Aptitude - Problems on Trains - Discussion

@Sarita.

As per relative motion law. If two particle moves in the same direction then you have to subtract their speed and if in opposite direction the add it.

Why the distance are substrated?

I mean 460-360. Please explain in detail.

The simple math is, 46km/1h - 36km/1h= 10km/1h. Now; it can be converted into meter/second that is 10km*1000m/3600sec. 10000m/3600s= 2.778m/1sec.


Now, for the difference of 2.778 meter/1sec the faster train cross the slower train in 36 sec. So, the distance the faster train cross 36sec * 2.778 meter= 100 meters.
This 100 meters distance/way is the length of two trains (faster&slower).
As if, two trains' length is same. So, one train's length is; 100/2= 50 meters.

The relative speed = (46-36)=10km/h.
converted into m/s = 10 * 5/18 = 25/9m/s -------------------------- 46km/h train 1 faster train
(x) 36 sec -------------------------- 36km/h train 2 slower train.
Find out the length of the train (distance) d = s * t.
25/9 * 36 = 100both train.
100/2 find the each train distance,
50.

46-36 = 10,
10*5/18 = 2.77,
2.77*36s = 100,
So, the length of train is 100/2 = 50m.

Length of both the trains is x+x = 2xkm/hr.
Relative speed = 46-36 = 10km/hr.
Time = 36 sec.
L= 10 * (5/18) * 36.
= 100,
2l= 100,
L= 100/2,
L= 50.

As per the given condition, the faster train crosses the slower train. So it means that.

First,
The fast train crosses the slow train of distance x metre.

Second,
Note (the fast train is not a point,). So it has a length of x meter to cross.

Finally x + x = 2X.

Hope you get it.

Here, two trains are equal length & same direction but speed is different that's why we calculate the relative speed = u - v.

Why have you taken the relative speed here? Please explain.

T = (L1 + L2)/(S1 + S2),
T * (S1 - S2) = L1 + L2,
36 * 10 * 5/18 = L1 + L2,
100 = L1 + L2.
Given that L1 = L2.
Then,
2L1 = 100.
L1 = 50.