Relative speed = | = (45 + 30) km/hr | |||||||
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We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
∴ Required time = | ❨ | 500 x | 6 | ❩ | = 24 sec. |
125 |
Speed of the train relative to man = | ❨ | 125 | ❩m/sec |
10 |
= | ❨ | 25 | ❩m/sec. |
2 |
= | ❨ | 25 | x | 18 | ❩km/hr |
2 | 5 |
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
∴ x - 5 = 45 ⟹ x = 50 km/hr.
Speed = | ❨ | 54 x | 5 | ❩m/sec = 15 m/sec. |
18 |
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
Then, | x + 300 | = 15 |
36 |
⟹ x + 300 = 540
⟹ x = 240 m.
Speed = | ❨ | 78 x | 5 | ❩ | m/sec | = | ❨ | 65 | ❩ | m/sec. |
18 | 3 |
Time = 1 minute = 60 seconds.
Let the length of the tunnel be x metres.
Then, | ❨ | 800 + x | ❩ | = | 65 |
60 | 3 |
⟹ 3(800 + x) = 3900
⟹ x = 500.
Relative speed | = (x + 50) km/hr | |||||||
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Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
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⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Total distance covered |
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∴ Time taken |
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= 3 min. |
Speed of the train relative to man | = (63 - 3) km/hr | |||||||
= 60 km/hr | ||||||||
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∴ Time taken to pass the man |
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= 30 sec. |
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
= | ❨ | 10 x | 5 | ❩m/sec |
18 |
= | ❨ | 25 | ❩m/sec |
9 |
∴ | 2x | = | 25 |
36 | 9 |
⟹ 2x = 100
⟹ x = 50.
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
∴ | 27x + 17y | = 23 |
x+ y |
⟹ 27x + 17y = 23x + 23y
⟹ 4x = 6y
⟹ | x | = | 3 | . |
y | 2 |
Then, | x | = 15 ⟹ y = | x | . |
y | 15 |
∴ | x + 100 | = | x |
25 | 15 |
⟹ 15(x + 100) = 25x
⟹ 15x + 1500 = 25x
⟹ 1500 = 10x
⟹ x = 150 m.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = | (120 + 120) |
12 |
⟹ 2x = 20
⟹ x = 10.
∴ Speed of each train = 10 m/sec = | ❨ | 10 x | 18 | ❩ | km/hr = 36 km/hr. |
5 |
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