11Let 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)^2×1\2
S=at1^2\2
Put the value of t1
S=at^2\3
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??? )
-
UP 0 DOWN 0 0 5
5 Answers
11
71Never 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.
158even 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?
1057@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...