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of$\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$until it reaches$\mathbf{95}\mathbf{km}\mathbf{/}\mathbf{h}$, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at$\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. Assume it stops at each intermediate station for 22 s.

The equation of motion is the fundamental concept for finding the behavior of the trains in the design of a transit system.

These equations of motion can be used to find the quantities which are required in any problem.

At the start of the train.

• The acceleration of the train is$a=1.1\mathrm{m}/{\mathrm{s}}^2$
• The final velocity of the train is $v_f=95\mathrm{km}/\mathrm{h}\left(\frac{1\mathrm{m}/\mathrm{s}}{3.6\mathrm{km}/\mathrm{hr}}\right)=26.388\mathrm{m}/\mathrm{s}$.
• The initial velocity of the train is $v_i=0\mathrm{m}/\mathrm{s}$.
• The constant velocity of the train is $v_f=v_c=26.388\mathrm{m}/\mathrm{s}$.

At the stoppage of the train.

• The deceleration of the train is $a=-2\mathrm{m}/{\mathrm{s}}^2$.
• The initial velocity of the train is $v_i=26.388\mathrm{m}/\mathrm{s}$.
• The final velocity of the train is $v_f=0\mathrm{m}/\mathrm{s}$.
• The total distance traveled by the train in the full trip is $x_1=15\mathrm{km}\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)=15000\mathrm{m}$.

In the given problem, consider the conditions of part (a) as the first situation and the conditions of part (b) as the second situation.

The distance moved by train at the final speed of $26.388\mathrm{m}/\mathrm{s}$can be given as.

$v_f^2=v_i^2+2a\Delta s_i$

Substitute the values 0 m/s for $v_i$, $26.388\mathrm{m}/\mathrm{s}$ for $v_f$, and $1.1\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{c}{\left(26.388\mathrm{m}/\mathrm{s}\right)}^2={\left(0\mathrm{m}/\mathrm{s}\right)}^2+2\times 1.1\mathrm{m}/{\mathrm{s}}^2\times \Delta s_i\\ \Delta s_i=316.512\mathrm{m}\end{array}$

The time taken by the train can be given as.

$v_f=v_i+a{t}_i$

Substitute the values 0 m/s for $v_i$, $26.388\mathrm{m}/\mathrm{s}$ for $v_f$, and $1.1\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{c}26.388\mathrm{m}/\mathrm{s}=0\mathrm{m}/\mathrm{s}+1.1\mathrm{m}/{\mathrm{s}}^2\times t_i\\ t_i=23.989\mathrm{s}\end{array}$

The distance moved by train at the initial speed of $26.388\mathrm{m}/\mathrm{s}$can be given as.

$v_f^2=v_i^2+2a\Delta s_f$

Substitute the values 0 m/s for $v_f$, $26.388\mathrm{m}/\mathrm{s}$ for $v_i$, and $-2\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{c}{\left(0\mathrm{m}/\mathrm{s}\right)}^2={\left(26.388\mathrm{m}/\mathrm{s}\right)}^2+2\times \left(-2\mathrm{m}/{\mathrm{s}}^2\right)\times \Delta s_f\\ \Delta s_f=174.082\mathrm{m}\end{array}$

The time taken by the train can be given as.

$v_f=v_i+a{t}_f$

Substitute the values 0 m/s for $v_f$, $26.388\mathrm{m}/\mathrm{s}$ for $v_i$, and $-2\mathrm{m}/{\mathrm{s}}^2$ for a in the above equation.

$\begin{array}{c}0\mathrm{m}/\mathrm{s}=26.388\mathrm{m}/\mathrm{s}+\left(-2\mathrm{m}/{\mathrm{s}}^2\right)\times t_f\\ t_f=13.194\mathrm{s}\end{array}$

Determine the distance covered by the train at a constant velocity in the first situation.

The stations can be depicted as.

Here, $x_1$is the total distance covered by the train in the full trip.

The total distance traveled by the train can be calculated as.

localid="1643114175686" $x_1=\left(\Delta s_i+\Delta s_f\right)n+\Delta x\dots \left(i\right)$

Here,$n$is the number of intermediate distances between six stations whose value is 5, and$\Delta x$is the distance covered at a constant velocity.

The total number of stations is six, and the total intermediate distance between six stations is 5.

Substitute the values as 5 for n,$316.512\mathrm{m}$for$\Delta s_i$,$174.082\mathrm{m}$for$\Delta s_f$, and$15000\mathrm{m}$for$x_1$in equation (i).

