A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of –0.5 m s-2. Find how far the train will go before it is brought to rest.
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The train will go 625 m before it is brought to rest.
Explanation:
The train is traveling at a speed of 90 km/h. And it is brought to rest by applying brakes.
Initial velocity of the train, u = 90 Km/h = 25 m/s
Final velocity of the train, v = 0 m/s
Acceleration of the train, a = -0.5 m/s²
The distance traveled by train before it is brought to rest can be found from the 3rd equation of motion.
v² - u² = 2aS
⇒ S = (v² - u²) / 2a
S = (0² - 25²) / (2 × -0.5)
S = -625 / -1
∴ S = 625 m
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