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PROBLEMS ON TRAINS

PROBLEMS ON TRAINS -> IMPORTANT FORMULAE

1. a km/hr = [a * 5/18]m/s.
2. a m/s = [a * 18/5] km/hr.
3. Time taken by a trian of length l metres to pass a pole or a 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 stationary object of length b metres is the time taken by the train to cover (l + b) metres.
5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relatives speed = (u - v) m/s.
6. Suppose two trains or two 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
8. If two trains of length a metres and b metres are moving in the same direciton 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.
9. If tow trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then
(A’s speed) : (B’s speed) = (√b : √a).

PROBLEMS ON TRAINS -> SOLVED EXAMPLES

1. Two tain 100 metres and 120 metres long are running in the same direction with speeds of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?
  Sol. Relative speed of the train = (72 - 54) km/hr = 18 km/hr
= [18 * 5/18] m/sec = 5 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (100 + 120) m at 5 m/sec = [220/5]sec = 44 sec.
2. A train 220 m long is running with a speed of 59 kmph. In what time will it pas a man who is running at 7 kmph in the direction opposite to that in which the tain is going?
  Sol. Speed of the train relative to man = (59 + 7) kmph
= [66 * 5/18] m/sec = [55/3] m/sec.
Time taken by the train to cross the man
= Time taken by it to cover 220m at [55/3] m/sec
= [220 * 3/55] sec = 12 sec.
3. A man siting in a trian which is travelling at 50 kmph observes that a goods trian, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.
  Sol.
Relative speed = [280/9] m/sec = [280/9 * 18/5] kmph = 112 kmph.
∴ Speed of goods train = (112 - 50) kmph = 62 kmph.

PROBLEMS ON TRAINS -> EXERCISE

19. Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at speed of 25 kmph. At what time will they meet?
 
  • A. 9 a.m.
  • B. 10 a.m.
  • C. 10.30 a.m.
  • D. 11 a.m.
Ans: B.
Sol.
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20 x 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.
 
20. The length of the bridge, which a train 130 metres long and travelling at 45 km / hr can cross in 30 seconds, is
 
  • A. 200 m
  • B. 225 m
  • C. 245 m
  • D. 250 m
Ans: C.
Sol.
Speed = [45x5/18] m/sec = (25/2)m/sec; Time = 30 sec.
Let the length of bridge be x metres.
Then, 130 + x / 30 = 25 / 2
⇔ 2(130 + x) = 750
⇔ x = 245 m.
 
 
21. In what time will a train 100 metres long cross an electic pole, if its speed be 144 km/hr?
 
  • A. 2.5 sec
  • B. 4.25 sec
  • C. 5 sec
  • D. 12.5 sec
Ans: A.
Sol.
Speed = [144x5/18] m/sec = 40 m / sec.
Time taken = (100/40)sec = 2.5 sec.