$\begin{array}{c}15000\mathrm{m}=\left(316.512\mathrm{m}+174.082\mathrm{m}\right)5+\Delta x\\ \Delta x=12547.03\mathrm{m}\end{array}$

The distance covered at a constant velocity can be expressed as.

$\Delta x=x_0+v_i{t}_c\dots \left(ii\right)$

Here,$x_0$is the initial distance traveled by the train whose value is 0 m, and$t_c$is the time taken by the train at constant velocity.

Substitute the values as$12547.03\mathrm{m}$for$\Delta x$, 0 m for$x_0$, and$26.388\mathrm{m}/\mathrm{s}$for$v_i$in equation (ii).

$\begin{array}{c}12547.03\mathrm{m}=0 \mathrm{m}+26.388\mathrm{m}/\mathrm{s}\times t_c\\ t_c=475.482\mathrm{s}\end{array}$

The train stops at the intermediate stations for 22 s. From the above figure, it can be observed that the train stops at stations 2, 3, 4, and 5. So, the total time taken by train to stop (take rest) at each station can be given as.

$\begin{array}{c}\Delta t=4\times 22\mathrm{s}\\ =88\mathrm{s}\end{array}$

The total distance traveled by the train can be calculated as.

role="math" localid="1643114580968" $t_1=\left(t_i+t_f\right)n+t_c+\Delta t\dots \left(iii\right)$

Here,$\Delta t$ is the total time taken by the train to stop at each station.

Substitute the values as$23.989\mathrm{s}$for$t_i$,$13.194\mathrm{s}$for$t_f$, 5 for n,$475.482\mathrm{s}$for$t_c$, and$88\mathrm{s}$for$\Delta t$in equation (iii).

$\begin{array}{c}{t}_1=\left(23.989\mathrm{s}+13.194\mathrm{s}\right)5+475.482\mathrm{s}+88\mathrm{s}\\ =749.397\mathrm{s}\end{array}$

Thus, the total time taken by the train to make a 15 km trip in the first situation is$749.397\mathrm{s}$.

Determine the distance covered by the train at a constant velocity in the second situation.

The stations can be shown as.

Here,$n_1$is the number of intermediate distances between four stations whose value is 3.

The total distance traveled by the train$\left(\Delta x\right)$when moving at a constant velocity from equation (i) can be calculated as.

$x_1=\left(\Delta s_i+\Delta s_i\right)n_1+\Delta x$

Substitute the values as 3 for n₁,$316.512\mathrm{m}$for$\Delta s_i$,$174.082\mathrm{m}$for$\Delta s_f$, and$15000\mathrm{m}$for$x_1$in the above equation.

$\begin{array}{c}15000\mathrm{m}=\left(316.512\mathrm{m}+174.082\mathrm{m}\right)3+\Delta x\\ \Delta x=13528.218\mathrm{m}\end{array}$

From equation (ii),$t_c$is the time taken by the train at a constant velocity; it can be calculated as.

$\Delta x=x_0+v_i{t}_c$

Substitute the values as$13528.218\mathrm{m}$for$\Delta x$, 0 m for$x_0$, and$26.388\mathrm{m}/\mathrm{s}$for$v_i$in the above equation.

$\begin{array}{c}13528.218\mathrm{m}=0 \mathrm{m}+26.388\mathrm{m}/\mathrm{s}\times t_c\\ t_c=512.665\mathrm{s}\end{array}$

The train stops at the intermediate stations for 22 s. From the above figure, it can be observed that the train stops at stations 2 and 3. So, the total time taken by the train to stop (take rest) at each station can be given as.

$\begin{array}{c}\Delta t=2\times 22\mathrm{s}\\ =44\mathrm{s}\end{array}$

From equation (iii), the total distance traveled by the train can be calculated as.

$t_1=\left(t_i+t_f\right)n_1+t_c+\Delta t$

Substitute the values as$23.989\mathrm{s}$for$t_i$,$13.194\mathrm{s}$for$t_f$, 3 for n₁,$512.665\mathrm{s}$for$t_c$, and$44\mathrm{s}$for$\Delta t$in the above equation.

$\begin{array}{c}{t}_1=\left(23.989\mathrm{s}+13.194\mathrm{s}\right)3+512.665\mathrm{s}+44\mathrm{s}\\ =668.214\mathrm{s}\end{array}$

Thus, the total time taken by the train to make a 15 km trip in the second situation is$668.214\mathrm{s}$.