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If the greatest admissible acceleration or retardation of a train be 3 feet/sec2.find the least time from one station to another at a distance of 10 metre the maximum speed being 60 miles per hour: (A) 500 sec (b) 58.67 sec (c) 400 sec this question is from arhiant bitsat 2014 book from motion chapter question no. Is 21 and the ans of question given is B but not solutionof problem. ?
a = 3 feet / sec^2since 1 feet = 0.3048 mso a = 3 *0.3048 = 0.9144 m/sec^2s = 10 mu = 60 miles/hr1 miles = 1.61 kmso u = 60*1.61 km/hr = 60*1.61*5/18 km/hru = 26.8 m/sechence s = ut -1/2at^210 = 26.8t - 0.5*0.9144t^2so 0.4592t^2 - 26.8t+10 = 0t = 26.8+sqrt(26.8^2 - 4*0.4592*10) / 2*0.4592t = 58.67 secRegards,Nirmal SInghAskiitians Faculty
a = 3 feet / sec^2since 1 feet = 0.3048 mso a = 3 *0.3048 = 0.9144 m/sec^2s = 10 mu = 60 miles/hr1 miles = 1.61 kmso u = 60*1.61 km/hr = 60*1.61*5/18 km/hru = 26.8 m/sechence s = ut -1/2at^210 = 26.8t - 0.5*0.9144t^2so 0.4592t^2 - 26.8t+10 = 0t = 26.8+sqrt(26.8^2 - 4*0.4592*10) / 2*0.4592t = 58.67 sec
Regards,Nirmal SInghAskiitians Faculty
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