An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road. If the train's speed is 75 km/h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2–36. What are the results if the car and train are traveling in opposite directions?

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be [tex]v_{AT}=v_A-v_T[/tex] (the train must see the car advancing at a lower speed), where [tex]v_A[/tex] is the speed of the automobile and [tex]v_T[/tex] the speed of the train.

So we have [tex]v_{AT}=(95km/h)-(75Km/h)=20Km/h[/tex].

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

[tex]t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s[/tex]

And in that time the car would have traveled (relative to the ground):

[tex]d=v_At=(95Km/h)(0.065h)=6.175Km[/tex]

If they are traveling in opposite directions, we have to do all the same but using [tex]v_{AT}=v_A+v_T[/tex] (the train must see the car advancing at a faster speed), so repeating the process:

[tex]v_{AT}=(95km/h)+(75Km/h)=170Km/h[/tex]

[tex]t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s[/tex]

[tex]d=v_At=(95Km/h)(0.00765h)=0.73Km[/tex]