Question

Two trains travel directly toward each other. One of the trains travels at a rate of 12 km/h while the other travels at a rate of 20 km/h. When the trains are 72 km apart a conductor at the front of one of the trains releases the insane pigeon Hyde. Hyde flies first from the slower of the two trains to the faster train at which point Hyde doubles back toward the slower train. Hyde continues to fly back and forth between the trains as they approach, always at a constant speed of 48 km/h. Assuming the trains never change speed until they meet and magically stop, how many kilometers has Hyde flown when the trains meet?

Answer 1

Write the equation of the first train: x = 12t
Write the equation of the second train: y = -20t+72, assuming the train one is at the origins when t = 0. 
Now solve the equation: 12t=-20t+72, 
32t = 72, then t = 2.25 hours. The two train meet after 2.25 hours. 
Now the equation of the pigeon is like this: x = 48t. 
Then for t = 2.25, we get
 $x=2.25\times48=108 km$

The Hyde has flown 108 km when the trains meet.