# Kinematics

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)at24 (b)at23 (c)at22 (d)at28
(According to me,
since the particle first accelerates uniformly and then decelerates uniformly,
total acceleration=a-2a
=-a
also,s=ut+at22
Since the lift should start from rest,u=0 ms-1
thus,s=-at22
but the depth of the shaft is the distance travelled,and hence magnitude of displacement..
thus, depth of shaft=at22....but the given correct answer is (b) ,...where have i been wrong??? )

• Manish gupta ·

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

• Vivek @ Born this Way ·

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.

• Anik Chatterjee ·

even if we do it that way......
when the lift is accelerating......let it accelerate for to s
s1=ato22
and,v=ato...this is also the initial velocity when the lift is retarding
when the lift is retarding,time is t-tos
s2=ato(t-to)-a(t-to)2
=atot-ato2-at2-ato2+2atot
=3atot-2ato2-at2

s1+s2=3atot-3ato22-at2
which is nowhere in the option....am i wrong somewhere?

• Ketan Chandak ·

@anik....
try to find the relation between t and to using the result that to is the time when the acc is a and (t-to) is the time where deacceleration is 2a..
turns out that to=2t/3
put it in your last equation and you will get the answer...

• Anik Chatterjee ·

@ketan...ya...got it ...thanks

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