1. km/hr to m/s conversion
y km/hr = y x 5/18m/s.
2. m/s to km/hr conversion
y m/s = y x 18/5km/hr.
3. Time taken by a train of length l metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres.
4. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.
5. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u  v) m/s.
6. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.
7. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:
The time taken by the trains to cross each other = (a + b)/(u + v) sec.
8. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:
The time taken by the faster train to cross the slower train = (a + b)/(u  v) sec.
1. 

Speed of train = 54 kmph = 54 x 5/18 = 15 m/s.
Distance = length of the train = 180 m.
∴ Time = 180/15 = 12 seconds.
2. 

Distance = length of the train + length of the platform.
distance = 100 + 150 = 250 m.
⇒ 45 x 5/18 = 12.5 m
∴ Time = 250/12.5 = 20 seconds.
3. 

Speed of the train = 54 kmph.
⇒ 54 x 5/18 = 15 m/s
∴ distance covered in 30 seconds.
⇒ 15 x 30 = 450 m
length of bridge = distance covered – length of train.
Hence, length of bridge = 450 – 120 = 330 m
4. 

Total distance covered = sum of length of the two train = 100 + 150 = 250 m
Relative speed of the two trains = 63 – 45 = 18 kmph
(since the trains are running in the same direction the relative speed will be the difference in the speeds).
⇒ 18 x 5/18 = 5 m/s
∴ Time = 250/5 = 50 seconds
5. 

Let the speed of train is s kmph
Relative speed of overtaking first cyclists = (s – 9) kmph
Time took to overtake the first cyclist = 12 seconds
∴ length of train = 12 x (s – 9) x 5/18 →(i)
Similarly, considering the case of overtaking the second cyclist,
∴ length of train = 18 x (s  18) x 5/18 →(ii)
Equating (i) and (ii),
⇒ 12 x (s – 9) x 5/18 = 18 x (s – 18) x 5/18
⇒ 2s – 18 = 3s – 54
⇒ s = 36 kmph.
Now, length = 12 x (s – 9) x 5/18 = 12 x 27 x 5/18
∴ length = 90 m
6. 

Speed = (72 x 5/18) m/s = 20 m/s.
Total distance covered = (120 + 130) m = 250 m
∴ Required time = (250/20) sec = 12.5 sec.
7. 

Speed = 45 x (5/18) m/s = 25/2 m/s
Total distance covered = (260 + 240) m = 500 m
∴ Required time = 500 x (2/25) sec = 40 sec.
8. 

Speed = 60 x (5/18) m/s = (50/3) m/s.
Length of the train = (Speed x Time) = (50/3 x 18) m = 300 m.
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