Aptitude - Problems on Trains - Discussion

This is the concept of reference of frame.

Train speed is (x)
Let's say A person in a train says that a man(outside of the train(or) beside of train running at a speed of 5 km/ hr) running slow as (x-5)km/hr.

This means if the train speed is 20 km/hr. Then the man's relative speed is as slow as 15 km/hr when compared to train speed.
This relation is mentioned in the mathematical formula as( x-5)km/hr or ( example:20-5 km/hr).
So, the relative speed is;
X - 5 = (120/10)*(18/5)
So x = 50 km/ hr.
*18/5 is used to convert m/s into km/ hr.

First we write Data:

We Have D=125m t=10s,
S=D/t =125/10=25/2 m/s.

We want to convert from m/s to km/kr.
so we multiply 25/2 into 18/5,
@When we convert km/hr to m/s we multiply by 5/18,

@When we have to convert m/s to km/hr we multiply by 18/5,

Our answer are given in km/hr so we convert according_to by multiplying 18/5.

=25/2*18/5=45 Km/hr,

let we assume train speed be X.
And Man 5 km/hr.

If given in opposite direct then we add,

Here have in the same direction so we subtract,

x-5=45 So Answer is 50.

A Train 125m long passes a man,running at 5km/hr.

It means that train speed =5000m/3600sec.------(1hr=60*60=3600sec & 1km=1000meters).

& Second Train running on same Direction & Passed in 10sec's.

It means Second train cross distance within 10sec's.

Speed of Train for meters/10sec = 5000m/3600sec*10sec=1.388888888889m
So,that train speed for km/hr=speed of train for 10 sec * remaining 3590 Seconds(3600-10).
=1.3888888889*3590=49.8611111111111111=50km/hr.

Therefore, A speed of that train=50km/hr.

For relativity the formula is divided as two that depends upon the question.

When question has same direction:

t=(l1+l2)\(u1-v1).

when question has opposite direction ;
t=(l1+l2)\(u1+v1).

In our question we has same direction, so we going to use 1st formula:
Here,
u1 = x;
v1 = 5km\hr;
t = 10 seconds;
l1 = 125m;
l2 = 0; These are the given data's.

By substitute the data's in formula;
u1-v1 = (x-5)(5\18),
10 = 125\(x-5)(5\18),
10 (x-5) = 125(18\5),
10x-50 = 450,
10x = 450+50,
10x = 500,
x=50.

Let the speed of the train is x km/hr
Then the relative speed is (x-5)km/hr
Therefore,
Time = 10sec
Distance = 125m
Speed = (x-5)km/hr
As we are calculating in m/sec so we have to convert km/hr to m/sec.
(x-5)*1000/3600 m/sec or (x-5)*5/18 {by cancelling by 200.
200*5=1000,200*18=3600}

Therefore ,

Speed = distance/ time
(x-5)5/18 = 125/10
x-5 = 125*18/10*5
x-5 = 2250/50
x-5 = 45
x = 45+5
x = 50
Therefore x is 50km/hr (speed of the train)

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.

Let the speed of the train is 'x'
Length=125m
Speed=10s
--> as per "Length of the train = speed X time"
--> 125m = 10s.x
--> x=(125m/10s) X (360/360)=(360 X 125m)/3600s
--> =45000m/3600s = 45km/1hr

As the above value is relative with the speed of man whose speed of running is 5km/1hr also to be added for the exact speed of the train.

There fore..
-->exact speed of the train is = (45km/1hr)+(5km/1hr)=50km/1hr

I believe its simple and right answer.

The best formula is:

Step1: distance = speed x time.
Then, speed = distance /time (125/10)m/sec.
=> (125/10)m/sec = (25/2)m/sec.

Step 2: convert m/sec to km/hr.
Note: 1000m - 1km.
3600sec - 60min - 1hr.
25m - 0.025km.
2sec - 0.0005hr.

Now (25/2)m/sec = (0.025/0.0005)km/hr. (0.025/0.0005)km/hr = 50km/hr.

How it 18÷5 will become? Please explain.

Here is the solution for this question:

The length of a train is given as 125m (here it is mentioned as m/s), and the speed is given as 5km/hr (here it is mentioned as km/hr).

While converting km/hr into m/s, we are multiplying the speed by 5/18; therefore while converting m/s into km/hr, we have to take the reciprocal of 5/18, that is 18/5. The speed of the train should be calculated in km/hr; therefore, we take 18/5.

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

====> dist. travelled by man in 10 sec. = (5*5/18) *10=250/18
A train will have to travel its own length + dist.travelled by man. (when train will pass the man he would have travelled 125/9 m in same dir.)

i.e. 125+125/9 meters = 1250/9
speed of train = (1250/9)/10 = 125/9 m/s=125/9 *18/5 = 50 km/hr