if the maximum accleration or retardation that a train can have is a ..

find the minimum time in which it can travel distace s.....

WELL NOW  HERE CONSTANT accn = a , distance = s

so minimum time is required  , hence  train will accelerate and then deaccelerate

the body will not have constant vel at any of pt so for minimum time

the body will acce and deacce to same time

so for acceleration let T = TIME THEREFORE

X1 =

V = aT

NOW IT DEACCE till final velocity = 0

so x2 =  

SO x1 + x2 = s =

T =

therefore total time = 2T =

PLZ RATE

BEST OF LUCK

Read my solution to Q5) here, it might help you.



http://www.goiit.com/posts/list/mechanics-1d-motion-81465.htm

Q) If the maximum acceleration or retardation that a train can have is A, find the minimum time in which it can travel a distance S from rest to rest.

A) Let x1 + x2 = S

[v^2 - u^2 / 2A]        +  [ V^2 - U^2 / 2A ]   = S

v^2 / 2A + (-v^2 / -2A) = S



v^2 = AS



v = sqrt(AS)

Also,

v = A.t1



t1 = sqrt(AS) / A  = sqrt(S/A)

t2 = -v / -A = -A.t1 / -A = t1 = sqrt(S/A)

So total time = t1 + t2 = 2.sqrt(S/A)

(The answer you have is incorrect)

you should check the ans

I mgetting that ans only

PLZ RATE IF CORRECT