a train starts from rest and accelerates uniformly at
100m/minutesquare for 10 minutes . it then maintains a constant velocity for 20 minutes. the brakes are applied and then the train is uniformly retarded. it comes to rest in 5 minutes.
find the distance .
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Answers
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ANS.
a)Initial motion with acceleration 2m/s2 for 10sec.
Distance travelledS1=u×t+(1/2)×a×t2,where u is initial spee(u=0 starts from rest),a acceleration and t is time of travel .
S1=(1/2)×2×10×10=100m..................(1)
speed after 10 sec. :-v=u+a×t=0+2×10=20m/s
b)Motion with constant velocity
Distance travelled S2=velocity×time = 20×40=800m.................(2)
(c)break applied
stops after 20sec. ; hence radiation will be obtained from "v = u - a×t" with v=0
retardiation a =u/t20/20=1 m/S2
Distance travelled S3=u×t-(1/2)a×t2=20×20-(1/2)×1×20×20=200m. .........(3)
(1)maximum velocity reached =20m/s=20×(18/5)=72 km/hr
(2)Retardation in last 20s =1m/S2
(3)Total distance travelled =S1+ S2+ S3=100+200+800=1100m
(4)Average velocity=distance/time=1100/70= 110/7 m/s=110/7×(18/5)=56.6 km/hr.
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