Let the length of each train be L m.
Relative speed = 46 - 36 = 10 km/h = 10 x 5/18 m/s = 25/9 m/s
Sum of length of train / Relative speed of train = Time taken
? 2L/(25/9) = 36
? 2L = (36 x 25)/9 = 100
? L = 50 m.
Speed = 60 km/h = 60 x (5/18) m/s = 50/3 m/s
Required time = (110 + 170) / (50/3) = (280 x 3)/50 = 16.8 s
Total length of the train = 105 + 90 = 195 m
Relative speed = 72 + 45 km/h = 117 km/h = 117 x (5/18) = 585/18 m/s
? Required time = (195 x (18/585) s = 6 s
Relative speed of train = 400/36 m/s = 400/36 x (18/5) = 40 km/h
Relative speed of train = Speed of train + Speed of man
? 40 = Speed of train + 20
? Speed of train = 40 - 20 = 20 km/h
According to the formula .
Required time = (x + y) / (u + v)
Here, x = 70 m, y = 90 m, u = 10 m/s and v = 6 m/s
? Required time = (70 + 90) / (10 + 6) = 160/16 = 10 s
Let length of the train = L
According to the question,
(L + 300)/21 = (L + 240)/18
? (L + 300)/7 = (L + 240)/6
? 6L + 1800 = 7L + 1680
? L = 120 m
Taking the length of the 2nd bridge into consideration,
Speed of train = (L+ 240)/18 = (120 + 240)/18 m/s
= (360/18) x (18/5) km/h
= 72 km/h
Let length of each train be L m.
Then. (65 + 85 ) x 5/18 = (L + L)/6
? (150 x 5 x 6)/18 = 2L
? 2L = 250
? L = 125 m
Given that, T1 = 9 h and T 2 = 4 h
According to the formula.
(1st train's speed) : (2nd train's speed) = ?4 : ?9
= 2 : 3
Let speed of 1st train be 8S and speed of 2nd train be 9S
According to the question.
Speed of second train = 360/4 = 90 km/h
9S = 90
? S = 10
So, speed of 1st train = 8 x 10 = 80 km/h
? Required distance = 80 x 3 = 240 km
Let the trains meet after time t at a distance x from station N.
Then another train coming from station M covers a distance of (x + 50)
For station M,
(x + 50) = 125t
? x = 125t - 50 ..(i)
For station N,
x = 75t .....(ii)
From Eqs. (i) and (ii), we get
75t = 125t - 50
? t = 1h
Distance between station M and N = 125 + 75t = 200 x 1 = 200 km
Because of stoppages, train covers 18 km less per hour .
? Time taken to cover 18 km = 18/108 = 1/6 h = (1/6) x 60 = 10 min
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