Relative speed = (40 - 20) km/hr = | ❨ | 20 x | 5 | ❩ | m/sec = | ❨ | 50 | ❩ | m/sec. |
18 | 9 |
∴ Length of faster train = | ❨ | 50 | x 5 | ❩ | m = | 250 | m = 27 | 7 | m. |
9 | 9 | 9 |
Formula for converting from km/hr to m/s: X km/hr = | ❨ | X x | 5 | ❩ | m/s. |
18 |
Therefore, Speed = | ❨ | 45 x | 5 | ❩m/sec | = | 25 | m/sec. |
18 | 2 |
Total distance to be covered = (360 + 140) m = 500 m.
Formula for finding Time = | ❨ | Distance | ❩ |
Speed |
∴ Required time = | ❨ | 500 x 2 | ❩sec | = 40 sec. |
25 |
Speed = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
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.
Relative speed | = (x + 50) km/hr | |||||||
|
||||||||
|
Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
|
⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Then, | x | = 8 ⟹ x = 8y |
y |
Now, | x + 264 | = y |
20 |
⟹ 8y + 264 = 20y
⟹ y = 22.
∴ Speed = 22 m/sec = | ❨ | 22 x | 18 | ❩ | km/hr = 79.2 km/hr. |
5 |
2 kmph = | ❨ | 2 x | 5 | ❩ | m/sec = | 5 | m/sec. |
18 | 9 |
4 kmph = | ❨ | 4 x | 5 | ❩ | m/sec = | 10 | m/sec. |
18 | 9 |
Let the length of the train be x metres and its speed by y m/sec.
Then, | ❨ | x | ❩ | = 9 and | ❨ | x | ❩ | = 10. |
|
|
∴ 9y - 5 = x and 10(9y - 10) = 9x
⟹ 9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
∴ Length of the train is 50 m.
Speed of the train relative to man | = (63 - 3) km/hr | |||||||
= 60 km/hr | ||||||||
|
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|
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∴ Time taken to pass the man |
|
|||||||
= 30 sec. |
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
∴ | (100 + 100) | = 3x |
8 |
⟹ 24x = 200
⟹ x = | 25 | . |
3 |
So, speed of the faster train = | 50 | m/sec |
3 |
= | ❨ | 50 | x | 18 | ❩km/hr |
3 | 5 |
= 60 km/hr.
Relative speed = (40 - 20) km/hr = | ❨ | 20 x | 5 | ❩ | m/sec = | ❨ | 50 | ❩ | m/sec. |
18 | 9 |
∴ Length of faster train = | ❨ | 50 | x 5 | ❩ | m = | 250 | m = 27 | 7 | m. |
9 | 9 | 9 |
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