1a is the ans.
, atleast at some point, the acc. has to be≥4 for the given condn to be satisfied....(by using calculus)
If (c) would have been \left|a \right|≥4,then it would have been true....
A particle of mass m moves on the axis as follows...it starts from rest at t=0from the point x=0 and comes to rest at t=1at x=1.no other info is available abt its motion....a is its acceleration.then:
a)a cannot remain positive for t=0 to t=1
b)|a|<2 for all points in the path
c)|a|>4 at some point(s)
d)a must change signs else no assertion can be made!!!!
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13 Answers
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1@rahul then u should tell a,c,d sholud be the answer....show tje solution too
@shreya.....its obviously seems that the acceleration is first +ve and then negative but can u tell me how diid u conclude that b) is also the answer....i.e. highest acceleration is less than 2
1a and d is the answer....nothong can be said about option b and c because...magnitude of acceleration cannot be determined
1uh!!! i was wrong!
b cannot be correct at all...
it must be only a and c.
11First of all we cant determine the value of a .. But when the questions says the particle starts from rest and comes to rest at x=1 and t=1 from x=0 and t=0 , then a must change sign....
So if a changes sign it can not remain +ve throught t=0 to 1... therefore a and d are the correct options..
Moreover i think the question is too vague
1The Question Is Not Vague... :) :P
we have,
t=0,x=0,u1=0 and t=1,x=1,v2=0
let acceleration a1 act for time t and a2 act for 1-t.
(see a1>0 and a2<0)
Now,
v1 = 0 + a1t
v2=0=u2 + a2(1-t) = v1 + a2(1-t) = a1t + a2(1-t)
so t= a2/a2-a1
Now
s1 = 0.5a1t2
s2=(a1t)(1-t) + 0.5a2(1-t)2
now, s1 + s2 = 1
so,
0.5a1t2 + a1t(1-t) + 0.5a2(1-t)2 = 1
substitute value of t to get,
a2 = 2a1/2-a1
Substitute this value to get
t=2/a1
if a1<2, t becomes greater than one, not possible...
now when mod(a2)>=4 then
2a1/mod(2-a1) >= 4
simplify to get a1<= 4
for mod(a2)<=4 then
2a1/mod(2-a1) <= 1
simplify to get a1>=4
thus we have that when a1>=4, the value of mod(a2)<=4 and vice versa.
hence, mod(a) must be >=4 at some points in the path.
so a and c only are correct...
49itna bara proof......check this out.......
considering when acceleration and deceleration is equal,
so, v=at/2
now, x=area of triangle=1/2ht*base
=1/2(at/2)(t)
so, 1=1/4a(1)2
so, a=4.....
but if accn and dccn were not same, either would be greater than 4.
thus mod(a)≥4
and by logic the accn must change sign. thus correct options are (a) and (c)
62no it is not..
but the thing here is that the maximum distance travelled will bee only if you apply the maximum allowed acceleration..
Otherwise your area of the graph will be lesser thatn that done by subhomoy.