InterviewSolution
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InterviewSolution
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15. A train starts from rest and accelerates uniformlyat a rate of 2 m s2 for 10 s. It then maintains aconstant speed for 200 s. The brakes are thernapplied and the train is uniformly retarded andcomes to rest in 50 s. Find: (i) the maximumvelocity reached, (ii) the retardation in the last50 s, (iii) the total distance travelled, and(iv) the average velocity of the train.Ans. (i) 20 m s-1, (ii) 0.4 m s-2,(iii) 4600 m, (iv) 17-69 m S-l |
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Answer» Train accelerate from start for 10 sso velocity reached at 10s is a×t= 2×10=20m/s1) 20m/s is max velocity as after this the train runs on constant speed2) the train is retarded for 50s from this speed to rest so retardation is velocity /time=20/50=-0.4m/s23) distance travelled is total of distance travelled with accelaration+distance with constant speed+ distance with retardationdistance with acceleration is ut+1/2at2=1/2*2*10*10=100mdistance with constant speed is s×t=20×200=4000mdistance with retardation is ut+1/2at220×50-1/2×.4×50×50=1000-500=500mtotal distance =100+4000+500=4600m4) average velocity= total distance/total time= 4600/260=17.69m/s |
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