length of tunnel + length of train = distances covered by train in 1 min
? length of tunnel + 700 = 72/60 km = 1.2 km = 1200 m
? length of tunnel = 1200 - 700 = 500 m
Distance covered = 220 + 280 = 500 = 0.5 km
Speed = 60 kmph.
Time taken = 0.5 / 60 hr = (0.5 x 60 x 60)/60 sec
= 30 sec
? Speed of the train = 150/12 m/sec.
Required time taken to cross the bridge = [(150 + 250) x (12/150)] = 32 sec.
Distance covered = 200 + 160 = 360.
Speed of train = 72 km = 72 x (5/18) = 20 sec.
Time = 360 / 20 = 18 sec.
? Speed of the train = 72 km/hr = 72 x (5/18) = 20 m/sec.
Sum of the length of the train and tunnel = 540 + 160 = 700 metres.
? Required time taken to pass the tunnel = Time taken to cover 700 metres at 20 m/sec.
= 700/20
= 35 sec.
Speed of the column of men = (50 x 75) / (100 x 60) = 5/8 m/sec.
Let the length of the street be L metres
? (L + 250) = (60 x 60 x 5) / 8 = 2250 metres
? L = (2250 - 250) m
? L = 2000 m = 2 km.
Speed of the first train = 90 km/hr = 90 x (5/18) m/sec. = 25 m/sec
Let the length of the train and bridge be L metre.
? L = 25 x 36 = 900 m.....(1)
Again Speed of the second train = 45 km/hr = 45 x (5/18) = 25/2 m/sec.
Now, putting the value from eqn. (1) we have
Time taken = (900 - 100) x 2/25 = 800 x 2/25 = 64 sec
Relative speed of the train = (32 + 40) = 72 km/hr = 72 x 5/18 = 20 m/sec.
Sum of length of the train = 132 + 108 = 240 m
time taken by the trains in passing each other = 240/20 = 12 sec .
Speed of the train = 72 x (5/18) = 20 m/sec
Let the length of tunnel = L meters
Then, (700 + L )/ 60 = 20
? 700 + L = 1200
? L = 500 meters
Speed = 36 x (5/18) = 10 m/sec
Let the length of the bridge = L m
Distance covered = (100 + L) m
Time taken = 25 sec
? (100 + L ) / 25 = 10 m/sec
? 100 + L = 250
? L = 150 meter
Let the length of the first train be x m.
Then, the length of second train is (x/2) m
? Relative speed = (48 + 42) km/h = (90 x 5/18) m/s = 25 m/s
According to the question. (x + x/2) / 25 = 12
? 3x/2 = 300
? x = 200 m
? Length of first train = 200 m
Let the length of platform be y m .
Speed of the first train = (48 x 5/18) m/s = 40/3 m/s
Time = Distance / Speed
? (200 + y) x 3/40 = 45
? 600 + 3y = 1800
? y = 400 m
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