Quadratics/Algebra word Problem (based on physics)

Question

There is a problem from an old high school math book I have. It appears in a chapter on quadratics and polynomials. It uses constant velocity, so hence it is physics based type of problem.

Over the years I have come back to this problem, and could never figure it out. I once took this problem to a university Masters student studying "Algebraic Geometry", but he gave me a very convoluted answer, or I should maybe say he gave me an abstract method to solve this, which later I could not figure out.

So here is the question:

Two runners start at the opposite sides of a $60$ meter field. One person runs at $4$ m/s, the other person runs at $5$ m/s. If they run back and forth for $12$ minutes, how many times with they pass each other.

Answer in the book is: They pass each other $54$ times.

I have no idea how to model this problem.

Hope someone knows how to figure this problem out.

Regards,

P

Answer 1

The faster runner crosses the field 60 times, and the slower runner 48 times. Since they each cross the field an even number of times, they end up where they started. Since the faster runner crosses the field more often, he must overtake the runner sometimes. (This is not counted as a cross.) In order for him to overtake a second time, the runners must first cross. So, the $12$ extra laps consist of $6$ overtakes and $6$ crosses, and there are $48 + 6 = 54$ crosses in all.

Answer 2

In the first minute the faster runner runs the field 5 times while the slower runners runs the field 4 times, thus they cross 4 times + 1 time and after a minute the start toghether from the same side. In the second minute they cross again 4 times and then at the end of the second minute the are again at opposite sides. Since the same cicle of 2 minute repeates 6 times,they cross $9\times6=54$ times.

Answer 3

Relative to one of the runners, the other runner approaches their next meeting at a speed of $4+5=9$m/s. So if you imagine one of the runners is stationary from the beginning, the other runner is doing all the running at $9$m/s.

In $12$ minutes this runner will cover a distance of $9\times 12\times 60$ metres and he meets the stationary runner every $120$ metres, so the number of times they meet is $$\frac{9\times 12\times 60}{120}=54$$