Speed = | ❨ | 72 x | 5 | ❩m/sec | = 20 m/sec. |
18 |
Time = 26 sec.
Let the length of the train be x metres.
Then, | x + 250 | = 20 |
26 |
⟹ x + 250 = 520
⟹ x = 270.
Total distance covered |
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∴ Time taken |
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= 3 min. |
Speed = | ❨ | 60 x | 5 | ❩m/sec | = | ❨ | 50 | ❩m/sec. |
18 | 3 |
Length of the train = (Speed x Time).
∴ Length of the train = | ❨ | 50 | x 9 | ❩m = 150 m. |
3 |
= | ❨ | 66 x | 5 | ❩m/sec |
18 |
= | ❨ | 55 | ❩m/sec. |
3 |
∴ Time taken to pass the man = | ❨ | 110 x | 3 | ❩sec = 6 sec. |
55 |
Relative speed = (60 + 40) km/hr = | ❨ | 100 x | 5 | ❩m/sec | = | ❨ | 250 | ❩m/sec. |
18 | 9 |
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = | ❨ | 300 x | 9 | ❩sec | = | 54 | sec = 10.8 sec. |
250 | 5 |
Speed of the first train = | ❨ | 120 | ❩ | m/sec = 12 m/sec. |
10 |
Speed of the second train = | ❨ | 120 | ❩ | m/sec = 8 m/sec. |
15 |
Relative speed = (12 + 8) = 20 m/sec.
∴ Required time = | [ | (120 + 120) | ] | sec = 12 sec. |
20 |
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.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
∴ 20x + 25(x - 1) = 110
⟹ 45x = 135
⟹ x = 3.
So, they meet at 10 a.m.
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
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 |
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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