Aptitude - Problems on Trains - Discussion

We know for same direction, relative speed = (V1-V2).

Now, V1 = 46 Km/hr & V2 = 36 m/hr or we can say;

V1 = 46x5/18 m/s & V2 = 36x5/18 m/s.

Now, Relative Speed = Length or Distance/Time,

i.e (46x5/18) - (36x5/18) = length/36.

Take 5/18 common we get,

5/18(46-36) = length/36.

Length = 5/18x10x36, on solving we get, length = 100 meters.

But since length of trains is same.

Hence length of each train will be, length = 100/2 => length = 50 meters.

Since the Trains are moving in the same direction The diff between their speeds has to be calculated.
46 kmph - 36 kmph = 10 kmph

convert into meters per second 10 x 1000 meters / 3600 seconds
We get 25/9.

Now apply formula distance = speed x time = 25/9 x 36
Cancelling 36 with 9 we get 4 multiply with 25 we get 100 meters, which is the length of two trains.
As per the question the two trains length is the same there fore each train length is 50 meters.

The correct answer is 100 m. because the distance travel by faster train in 36 sec is 460 m
and distance travel by the slower train in 36 sec is 360 sec
so the answer is the difference of distance travel by the both train
(460-360)=100m
see practical example following
&&&&&&&& faster train
@@@@@@@@ smaller train
after 36 sec

|......460 m..&&&&&&|
|..360 m @@@@@@..|
so actual result is 100.

If we consider relative speed = 46-36 = 10 (km/h), then at this speed the faster train needs to cover just the length of the slowest train (so we have just x not 2*x). Length of one train equals 100 m.

Also d1 = 460 m faster train in 36 s.

d2 = 360 m slower train in 36 s.

----x, d1.

---x, d2.

Let's suppose the length of train is x = 50 m, then d1-x = 460-50 = 410 m! = 360 m (distance covered by the slowest train).

So x = 100 m, length of one train.

We have formula:

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 is taken by the faster train to cross the slower train = (a + b)/(u - v) Sec.

- Since two trains are of same length, so distance or(a+b) = 2x
- Time taken to cross slower train = 36 sec
- Speed of first train(u) = 46 kmph
- Speed of second train(v) = 36 kmph
- By applying formula ::
36 = 2x/((46-36)*(5/18)).
x = 50m.

Here the length of first train is x meters
and the distance covered by the two train is 2x meters
{Here we have to convert km/hr into m/sec because the answer is in the meter}

Relative speed between them is ((46-36)*))5/18 then we get
25/9 m/sec

after that we must equating the two equations then we get the length of the first train


we are using the formula distance=time*speed
distance=2x
time=36 sec
speed=25/9

2x=25/9*36
we get x=50 m

Easiest process:

Let, length = x m.

Speed of 1st mvng train = 46 * (5/18) = 115/9 m/sec.

""slow"" = 36 * (5/18) = 10 m/sec.

Formula = time is taken by a first moving train to overtake (same direction) a slow moving train = (length of first moving train + length of slow moving train) / (speed of first moving train - speed of slow moving train).

Hence 36 = 2x / ((115/9) - 10) 2x is taken because both lenght are equal.

Hence, x = 50m.

As two trains are in parallel (same direction). So we can subtract the speeds of other trains. Relative speed is (25/9) m/sec.

NOTE: Time in contact of 2 trains at a time MEANS the time at which faster train engine meet the last part of slower train TO faster train last part meet the engine of slower train.

Both trains relative speed = Distance of 2 trains/(Time in contact of 2 trains at a time).

(25/9) m/sec = 2x/36.

You can easily understand by assuming that one train is a brigde and its length is same as the train, let it be x and the speed of train is 10km/h (46-36).

Now, the time taken to cross the bridge is 36sec.
Dis = speed * time.
= (10*5/18) * 36.
= 100. Is that clear?
We know that 100 is the length of train+length of bridge.
We know both are same so 100÷2 = 50.
Train length is 50 and bridge(another train) length is 50.

After relative speed one have '0' and other '10kmph'.

Then,
-----------> <=1st train with 46kmph relative speed is '10 kmph.
-----------> <=2nd train with 36kmph relative speed is '0' kmph.

After passing,
-----------> <=with relative speed '10' kmph.
-----------> <=with relative speed '0' kmph.

So only one train length speed to be calculated, then why it take "2x"?

Please explain.