Aptitude - Problems on Trains - Discussion

Step 1:

The relative speed of the train concerning man = Distance/Time =125m/10sec= 12.5m/s.

Step 2: Convert the answer of "step 1" from m/s to km/hr..
To convert it into km/hr multiply it with 18/5.
12.5 * 18/5 = 45km/hr.

Step 3:
Train as well as the man both are moving in the same direction so we will have to subtract the speed of train(x) from the speed of the man(5).
(Note: If the train and man both are moving in opposite directions then add both of their values (x+5) ).

Relative speed = (x-5)km/hr
45km/hr = (x-5),
x=50 km/hr.

Given:

Length of the train = 125 m.
Speed of the man = 5 km/hr.
Time taken to pass the man = 10 seconds.

First, we need to convert the speed of the man from kilometres per hour to meters per second:
5 km hr=5×1000 m3600 s=50003600≈1.39 m s.
5 km hr=5×3600 s1000 m​=36005000​≈1.39 m s.

Next, we calculate the speed of the train using the formula:
Speed=DistanceTime.
Speed=TimeDistance​.

The distance in this case is equal to the length of the train (125 m), and it passes the man in 10 seconds:
Speed of train relative to man =125 m10 s=12.5 m s.
Speed of train relative to man =10 s125 m​=12.5 m s.

Since this is the relative speed (train's speed minus man's speed), we can express it as:
Speed of train − Speed of man = 12.5 m s.
Speed of train − Speed of man =12.5 m s.

Substituting in the man's speed:
Speed of train−1.39 m s=12.5 m s
Speed of train−1.39 m s=12.5 m s.

Now, solving for the speed of the train:
Speed of train=12.5+1.39=13.89 m s
Speed of train=12.5+1.39=13.89 m s

Finally, converting this back to kilometres per hour:
13.89 m s=13.89×36001000=50 km hr
13.89 m s=13.89×10003600​=50 km hr.

Therefore, the speed of the train is 50 km/hr.

@All

Here they mentioned the relative time of the train to the man is 10 seconds which means the train's length(i.e 125m) has just passed the moving man in 10 seconds,so we need to find the relative speed for the train to the man as mentioned below.

i.e, 125/10 =25/2 m/s as it is in m/s,we need to convert into km/hr
25/2 x 18/5 = 45 km/hr (relative speed between train and man)

Next we need to find the original speed of the train that passes the man's speed with the help of the relative speed between the train and man i.e,45 km/hr(,if the Direction of the man and train is opposite to one another, then do addition, if both man and train running in same direction subtract both )
Here man and train are in the same direction.
So subtract;
Let x be the speed of the train and we know the man's speed as 5 km/hr and the relative speed is 45 km/hr.
Since same direction subtraction,
x = 45+5
x = 50 km/hr (answer)
By checking:
Train's speed =50 km/hr ,,man's speed = 5 km/hr, same direction so do subtraction;
50 -5 = 45 to get the relative speed of both.

125/10 = 12.5,
12.5 * 3.6 = 45,
45 + 5 = 50.

I am not getting this. Anyone, help me to understand this.

How it is 50? please explain to me.

3600sec then, 1sec = 1/3600hr.
1km =1000m then, 1m = 1/1000km.
So,
km/hr to m/s is : 1/1000km bigger divided by 1/3600hr.
So, they put 18/5.

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.

How answer is 50 km/hr? Please explain to me.

Sp of man = 5 * 5/18
= 25/18 m/sec.

Let Sp of train be x, then;
x - 25/18 = 125/10
So x = 125/10 + 25/18 = 125/9 m/sec.
So x = 125/9 * 18/5 =50 Km/h.