Question

# A train starts from rest and moves with a constant acceleration of $$2.0\ m/s^{2}$$ for half a minute. the brakes are then applied and the train comes to rest in one minute. Findthe total distance moved by the train,

Solution

## Train starts from rest, hence the initial velocity u = 0. It moves with acceleration = 2m/s2 for half minute (30 seconds). Distance covered in this time interval is given by: $$S = ut + ½ at^2$$ $$= 0 + ½ \times 2 \times 30 \times 30$$$$= 900m$$ Velocity attained by this acceleration after 30 seconds: $$v = u + at$$ $$=> v = 0 + 2 x 30$$$$=> v = 60 m/s$$ From this velocity, brakes are applied and train comes to rest in 60 seconds. The retardation is given by: $$v = u – at$$ $$=> 0 = 60 – a \times 60$$ $$=> a = 1 m/s^2$$ Distance covered in this time: $$V2= u2 + 2aS$$ $$=> 0 = (60)2 + 2 (-1) S$$ $$=> 0 = 3600 – 2S$$$$=> S = 3600/2 = 1800m$$. So, total distance moved =$$900m + 1800m = 2700m$$. Maximum speed of the train$$= 60 m/s$$. Position of the train at half its maximum speed. Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s. (I) When the train is accelerating with an acceleration of 2 m/s, time at which speed = 30m/s is: $$v = u + at$$$$=> 30 = 0 + 2 x t$$ $$=> t = 15s$$At 15s, distance covered from origin is: $$S = ut + ½ at^2$$ $$= 0 + ½ \times 2 \times 15 \times 15$$ $$= 225m$$ (II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 ,  time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics

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