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A train travelling at a speed of 60 km per are breaks are applied so as to produce a uniform acceleration of - 0.5 m per second square find how the train will go before it is brought to rest?
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A train travelling at a speed of 60 km per are breaks are applied so a...
Calculation of Distance Traveled by Train Before Coming to Rest
- Initial speed of the train, u = 60 km/h = 60 x 1000 m / 3600 s = 16.67 m/s
- Final speed of the train, v = 0 m/s (since the train is brought to rest)
- Acceleration of the train, a = -0.5 m/s^2
- Distance traveled by the train before coming to rest, s = ?

Using the equation of motion:
v^2 = u^2 + 2as

Substitute the values:
0 = (16.67)^2 + 2(-0.5)s
0 = 277.89 - s

Calculate the distance traveled:
s = 277.89 m
Therefore, the train will travel 277.89 meters before it is brought to rest.
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A train travelling at a speed of 60 km per are breaks are applied so as to produce a uniform acceleration of - 0.5 m per second square find how the train will go before it is brought to rest?
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