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The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop?
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The breakup light to train moving at 90 km per hour produces retardati...
To determine the distance a train will cover before coming to a stop, we can use the equations of motion. Here’s a detailed explanation:
Initial Velocity (u)
- The initial velocity of the train is given as 90 km/h.
- To convert this into meters per second (m/s), we use the conversion factor:
\[ u = 90 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 25 \, \text{m/s} \]
Retardation (a)
- The retardation (deceleration) is given as 5 m/s².
- Since the train is slowing down, we will consider this value as negative in our equations:
\[ a = -5 \, \text{m/s}^2 \]
Final Velocity (v)
- The final velocity when the train comes to a stop is:
\[ v = 0 \, \text{m/s} \]
Using the Equation of Motion
- We can use the equation:
\[ v^2 = u^2 + 2as \]
- Rearranging to solve for distance (s):
\[ s = \frac{v^2 - u^2}{2a} \]
- Plugging in the values:
\[ s = \frac{0 - (25)^2}{2 \times (-5)} \]
\[ s = \frac{-625}{-10} = 62.5 \, \text{m} \]
Conclusion
- The distance covered by the train before coming to a stop is:
62.5 meters.
Initial Velocity (u)
- The initial velocity of the train is given as 90 km/h.
- To convert this into meters per second (m/s), we use the conversion factor:
\[ u = 90 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 25 \, \text{m/s} \]
Retardation (a)
- The retardation (deceleration) is given as 5 m/s².
- Since the train is slowing down, we will consider this value as negative in our equations:
\[ a = -5 \, \text{m/s}^2 \]
Final Velocity (v)
- The final velocity when the train comes to a stop is:
\[ v = 0 \, \text{m/s} \]
Using the Equation of Motion
- We can use the equation:
\[ v^2 = u^2 + 2as \]
- Rearranging to solve for distance (s):
\[ s = \frac{v^2 - u^2}{2a} \]
- Plugging in the values:
\[ s = \frac{0 - (25)^2}{2 \times (-5)} \]
\[ s = \frac{-625}{-10} = 62.5 \, \text{m} \]
Conclusion
- The distance covered by the train before coming to a stop is:
62.5 meters.
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The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop?
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The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop?.
The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The breakup light to train moving at 90 km per hour produces retardation of 5 metre per second square what distance will it cover before coming to stop?.
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