A straight line connects City A at $(0, 0)$ to City B, 300 meters away at $(300, 0)$. At time $t=0$, a bullet train instantaneously sets out from City A to City B while another bullet train simultaneously leaves from City B to City A going on the same train track. Both trains are traveling at a constant speed of $50$ meters/second. 

Also, at $t=0$, a super 
y stationed at $(150, 0)$ and restricted to move only on the train tracks travels towards City B. The 
y always travels at 60 meters/second, and any time it hits a train, it instantaneously reverses its direction and travels at the same speed. At the moment the
trains collide, what is the total distance that the y will have traveled? Assume each train is a point
and that the trains travel at their same respective velocities before and after collisions with the y
1. **Determine the time until the trains collide:**
   - The two trains are traveling towards each other from City A and City B, which are 300 meters apart.
   - Each train travels at a constant speed of 50 meters per second.
   - The combined speed of the two trains is \(50 + 50 = 100\) meters per second.
   - The time \(t\) it takes for the trains to collide is given by:
     \[
     t = \frac{\text{distance}}{\text{combined speed}} = \frac{300 \text{ meters}}{100 \text{ meters/second}} = 3 \text{ seconds}
     \]

2. **Calculate the distance traveled by the y:**
   - The y starts at the midpoint \((150, 0)\) and travels towards City B at a speed of 60 meters per second.
   - The y will travel for the same amount of time it takes for the trains to collide, which is 3 seconds.
   - The distance \(d\) traveled by the y is given by:
     \[
     d = \text{speed} \times \text{time} = 60 \text{ meters/second} \times 3 \text{ seconds} = 180 \text{ meters}
     \]

3. **Conclusion:**
   - The total distance that the y will have traveled by the time the trains collide is 180 meters.

The final answer is \(\boxed{180}\) meters.