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. |
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 |
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
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.
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 = | ❨ | 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 |
4.5 km/hr = | ❨ | 4.5 x | 5 | ❩ | m/sec = | 5 | m/sec = 1.25 m/sec, and |
18 | 4 |
5.4 km/hr = | ❨ | 5.4 x | 5 | ❩ | m/sec = | 3 | m/sec = 1.5 m/sec. |
18 | 2 |
Let the speed of the train be x m/sec.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
⟹ 8.4x - 10.5 = 8.5x - 12.75
⟹ 0.1x = 2.25
⟹ x = 22.5
∴ Speed of the train = | ❨ | 22.5 x | 18 | ❩ | km/hr = 81 km/hr. |
5 |
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, | 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 |
Then, | x | = 15 ⟹ y = | x | . |
y | 15 |
∴ | x + 100 | = | x |
25 | 15 |
⟹ 15(x + 100) = 25x
⟹ 15x + 1500 = 25x
⟹ 1500 = 10x
⟹ x = 150 m.
Speed = | ❨ | 300 | ❩ | m/sec = | 50 | m/sec. |
18 | 3 |
Let the length of the platform be x metres.
Then, | ❨ | x + 300 | ❩ | = | 50 |
39 | 3 |
⟹ 3(x + 300) = 1950
⟹ x = 350 m.
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