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ICSE Class 9 Answered

A train starts from rest and accelerates uniformly at  a rate of 2 m/s2  for 10 seconds.It then maintains a constant velocity for 40s.On applying brakes, a train is uniformly retarded and stops in 20s.Calculate : (1) maximum velocity reached (2) retardation in last 20 secs (3) total distance travelled (4) average velocity of the train? 
Asked by nishola82 | 22 May, 2018, 11:39: PM
answered-by-expert Expert Answer
(a) Initial motion with acceleration 2 m/s2 for 10 seconds.
 
Distance travelled S1 = u×t+(1/2)×a×t2 , where u is initial speed (u=0, starts from rest), a acceleration and t is time of travel
 
S1 = (1/2)×2×10×10 = 100 m ...................(1)
 
speed v after 10 seconds :-  v = u+a×t = 0 + 2×10 = 20 m/s
 
(b) motion with constant velocity 
 
Distance travelled S2 = velocity×time = 20×40 = 800 m .................(2)
 
(c) break applied
 
stops after 20 seconds ; hence retardation will be obtained from "v = u - a×t " with v =0
 
retardataion a = u/t = 20/20 = 1 m/s2
 
distance travelled S3 = u×t - (1/2)×a×t2 = 20×20 -(1/2)×1×20×20 = 200 m. ...............(3)
 
(1) maximum velocity reached = 20 m/s = 20 × (18/5) = 72 km/hr
(2) retardation in last 20 s = 1 m/s2
(3) Total distance travelled = S1+S2+S3 = 100+800+200 = 1100 m
(4) average velocity = distance / time = 1100/70 = 110/7 m/s = 110/7 × (18/5) = 56.6 km/hr
Answered by Thiyagarajan K | 23 May, 2018, 01:59: PM

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