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. ?

Question

Nirmal Singh. · Answer

a = 3 feet / sec^2
since 1 feet = 0.3048 m
so a = 3 *0.3048 = 0.9144 m/sec^2
s = 10 m
u = 60 miles/hr
1 miles = 1.61 km
so u = 60*1.61 km/hr = 60*1.61*5/18 km/hr
u = 26.8 m/sec
hence s = ut -1/2at^2
10 = 26.8t - 0.5*0.9144t^2
so 0.4592t^2 - 26.8t+10 = 0
t = 26.8+sqrt(26.8^2 - 4*0.4592*10) / 2*0.4592
t = 58.67 sec

Regards,
Nirmal SIngh
Askiitians Faculty

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