P01R2020 – Trains Crossing

Train $A$ travels at constant speed and therefore is acceleration is zero. Thus, the position of train $A$ is given by the kinematic equation

x_A\left(t\right)=\frac{1}{2}a_At^2+v_At+x_{0A}=v_At

Similarly, train $B$ travels at constant speed and therefore is acceleration is zero. Thus, the position of train $B$ is given by the kinematic equation

x_B\left(t\right)=\frac{1}{2}a_Bt^2-v_Bt+x_{0B}=-v_Bt+d

where the negative sign is required to describe the fact that train $B$ moves to the left.

The trains then cross each other at time $t_1$ such that $x_A\left(t_1\right)=x_B(t_1)$ and we solve for $t_1$ as follows

\begin{aligned}
x_A\left(t_1\right)=x_B\left(t_1\right)\ \ \ \ &\Rightarrow \ \ \ \ \ v_At_1=-v_Bt_1+d \\
\\
&\Rightarrow \ \ \ \ \ \left(v_A+v_B\right)t_1=d \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{d}{v_A+v_B}
\end{aligned}

The trains cross at time $t_1$ equal to

\boxed{t_1=\frac{d}{v_A+v_B}=\frac{180}{20+12}=5\ s}

The location of the trains when they go past each other is given by

x_A\left(t_1\right)=v_At_1=20\cdot 5=100\ m

and their distance to the station is therefore equal to

\boxed{d_S=D+d-x_A\left(t_1\right)=320+180-100=400\ m}

Note: you could use $x_B\left(t_1\right)$ instead since $x_A\left(t_1\right)=x_B\left(t_1\right)$ when the trains cross.

Train $A$ is moving at an initial speed $v_A=20\ m/s$ when it goes past train $B$ and begins to apply the brakes. Its acceleration $a_A$ is then such that it stops at the station after traveling the $d_S=400\ m$ that separates it from the station.

Thus, the acceleration $a_A$ must satisfy the following equation and can be solved for as follows

\begin{aligned}
v^2_{fA}=v^2_A+2a_A\mathrm{\Delta }x\ \ \ \ &\Rightarrow \ \ \ \ \ 0=v^2_A+2a_A\left(d_S-0\right) \\
\\
&\Rightarrow \ \ \ \ \ a_A=-\frac{v^2_A}{2d_S}
\end{aligned}

The acceleration $a_A$ of train $A$ is therefore equal to

\boxed{a_A=-\frac{v^2_A}{2d_S}=-\frac{{20}^2}{2\cdot 400}=-0.5\ m/s^2}

where the negative sign is consistent with train $A$ slowing down to a stop.

Train $A$ begins to slow down from an initial velocity $v_A$ with an acceleration $a_A$ and its velocity at any time $t$ after beginning to slow down is therefore given by

v_A\left(t\right)=a_At+v_A=-0.5t+20

The time $t_S$ it takes train $A$ to come to a stop is such that $v_A\left(t_S\right)=0$ and can be derived as follows

\begin{aligned}
v_A\left(t_S\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -0.5t_S+20=0 \\
\\
&\Rightarrow \ \ \ \ \ t_S=\frac{20}{0.5}=40\ s
\end{aligned}

Thus, we conclude that train $A$ reaches the station, coming to a full stop, in an amount of time

\boxed{t_S=40\ s}

After passing train $A$, train $B$ continues at constant speed $v_B=16\ m/s$ for an additional $t_S=40\ s$ thus traveling an extra distance

d=v_Bt_S=16\cdot 40=640\ m

Adding the $v_Bt_1=80\ m$ traveled in the first $5\ s$, we conclude that the total distance traveled by train $B$ when train $A$ reaches the station is equal to

\boxed{d_B=v_Bt_S+v_Bt_1=16\cdot 40+16\cdot 5=720\ m}