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Homework Help: Kinematics - Trains crossing eachother

  1. Sep 13, 2006 #1
    ok so, i have a problem.. it says:

    two trains have the eame length L=150m, are traveling in a parallel itinerary. the first is going with 60 km/s, the second 90km/s.

    1) the trains are going in the oppsite way. how much time is it gonna take them to cross each other?
    2) same question, but this time, they're going on the same way.

    well, for the first question:
    the first trrain is gonna take t=0.15/60 = 0.0025 hour to pass an immobile point outside the train. the second train is gonna take 0.0017hour. I just substracted 0.0025 - 0.0017 to get 0.0008 hour. and that's the time the crossing is gonna take. am i right?
  2. jcsd
  3. Sep 13, 2006 #2
    Last edited: Sep 13, 2006
  4. Sep 13, 2006 #3
    please? !!
  5. Sep 13, 2006 #4


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    Going in opposite directions, the two trains have speed 90+ 60= 150 km/h relative speeds. I assume that to "cross each other" means from the time their noses meet until their tails pass. That means the nose of one must pass from the nose to the tail of the other, 150 m, then its tail move past, another 150 m a total of 300 m. At a relative speed of 150 km/h= 150000 m/h= 150000/3600 m/s= 41.7 m/s, it will 300/41.7= 7.2 seconds.

    Going in the same direction, they have a relative speed of 90- 60= 30km/h= 30000 m/h= 30000/3600= 8.3 m/s. At 8.3 m/s a train will cross the 300 m necessary to pass in 300/8.3= 36 seconds.
  6. Sep 13, 2006 #5
    What you have calculated is the time it takes for each train to get past a fixed point. After 0.0017hr, the second train is just past it. However, the first is only 2/3 past it, as it is travelling at 2/3 the speed. So they are not properly past each other.

    You want the time it takes for the back of one train to get past the back of the other. The clock starts when they are nose to nose- what is the distance between the backs of the trains then? If one train were stationary, what speed would the other be travelling at to compensate? You can use these values to calculate the time.
  7. Sep 14, 2006 #6
    yup!! I got it! thank you very much =)
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