Given, a = 45 km/h, y = ?, t1 = 4 h 48 min and t2 = 3 h 20 min
Using y = a?t1 / t2= 45?4 h and 48 min /3 h and 20 min
= 45?(24/5h) / (10/3 h) = 45?24 x 3/5 x 10
= 45 x ?1.44 = 45 x 12 = 54 km/h
Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec
? 3L / (2 x 25) = 12
? 3L = 600
? L = 200
? Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.
? 200 + M = 900
? M = 700 meters
Length of platform = 700 meters.
Relative speed =280/9 m / sec = (280/9*18/5) kmph = 112 kmph.
Speed of goods train = (112 - 50) kmph = 62 kmph.
Note : If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then: (A's speed) : (B's speed) = (b : a)
Therefore, Ratio of the speeds of two trains = = 3 : 2
Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.
Therefore, Man's rate against the current = (12.5 - 2.5) = 10 km/hr.
Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v + v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10 × 36/10 km/hr = 36 km/hr
108 kmph = 108*[5/18] m/sec = 30 m / s.
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