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A train takes 2 minutes to acquire its full speed 60kmphfrom rest and 1 minute to come to rest from the full speed.If somewhere in between two stations 1 km of the track beunder repair and the limited speed on this part be fixced to20kmph, find the late running of the train on account ofthis repair work, assuming otherwise normal at running ofthe train between the stations. |
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Answer» Let the distance between the two stations is s meter. =>given v = 60 km/hr = 60 x 1000/3600 = 16.67 m/s By v = u + at =>16.67 = a x 120 =>a = 0.14 m/s^2 s = ut + 1/2at^2 =>s1 = 0 + 1/2 x 0.14 x (120)^2 =>s1 = 1000 m By v = u + at =>0 = 16.67 + a x 60 =>a = -0.28 =>By s = ut + 1/2at^2 =>s2 = 17.67 x 60 - 1/2 x 0.28 x (60)^2 =>s2 = 560.10 m Thus if s = s1+s1+ 1000 1st case:-(no interruption) =>T = t1+t2+ t3 =>T = 120 + 60 + 1000/16.67 =>T = 239.99 sec 2nd case:-(with interruption) =>T = t1+t2+t3 =>T = 120 + 60 + 1000/[(20 x 1000)/3600] =>T = 270 sec =>Delay = T(1st) - T(2nd) =>Delay = 270 - 239.99 =>Delay = 30.01 sec |
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