Relative speed = | = (45 + 30) km/hr | |||||||
|
||||||||
|
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 |
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 | = (x + 50) km/hr | |||||||
|
||||||||
|
Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
|
⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
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 |
Speed in km/ hour =200
Speed in m/sec = 200 x 5/18 = 500/9 m/sec = 55.55 m/sec
Let the length of train be x meter and speed be y m/sec
Therefore, time taken to cross the pole = x/y sec
and time taken to cross the platform = ( Length of train + Length of platform) / speed of train
Therefore x/y = 10
And, (x+200) / y = 20
? 10y + 200 = 20 y
? speed of train, y = 20 m/sec
And length of rain = x = 10y = 200m
Relative speed, when the trains are running in the opposite direction = Sum of the speed
Therefore, Relative speed = 36 km/hr +72 km/hr = 90 km/hr
Relative speed in m/sec = 90 km/hr x 5/18= 25 m/sec
Distance covered = sum of the lengths of the train = 200 m+ 300 m = 500m
Therefore, time taken = 500m / 25 m/sec = 20 sec
Speed of the train = (60 x 5/18) m/sec = 50/3 m/sec
Distance covered in passing the standing person = 200m
Required time taken= =200/(50/3) = 200 x (3/50) = 12 sec
Speed of the train = (264 x 5/18) m/sec = 220/3 m/sec
Distance covered in passing the platform = (330+330) m= 660m
Time taken= ( 660 / (220/3) ) sec = (660 x 3/ 220 ) = 9 sec
Length of train = 264 m
Time taken to cross the pole = 12 sec
Speed of the train, in m/sec = Length of the train/ Time taken to cross the pole
Therefore, speed in m/sec= (264 / 12) m/sec = 22 m /sec
Calculating speed in km/ hour = (22 m/sec ) x 18/5 = 396/5 km/hr = 79.2 km/ hour
Copyright ©CuriousTab. All rights reserved.