Aptitude - Problems on Trains - Discussion

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.

Time(t) = Distance(s)/Speed.

S = ut + .5at^2
v = u + at
2as = (v^2 - u^2)

Where, a = acceleration, v = final vel, u = ini.vel, s = dist, t = time.

In these train equation, v and a are taken as zero.

What are the formulas used to find the time?

The given speed is 5 km/hr. How it comes 18/5?

Train Is 125 M.

Man Speed 5 KM/H.

Train Passed Man in 10 Seconds.

HR/KM = 3600 seconds/1000 MR = 36/10 divided both on 2 = 18/5.

Train Passed in 10 seconds with his tall 125 M so 125/10 divided on 5 = 25/2.

(18/5 * 25/2) = 45 KM/H speed + man speed 5 KM/H = 50 :).

@Jaydev Ramnath

To convert km/hr into m/s we should multiply with 18/5.

Since train and man run in the same direction,

So, in 10 seconds, train crosees 125 mts (train length) + 125/9 mts,

(5 km/hr => 5000/3600, 10 seconds => 125/9),

So train speed , 36(125 + 125/9) (10 seconds x 36 => 3600 seconds => 1 hr),

Answer : 50, 000 mts = 3600 secs => 50 km/hr.

Man speed is 5km/hr.

So, km = 1000metrs.
Per hour = 60mintes = 3600second.

3600/1000 = 36/10 = 18/5.

Trains cross the man in 10 seconds.

Length of the train is =125m.
Relative man trains speed = 125/10 = 25/2.

So 25/2 x 18/5 = 45km/hr.

We let train speed is = x km/hr.

Relative to man turn speed = x - 5 = 45.
x = 45 + 5 = 50km/hr.

Why it is 18/5? Please explain.

Thank you guys for your best explanations really easily understandable, for the non maths learners. You people are doing good, good things never stop. Please keep on posting.