Let the speed of the faster train be 'S' kmph
Then speed of the slower train will be '(S-5)' kmph
Time taken by faster train = 350/S hrs
Time taken by slower train = 350/(S-5) hrs
=> S = 30 km/hr.
Speed =[ 72 x (5/18) ]m/sec= 20 m/sec.
Time = 26 sec.
Let the length of the train be x metres.
Then,[ (x+250)/26 ]= 20
=> x + 250 = 520
=> x = 270.
Let 'd' be the distance and 's' be the speed and 't' be the time
d=sxt
45 mins = 3/4 hr and 48 mins = 4/5 hr
As distance is same in both cases;
s(3/4) = (s-5)(4/5)
3s/4 = (4s-20)/5
15s = 16s-80
s = 80 km.
=> d = 80 x 3/4 = 60 kms.
Let the length of the 1st train = L mts
Speed of 1st train = 48 kmph
Now the length of the 2nd train = L/2 mts
Speed of 2nd train = 42 kmph
Let the length of the bridge = D mts
Distance = L + L/2 = 3L/2
Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)
Time = 12 sec
=> 3L/2x25 = 12
=> L = 200 mts
Now it covers the bridge in 45 sec
=> distance = D + 200
Time = 45 sec
Speed = 48 x5/18 = 40/3 m/s
=> D + 200/(40/3) = 45
=> D = 600 - 200 = 400 mts
Hence, the length of the bridge = 400 mts.
Given speed = 63 km/hr = m/s
Let the length of the bridge = x mts
Given time taken to cover the distance of (170 + x)mts is 30 sec.
We know speed = m/s
--> x = 355 mts.
Distance = 70 x 1 ½ = 105 km
Relative Speed = 85 ? 70 = 15
Time = 105/15 = 7 hrs
4:30 + 7 hrs = 11.30 p.m.
We know that
distance= speed * time
d= 100 mts.
As Trains are moving in same direction,
Relative Speed = 60-20 = 40 kmph
--> m/sec
Length of Train= Speed * Time
Length =
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