Let the speed of the slower train = p kmph
ATQ,
Speed of the faster train = (p + 6) kmph
Then,
(p + p + 6) x 5 = 160
10p + 30 = 160
10p = 130
p = 13 kmph
Then, speed of the faster train = p + 6 = 13 + 6 = 19 kmph.
Given L1 = 140 m
L2 = 160 m
S1 = 60 km/hr
S2 = 80 km/hr
From the question we get,
S1 + S2 = (L1 + L2) / T
=> (60 + 80) 5/18 m/s = 140 + 160/T
=> T = 54/7 = 7.71 sec
Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.
[66*(5/18)] m/sec = (55/3) m/sec
Time taken to pass the man = [110*(3/55)]m/sec = 6 sec.
Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec.
[(25/2) x (18/5)] km/hr = 45 km/hr.
Let the speed of the train be 'S' km/hr.
Then, relative speed = (S - 5) km/hr.
S - 5 = 45 => S = 50 km/hr.
It is given train X leave station A at 6:30 am, here it is asked to calculate the distance from A when the trains meet, the
Distance traveled by train left at 6:30 am upto 7:40 am i.e. in 1 hr. 10 min. or 7/6 hours = 30 x 7/6 = 35 km
So train leaving at 7:40 am will meet first train after covering a distance of 35 km. with relative speed of 40-30=10 km/hr.
Hence time taken = 35/10 = 3.5 hours or 3 hours 30 minutes
So distance from A = Distance traveled by 2nd train in 3 hr. 30 min
= 40 x 3.5 = 140 km.
Relative speed = 42 + 36 = 78 km/hr = m/s
Distance = (520 + 520) =1040 mts.
Time = = 48 sec
Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec
x = 31.4208 km/hr and 2x = 62.8416 km/hr.
1h ----- 5 kms
? ------ 60 kms
Time = 12 hrs
Relative Speed = 16 + 21 = 37 kmph
T = 12 hrs
D = S x T = 37 x 12 = 444 kms.
Relative Speed = 60 -40 = 20 x 5/18 = 100/18
Time = 50
Distance = 50 x 100/18 = 2500/9
Relative Speed = 60 + 40 = 100 x 5/18
Time = 2500/9 x 18/500 = 10 sec.
Let the length of each train be x mts.
Then, distance covered = 2x mts.
Relative speed = 36 - 26 = 10 km/hr.
= 10 x 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50 mts.
Let the speed of the faster train be 'X' kmph,
Then their relative speed= X - 48 kmph
To cross slower train by faster train,
Distance need to be cover = (400 + 600)m = 1 km. and
Time required = 180 sec = 180/3600 hr = 1/20 hr.
Time = Distance/Speed
=> 1/20 = 1/(x-48)
X = 68 kmph
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