Never take shortcuts in Physics. When the lift first performs it's motion with acc. a , it's initial velocity is 0, but when it starts retardation the lift has some velocity. So you cannot merge the two.
A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation 2a.If t is the time of ascent,find the depth of the shaft.
(a)at^{2}4 (b)at^{2}3 (c)at^{2}2 (d)at^{2}8
(According to me,
since the particle first accelerates uniformly and then decelerates uniformly,
total acceleration=a2a
=a
also,s=ut+at^{2}2
Since the lift should start from rest,u=0 ms^{1}
thus,s=at^{2}2
but the depth of the shaft is the distance travelled,and hence magnitude of displacement..
thus, depth of shaft=at^{2}2....but the given correct answer is (b) ,...where have i been wrong??? )

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5 Answers
Let the total time is t .and when lift accilirate with accn a is time t1 and when lift ritardiate with accn 2a then time will be t1\2.
t=t1+t1\2
t=2t\3
By formula
S=0.t+2a.(t1/2)^2Ã—1\2
S=at1^2\2
Put the value of t1
S=at^2\3
even if we do it that way......
when the lift is accelerating......let it accelerate for t_{o} s
s_{1}=at_{o}^{2}2
and,v=at_{o}...this is also the initial velocity when the lift is retarding
when the lift is retarding,time is tt_{o}s
s_{2}=at_{o}(tt_{o})a(tt_{o})^{2}
=at_{o}tat_{o}^{2}at^{2}at_{o}^{2}+2at_{o}t
=3at_{o}t2at_{o}^{2}at^{2}
s_{1}+s_{2}=3at_{o}t3at_{o}^{2}2at^{2}
which is nowhere in the option....am i wrong somewhere?
@anik....
try to find the relation between t and t_{o} using the result that t_{o} is the time when the acc is a and (tt_{o}) is the time where deacceleration is 2a..
turns out that t_{o}=2t/3
put it in your last equation and you will get the answer...