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The greatest acceleration or deceleration that a train may have a. The minimum time in which the train may reach from one station to the other separated by a distance d is
The total distnace train has to cover is d.The first half it will accelerate and the second half it’ll deaccelerate, and it’ll take half-half time for both acceleration and deacceleration. Hence, d/2 = a/2(t/2)2 (u = 0 ,because the train was at rest initially)4d/a = t2t = (4da)½.
Accn and declrcn each have taken time = t/2 S= 1/2(at/2)*t= at^2/4 t= 2√d/a .....................................
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