# To and fro motion of two people on a straight track.

Question: Two persons start from opposite ends of a $$90 \,\text{km}$$ straight track and run to and fro between the two ends. The speed of the first person is $$108 \; \text{km/hour}$$ while that of the second person is $$75\; \text{km/hour}$$. They continue their motion for $$10$$ hours. How many times do they pass each other?

My Attempt: It is easy see that for the first time they will meet after $$\frac{90}{108+75}$$ hours, after that they collectively would travel $$2 \cdot 90$$ km further at the speed of $$183$$ km/h to meet for the second time. Using this we can say that they meet for the $$n^{\text{th}}$$ time after $$\frac{(2n-1) \cdot 90}{108+75} \text{hours}$$ By putting in some values we see that to meet for the $$10^{\text{th}}$$ time, the two guys would require $$\approx9.344$$ hours and to meet for the $$11^{\text{th}}$$, they would require $$\approx 10.327$$ hours. This ultimately means that they meet $$10$$ times during their motion.

However, my friend used another method and got $$12$$ as the answer. He said that in $$10$$ hours the first person would have travelled $$1080$$ km which ultimately means that he would have traversed the distance of $$90$$ km track $$12$$ times and hence would have met the other guy $$12$$ times.

I want to know why our answers are differing and which one is right?

Intuitively I expected you to be correct. However after some messy graphical analysis it turns out your friend is.

The crossings are at [0.49, 1.48, 2.46, 2.73, 3.44, 4.43, 5.41, 6.39, 7.38, 8.18, 8.36, 9.34] hrs (2 d.p.). There are 12 in total. Your equation misses Crossings 4, 10 and 12 at 2.73, 8.18 and 9.34 respectively.

I think the reason for this is that sometimes they do not have to cover 180 miles for their next meeting.

I expect the distance (90 km) being an even factor of the total time (10 hours) multiplied by the faster speed (108 kmph) is also a reason why your friends solution works. (1080/90 = even whole number). However I'm not sure.

The fourth crossing

If you want to make this yourself the inputs were

u(x)=90+ (-1)^(m+1)floor((m)/(2))180+ (-1)^(m)108x

and

t(x)= (-1)^(n)floor((n)/(2))180+ (-1)^(n+1)75x

set up sliders and set n=3. flip between m=3 and m=4 to see the first 'short' crossing.