Aptitude - Problems on Trains - Discussion

Length of the train = 125 m.
Speed of the man = 5 km/hr.
Speed of the man in meters = (5*1000)/(60*60)=(50/36)=(25/18).
=1.388 m/sec.

Since train has covered the distance in 10 secs.
Let us calculate the distance covered by the man in 10 secs.
distance covered by the man in 10 secs = 13.88 m.

Therefore the displacement(d) = 125 m+13.88 m = 138.88 m.

Speed of the train = (138.88/10) = 13.88 m/s.

Therefore The Speed of the train in Hours = 13.88 * 60 * 60 = 50000 m/sec = 50 Km/hr.

Train is same dxn and man is also some dxn it means it means speed is minus x-5 = 125/10*18/5 = 50.

Why we relate the speed of man and train length?

Why relative speed? They asked only speed?

1 km = 1000 m/1 hr = 3600 sec = 5/18 but we want to convert just opposite.

So numerator and denominator will be interchange.

We have added the 5 km/hr to the final answer because the speed that comes out is with respect to man and to calculate the actual speed of the train we have added 5 km/hr to the final answer.

X-5 = 45 is right because train and man same side. In same side it is correct.

How he equated 45 to x-5?

What is 5/18 and where its come in?

18/5 is to convert m/s to km/hr.

5/18 is to convert km/hr to m/sec.

1 km = 1000 m
1 hr = 3600 s.

M = 1 km/1000
s =1 hr/3600.

M/s = (3600/1000) km/hr.

= (5/18) km/hr.
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Km/hr = (1000/3600) m/s.

= (5/18) m/s.