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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.
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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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