1.

A train is travelling at a speed of 90 kmh-1 Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest.

Answer»

Let the initial speed of the train be u = 90 Km/h

= 25 m/s.

Final speed of the train, v = 0 (train comes to rest)

acceleration a = 0.5 ms-2

As per 3rd law of motion

v2 = u2 + 2as

(0)2 = (25)2 + 2(0.5)s

s = train travelled distance

∴ s = \(\frac{(25)^2}{2(0.5)}\) = 625 m

Train will travel 625 km before it is brought to rest.



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