so lets say 2 trains are moving in opposite directions with one going x m/s while train two is going y m/s and they are d distance away from each other. At what point do they cross each other? Just for example, pretend x is 5 m/s while y is 3 m/s or something and d is 100m. What is the equation you need to use to find the answer to these problems?
+130466 Let the speed of one be x m/s .......and let the other be y m/s......and let the distance between them be d
Tnen ......Rate x Time = Distance......and each of them travels the same amount of time
So..... x * t + y * t = d or, solving for t, we have t (x + y) = d → t = d / (x + y)
So....the distance that the train moving at x m/s will tavel is just x * (d / [ x + y] )
And the distance that the train moving at y m/s will travel is just y * (d / [x + y])
Let's use your example:
3*t + 5*t = 100
t = 100/(3 + 5) = 100/8 = 12.5 sec
So....this is how long before the trains meet
The train moving at 3m/s travels 3(12.5) = 37.5m
And the train moving at 5m/s travels 5(12.5) = 62.5m
So.....they meet at a point either 37.5m from where the slower train starts or at a point 62.5m from where the faster train starts, depending upon your preference.....
+130466 Let the speed of one be x m/s .......and let the other be y m/s......and let the distance between them be d
Tnen ......Rate x Time = Distance......and each of them travels the same amount of time
So..... x * t + y * t = d or, solving for t, we have t (x + y) = d → t = d / (x + y)
So....the distance that the train moving at x m/s will tavel is just x * (d / [ x + y] )
And the distance that the train moving at y m/s will travel is just y * (d / [x + y])
Let's use your example:
3*t + 5*t = 100
t = 100/(3 + 5) = 100/8 = 12.5 sec
So....this is how long before the trains meet
The train moving at 3m/s travels 3(12.5) = 37.5m
And the train moving at 5m/s travels 5(12.5) = 62.5m
So.....they meet at a point either 37.5m from where the slower train starts or at a point 62.5m from where the faster train starts, depending upon your preference.....
+118696 I was already lookin when Chris answered so I am going to finish my answer too :)
I am going to let the train that is travelling at x km/hour be at point 0
and I am going to let the train that is travelling at -y km/hour be at d
AT any point T in time the trains will be at point xT and d-yT
They will be at the same point when
xT=d-yT
xT+yT=d
T(x+y)=d
T=d/(x+y)
At this point in time the trains will be at dx / (x+y)
It is easier to work out this formula by doing it with the numbers first. 
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Using the fromula for you situation
x=5 y=3 and d=100
100*5 / 8
=500/8
= 62.5 km from the x start point
Just as Chris said :))
