A train 120 m long, travelling at 90 km/hr overtakes another train travelling in the same direction at 72 km/h

130 m 2.78 sec

250m......and with respect to first train 5s and second train around 3s

90 km/hr is 25 mps (90,000 meters / 3600 seconds). This means that your train would pass a staionary point in 4.8 seconds. The second train is traveling at 20 mps (72,000 meters / 3600 seconds). The differential is all that matters at this point. The difference between the two is 5 meters per second, so eliminate all the fancy stuff by subtracting the differential velocity. Now the question becomes "if a 120 m train passes a stationary train at 5 mps in 50 seconds, how long is the stationary train?" Since a 120 meter train would pass a single point at 5 mps in 24 seconds, you then subtract the 24 seconds and you now start timing "how long it takes the end of your train to reach the front of the stationary train". You know that will be 26 seconds, so 26 seconds x 5 mps is 130 meters. So your train will take 24 seconds to completely pass the end of the other train, and 26 more seconds for the end of your train to pass the front of the other train. There is your 50 seconds, and the total length must be 250 meters. Alternative method: 50 seconds times 5 mps differential velocity yields 250 meters. 250 - 120 (length of your train) yields 130 (length of other train). Given this answer to the first part, you can easily figure out the second part.

(1): v1 = 90 km/h = 25 m/s. v2 = 72 km/h = 20 m/s. So the relative speed of the first train is v = v1 - v2, v = 25 - 20, v = 5 m/s. If L1 is the length of the first train and L2 is the length of the second train then: L1 + L2 = vt, 120 + L2 = 5x50, 120 + L2 = 250, L2 = 250 - 120, L2 = 130 m. (2): Now the relative speed is v' = v1 + v2, v' = 25 + 20, v' = 45 m/s. t' = (L1 + L2)/v', t' = (120 + 130)/45, t' = 250/45, t' = 50/9 s or t' = 5,55 s.