Two trains are moving on straight tracks in opposite directions. They are approaching a single-track section of length $L$ where only one train can drive at a time. A cargo train is coming from the left at speed $v_0$, and reads the sign about the upcoming single-track section at distance $L$. It starts decelerating immediately and loses half its speed over the distance $L/2$. Simultaneously, a passenger train is coming from the right. It has priority to use the single-track and maintains constant speed of $2v_0$. Both trains are a distance $L$ away from the single-track section at time $t=0$ as shown in the figure below.

As a first approximation, both trains can be considered as point masses.

1. Without any calculation, predict whether or not there would be a collision between the two trains. Explain why.

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After a distance $L/2$, the cargo train has halved its speed. As long as its acceleration remains constant, we can conclude that its speed will drop to zero after traveling a distance $L/2$ again. Thus, the cargo train will come to a full stop after traveling a distance $L$ which means that it will not enter the single-track section and there is no risk of collision between the trains.

2. Determine the acceleration of the cargo train, right after it starts decelerating.

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The acceleration $a_1$ of the cargo train is such that over the distance $L/2$, its initial velocity $v_0$ decreases to $v_0/2$. Thus, we solve for $a_1$ as follows

\begin{aligned} v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{v^2_0}{4}=v^2_0+2a_1\cdot \frac{L}{2} \\ \\ &\Rightarrow \ \ \ \ \ a_1L=\frac{v^2_0}{4}-v^2_0 \\ \\ &\Rightarrow \ \ \ \ \ a_1=-\frac{3v^2_0}{4L} \end{aligned}

The acceleration of the cargo train is equal to

\boxed{a_1=-\frac{3v^2_0}{4L}}

where the negative sign is consistent with the cargo train slowing down.

3. Determine how long it takes until the passenger train exits the single-track section.

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To clear the single-track section, the passenger train must travel a distance $2L$ at a constant speed $2v_0$ which requires an amount of time $t_f$ equal to

\boxed{t_f=\frac{2L}{2v_0}=\frac{L}{v_0}}

4. During the time found in part 3., what is the distance traveled by the cargo train? Does that confirm your prediction made in part 1.?

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The position of the cargo train as it slows down is given by

x\left(t\right)=\frac{1}{2}a_1t^2+v_{0x}t=-\frac{3v^2_0}{8L}t^2+v_0t

In the time $t_f$ it takes the passenger train to clear the single-track section, the cargo train will travel a distance equal to

\boxed{x\left(t_f\right)=-\frac{3v^2_0}{8L}t^2_f+v_0t_f=-\frac{3v^2_0}{8L}{\left(\frac{L}{v_0}\right)}^2+v_0\cdot \frac{L}{v_0}=-\frac{3L}{8}+L=\frac{5L}{8}\ <\ \ L}

This distance is less than the distance $L$ separating the cargo train from the single-track section and therefore there is no risk of collision since the passenger train will clear the single-track section before the cargo train even reaches it.

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