Is this question of IIT JEE / AIEEE STANDARD OR LESS ? ...PLEASE SOLVE IT..IT SEEMS EASY..BUT I THINK IT IS NOT..
Q: 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 km , the maximum speed of train being 60 mile/hr ?
(i) 500 sec [ii] 58.67 sec
[iii] 400 sec [iv] 314 23 sec
ANS : [ii]
29This is a simple question..but dont expect any such question neither in AIEEE nor in JEE..it's just a question for testing the calculation part..so many conversions..feet to metre and miles to km...the question could have been asked in a better way by using the standard SI units..
29see first convert feet into metres.. i guess 3 feet = 0.9m
and miles into km/hr and then m/s
1 mile = 1.6 km
so the train is at 96km/hr or 26m/s
now time taken to achieve top speed with a = 0.9m/s2
t≈ 30 seconds
now we have
s = 1/2 a t2
so s ≈ 400m
so the same distance will be covered during retardation
so the distance which the train travels at constant velocity = 9.2 km
time taken to travel it 350 seconds
so total time 350 + 30 + 30 = 410 ≈ 400 seconds..
PS : i manually calculated the abv values..so there may be some error ± 5% in the calculalations..
1dude always use graphs to solve these types of sum it saves time
