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The brakes applied to a train moving at 90km/h produces a retardation of 5m/s What distance will it cover before coming to stop?
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The brakes applied to a train moving at 90km/h produces a retardation ...
Calculation of Distance Covered by a Train Before Coming to a Stop

Given:

  • Initial velocity (u) = 90 km/h = 25 m/s (converted using conversion factor 1 km/h = 1000 m / 3600 s)

  • Retardation (a) = 5 m/s^2

  • Final velocity (v) = 0 m/s (as the train comes to a stop)



Formula:

  • v^2 = u^2 + 2as

  • s = (v^2 - u^2) / 2a



Calculation:

  • s = (v^2 - u^2) / 2a

  • s = (0 - (25)^2) / (2 x 5)

  • s = -625 / 10

  • s = -62.5 m



Explanation:
The above calculation shows that the train will cover a distance of -62.5 m before coming to a stop. The negative sign indicates that the train will move in the opposite direction to its initial motion. The actual distance covered by the train will be positive, which is 62.5 m. This is because the distance formula calculates the magnitude of the distance irrespective of the direction.

Conclusion:
Therefore, the train will cover a distance of 62.5 m before coming to a stop when the brakes are applied to a train moving at 90 km/h with a retardation of 5 m/s^2.
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The brakes applied to a train moving at 90km/h produces a retardation ...
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The brakes applied to a train moving at 90km/h produces a retardation of 5m/s What distance will it cover before coming to stop?
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