A train starts from rest and moves with a constant acceleration of 2.0 metre per second square for half a minute. The breaks are then applied and train comes to rest in 1 minute after applying brakes.find (A) the total distance moved by the train (B) the maximum speed attained by the train and (C) the position of train at half maximum speed (assume retardation to be constant)

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Arun · Answer

Initial velocity u = 0 Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks) t = 30 sec v = u + at ⇒ 0 + 2 × 30 = 60 m/s a) S1 = ut + ½ at2 = 900 m when breaks are applied u’ = 60 m/s v’ = 0, = 60 sec (1 min) Declaration a’ = (v-u)/t = = (0 – 60 ) / 60 = - 1 m/s2. S2 = (V^'2-u^(;2))/(2a^' ) = 1800 m Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = (V^2-u^2)/2a = (〖30〗^2-0^2)/(2 x 2) = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. ∴ u = 60 m/s, v = 30 m/s, a = –1 m/s2 Distance = (V^2-u^2)/2a = (〖30〗^2-〖60〗^2)/(2(-1)) = 1350 m Position is 900 + 1350 = 2250 = 2.25 km from starting point

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