UPSC Exam > UPSC Questions > If train travelling at 72 km\h is to be broug...
Start Learning for Free
If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be?
Most Upvoted Answer
If train travelling at 72 km\h is to be brought to rest in a distance ...
Understanding the Problem
To determine the retardation required to bring a train traveling at 72 km/h to rest in a distance of 200 meters, we first need to convert the speed into meters per second.
Conversion of Speed
- 72 km/h = 72 * (1000 m / 3600 s) = 20 m/s
Using the Equation of Motion
We can apply the equation of motion that relates initial velocity, final velocity, distance, and acceleration:
- v² = u² + 2as
Where:
- v = final velocity (0 m/s, as the train comes to rest)
- u = initial velocity (20 m/s)
- a = acceleration (retardation in this case)
- s = distance (200 m)
Rearranging the Equation
Since we know v, u, and s, we can rearrange the equation to solve for a:
- 0 = (20)² + 2 * a * 200
This simplifies to:
- 0 = 400 + 400a
Solving for Retardation
- 400a = -400
- a = -1 m/s²
Conclusion
The required retardation to bring the train to a stop in 200 meters is 1 m/s². This means that the train must decelerate at a rate of 1 meter per second squared to safely come to a halt within the specified distance. Understanding these calculations is crucial for ensuring passenger safety and efficient train operations.
To determine the retardation required to bring a train traveling at 72 km/h to rest in a distance of 200 meters, we first need to convert the speed into meters per second.
Conversion of Speed
- 72 km/h = 72 * (1000 m / 3600 s) = 20 m/s
Using the Equation of Motion
We can apply the equation of motion that relates initial velocity, final velocity, distance, and acceleration:
- v² = u² + 2as
Where:
- v = final velocity (0 m/s, as the train comes to rest)
- u = initial velocity (20 m/s)
- a = acceleration (retardation in this case)
- s = distance (200 m)
Rearranging the Equation
Since we know v, u, and s, we can rearrange the equation to solve for a:
- 0 = (20)² + 2 * a * 200
This simplifies to:
- 0 = 400 + 400a
Solving for Retardation
- 400a = -400
- a = -1 m/s²
Conclusion
The required retardation to bring the train to a stop in 200 meters is 1 m/s². This means that the train must decelerate at a rate of 1 meter per second squared to safely come to a halt within the specified distance. Understanding these calculations is crucial for ensuring passenger safety and efficient train operations.
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Similar UPSC Doubts
Top Courses for UPSCView all
If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be?
Question Description
If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be?.
If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be?.
Solutions for If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? defined & explained in the simplest way possible. Besides giving the explanation of
If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be?, a detailed solution for If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? has been provided alongside types of If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? theory, EduRev gives you an
ample number of questions to practice If train travelling at 72 km\h is to be brought to rest in a distance of 200 m, then it's retardation should be? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup