To have escaped 'kruschev's shoe' only to be buried by the rising generation{s} -- It's almost funny -- in a fatalistic sort of way...

Isn't it, though. But then since I long ago concluded that civilization is a self-defeating concept, I'm trying to adopt the attitude of a disinterested bystander enjoying the show as best I can (which, alas, is not all that well). I would find it almost amusing, if it weren't for the knowledge that my daughter will have to live in the increasingly dysfunctional mess that we are leaving her. One the bright side, if we can raise her right, there won't be anyone that can compete with her. But, with the way things are going, it will become illegal for her to compete and win since that wouldn't be fair to people like.... well, enough said.

 

If a train is 500 meters long and is traveling 100 km/hour, how many seconds does it take it to pass a given point?

Can you figure that much out?

the speed is 27.77meter/sec
d=sxt
t=18sec

 

I'm trying to adopt the attitude of a disinterested bystander enjoying the show as best I can (which, alas, is not all that well). I would find it almost amusing, if it weren't for the knowledge that my daughter will have to live in the increasingly dysfunctional mess that we are leaving her

That's just it - But for the unenviable plight of our kids....

 

the speed is 27.77meter/sec

Why are you contradicting the stipulation? The train's velocity is given as 54km/hr = 15 m/s

d=sxt

I give up -- What is 'sxt'???

t=18sec

Does that make any sense? We know that it requires fully 20 seconds to pass the observer -- Granting even a 'dimensioned' observer we wouldn't expect him to be larger than the platform?

Anyway you're supposed to be solving for platform length -- you already know the 'apposition times' (To wit 36s Platform and 20s Observer)
Best regards
HP

 

Why are you contradicting the stipulation? The train's velocity is given as 54km/hr = 15 m/s


I give up -- What is 'sxt'???


Does that make any sense? We know that it requires fully 20 seconds to pass the observer -- Granting even a 'dimensioned' observer we wouldn't expect him to be larger than the platform?

Anyway you're supposed to be solving for platform length -- you already know the 'apposition times' (To wit 36s Platform and 20s Observer)
Best regards
HP

He was answering the simpler question I gave him.

I'm assuming that sxt is s·t for (speed x time).

 

the speed is 27.77meter/sec
d=sxt
t=18sec

Okay. So now let's turn this around.

If a train is traveling 54 km/hr and takes 20 seconds to pass a certain point, how long is the train?

 

If a train is 500 meters long and is traveling 100 km/hour, how many seconds does it take it to pass a given point?

Can you figure that much out?
the speed is 27.77meter/sec
distance =speed x time
t=18sec

is that right answer?

 

If a train is traveling 54 km/hr and takes 20 seconds to pass a certain point, how long is the train?

Distance = speed x time
speed in mtr/sec 15meter/sec
so, 15x20 =300meters

 

Can you figure that much out?
the speed is 27.77meter/sec
distance =speed x time
t=18sec

is that right answer?

Correct! -- Now please see post #66

 

Is this correct?
Distance = speed x time
speed in mtr/sec 15meter/sec
so, 15x20 =300meters

 

Can you figure that much out?
the speed is 27.77meter/sec
distance =speed x time
t=18sec

is that right answer?

Yes. I wouldn't have moved on to the next part if it hadn't been.

 

ok thats great.

 

Distance = speed x time
speed in mtr/sec 15meter/sec
so, 15x20 =300meters

Good! so you know that the length of the train (300m) + the length of the platform (X) are passed in 36 seconds at 15m/s So....?

 

Is this correct?
Distance = speed x time
speed in mtr/sec 15meter/sec
so, 15x20 =300meters

Yes, but you really need to track your units, particularly given how weak your math and reasoning skills are.

\(
Length_{train} \: = \: Speed_{train} \, \times \; Time_{train}
Length_{train} \: = \: \( \frac{54 \, km}{1\, hr} \)\( 20 \, s \)\( \frac{1000 \, m}{1 \, km} \)\( \frac{1 \, hr}{3600 \, s} \) \: = \: 300 \, m
\)

If I can take the time to do it, why can't you?

 

So now you know how long the train is and how fast it is moving.

You also know that it takes 20 seconds between the time that the nose of the train reaches a point and when the tail of the train reaches that same point. How much further does the tail of the train go in the remaining 16 seconds that it takes for the train to completely pass the platform?

 

Distance = speed x time
speed in mtr/sec 15meter/sec
D=36x15=540meter

 

@RRITESH KAKKAR

If you were any closer to the solution it would snap your nose! -- Please a deep breath and think!

 

Distance = speed x time
speed in mtr/sec 15meter/sec
D=36x15=540meter

Yes! --- That's the total distance -- So what is the length of the platform alone?

 

If I can take the time to do it, why can't you?

Because i can't figure out the scene.

 

Distance = speed x time
speed in mtr/sec 15meter/sec
D=36x15=540meter

Draw a picture of what is happening and identify the length of the train, the length of the platform, and the combined length of them. Which of these is 540 meters? You are (finally) very close.