viagra generico 200 mg prezzo a Bologna Main-Topic: Time, Speed Distance (TSD)
http://cinziamazzamakeup.com/?x=vardenafil-originale-miglior-prezzo-garanzia Subtopic: Trains and Platforms.
you’ll be given some data and asked to find the length of railway platform or speed of train etc.
- Visualizing the Concept: Train and Pen
- Case 1: One Train One Man (or Tree, Pole)
- Case 2: One Train One Platform (or bridge, tunnel)
- Case 3: Two Trains
- Unit Conversions
- Previous Articles on Aptitude
Before going into Trains concept, try to visulize following. (ofcourse technically incorrect but for concept clarity)
You must have seen the advertizement of a ballpen “Flair Writometer”. They claim it can write for 2,500 metres and even provide a marking scale on the ballpen refill to measure the ‘distance’! See the photo
So You’re standing on a Railway platform, holding that writometer pen in your hand, pointing it straight towards the upcoming train’s side, like this photo
The train passes, and your pen makes a straight line on the entire body of this train.
Now you open the pen’s refill and check the markings, obviously few mili-litres of ink is used and marking would show that you’ve drawn a line of 200 meters.
Meaning length of this train is 200 m.
At the same time, your friend was sitting in the window seat of the same railway and he too held a writometer pen in his hand pointing it out of the window and towards the platform.So as the train passed, his pen also drew a straight line on the entire platform. Later he checks the refill of his pen: it drew a line of 1000 meters.
Meaning length of this railway platform is 1000 m.
so what is the total distance or length of lines drawn by the two pens?
200m of train + 1000m of platform = 1200m.
A railway train running at the speed of 20 meter per second, crosses a man in 10 seconds. What is the length of the train?
For all the problems about train, boats, cars, scooters, men required to complete a job, tata nano and kingfisher airplane, in short TSDW (time,speed, distance, work) you need to remember only one “STD” formula
speed (s) x time (t) = distance (d)
Back to the question, what is given?
Visualise the situation: you are the man, holding the writometer pen and the train crossed you in 10 seconds. What is the distance of the line drawn by your pen on the body of the train?
Speed (s)= 20 m/s (given)
Time (t)= 10 seconds (given)
s x t = d
20 x 10=200 m.
final answer: length of the train is 200 m.
Instead of man, they can rephrase and ask: Train crossed a “Tree, Telephone-pole, Tiger, Kalmadi, Raja..” The concept remains the same. Everybody is holding a pen.
Question: a train running at the speed of 20 m/s, crosses a railway platform in 20 seconds. If the length of the train is 100 m, what is the length of this railway platform?
So in this case, you were standing on a platform holding a http://cinziamazzamakeup.com/?x=comprare-viagra-generico-50-mg-a-Verona blue pen and I am sitting in the railway also holding a http://cinziamazzamakeup.com/?x=viagra-generico-25-mg-pagamento-online-a-Roma purple pen.
What is the total length of lines drawn by both of us?
So total distance covered by the train is (100+x)
speed is given: 20m/s
time is also given: 20 seconds
So apply the STD formula
s x t = D
20ms x 20 seconds = (100+x)
400m = 100+x
Final answer: length of this platform is 300 metres.
Alternative: instead of railway platform, they can use the word “Bridge, Tunnel” but the principle remains the same.
http://acrossaday.com/?search=viagra-professional-canadian-drugstore Two trains are coming from opposite directions. Both of them are running at the same speed 20 meter per second.
It takes 20 seconds for them to completely cross each other. If the length of first train is 90 m, what is the length of the second train?
Visualise: you are holding a pen sitting in first train, I am holding a pen sitting in the second.
The total length of line drawn is 90+x m
Since both of the trains are moving, and moving in opposite directions. Meaning it will take less time to draw these lines.
Final speed (s)
= speed of first train+ speed of second train.
= 40 m/s
Now it can be solved just like the case #2.
In the case #1 and #2, Man and Platform were stationary (unmoving) objects so their speed was zero, hence we took train’s speed as our final speed. But here second train is also moving so we’ve to consider it.
same question, but instead of two trains , it is one man riding on a bike and another train coming from the opposite direction.
Here’s the length of length of line
=Line drawn on Train + Line drawn on Bike
= 90m +zero meter. (because length of bike is negligible, we take it zero)
So Distance (d)= 90.
but for the speed we have to add
=Train + Bike
After that, principle remains the same as Case #3
It is essential that before applying the STD formula, all the unit is must be in the same system, else you will get incorrect answer.
and they will give that incorrect answer in one of the four options [a/b/c/d] so you will confidently tick it but end up loosing in negative marking.
So in the STD formula. Everything must be
either in meter,second,
or in kilometer, hour.
1 km per hour = 5/18 meter per second.
How did we derive this?
1 kilometer per hour
=1000 meters per 60 minutes
=1000 meters per (60×60) seconds
=1000 meters per 3600 seconds
=10 meters per 36 seconds (cut the zeros)
=5 meters per 18 seconds. (half of 10 and 36)
if a train is running a 36 kms per hour then, its speed in meter per second is:
1 kmph = 5/18 ms
36 kmph= 36x (5/18) = 10 m/s.
If the reverse is asked, (10 m/s then how much in kmph?),
simply multiply with reverse 18/5, because
5/18 ms = 1 kmph
1 ms= 18/5 kmph
10 ms= 10 x (18/5) kmph = 36 kmph.