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Blazing goIITian

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28 Jul 2008 23:04:12 IST

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Kinematics IIT Question:

None

Please give the answer with proper explanation:(rates for each effort)

MORE THAN ONE OPTION CORRECT:

Q. A particle of mass m moves on the x-axis as follows: it starts from rest at t=0 from the point x=0 and comes to rest at t=1 at the point x=1. N other information is available about its motion at intermediate times (0<t<1)

If "Z" denotes the instantaneous acceleration of the particle, then:

a) Z cannot remain positive for all "t" in the interval 0 to 1

b) |Z| cannot exceed 2 at any point in its path

c)|Z| must be greater than or equal to 4 at some point or points in its path

d) Z must change sign during the motion, but no other assertion can be made with the information given.

THE ANSWERS ARE GIVEN AS (a),(c)...which i feel are wrong....

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Comments (9)

BALGANESH

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28 Jul 2008 23:14:01 IST

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see it starts from rest and again comes to rest therefore fist it must have accelerated and then it would have retarded therefore potion a, d are correct

Dave

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29 Jul 2008 02:07:17 IST

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ans is a,d

where is this queston taken from???

svj 29

Blazing goIITian

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29 Jul 2008 02:55:03 IST

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Its an IIT 1992 or 1993 question...it came in my coachings test...

anchit saini

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29 Jul 2008 09:13:37 IST

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not as straightforward as it seems at first glance

Lets consider the case of one turning point only (all other cases can be shown equivalent to it )->

In this also , lets make 2 cases

Case 1 ->

In this case, lets consider the most symmetric case , ie when it turns at t=1/2 sec and it has same acceleration throughout

At t=1/2

s1=at^2 /2 =a/8

v1=at=a/2

t=1/2 to t=1

v2= 0 = u -At =a/2 - At =a/2- A/2

implies a = A

s2=ut - at^2/2 =a/4 - a/8 =a/8

but s1 + s2 = 1 = a/4

implies a = 4 m/s^2

the velocity time graph in this case is as shown in graph 1

------------------------------------------------------------------------------------------------------------------------------------------------------

Case 2- >

Now , suppose instead of t=1/2 , the turning point comes at some other time , then it is obvious that if acceleration is less than 4 for interval before turning point , acceleration would be greater than 4 for for interval after turning point and vice versa

------------------------------------------------------------------------------------------------------------------------------------------------------

Now , we have to proove that instead of 1 turning point suppose there are infinitely many turning points yet the above condition holds true .

For it , consider the graph 2 in which there are many turns .

Now we can consider any point P , and then we can see the the part before P will correspond to a certain average acceleration and part after P will correspond to a certain average acceleration , and hence the corresponding graph would come to be of either case 1 or case 2.

Now suppose , before P ,the average acceleration is greater than 4 , then since there are many turning points its obvious that to maintain average acceleration of more than 4 , at some points instantaneous accelerations would have to be greater than 4 .Same would be true for interval after P , if it has average acceleration greater than 4.

Hence , option a and c are correct

Srujana

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29 Jul 2008 14:17:35 IST

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Apologies fr not going thru anchits ans...it seemed too scary to me....here goes my attempt...

m assuming the particle to execute SHM ( though i dun have enough reasons to come to tat conclusion...it makes things easier :D )

The particle starts from rest at x=0 and again comes to rest at x=1 ....so surely...x=0 and x=1 are the extreme positions ,

also amplitude=(1-0)/2=1/2

time taken to complete half oscillation = 1sec...so T= 2 sec

now w =pi rad/sec (w=omega)

so the eq of SHM is x=1/2 sin (pi t+c)

so max acceleration can be found which will be greater than or equal to 4 (w^2A)

obviously Z cant hold the same sign throughout...

so ans =a,c

Hari Shankar

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29 Jul 2008 15:52:14 IST

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@ srujana, the point here is to prove that for every kind of motion between the two points, you will always have the a_max> 4

Srujana

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29 Jul 2008 15:56:43 IST

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sir..i have already mentioned tat my soln makes no sense as im unable to prove tat it is a SHM ....it is just one of the possibilites but not essentially a sufficient condition...

Hari Shankar

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29 Jul 2008 17:15:02 IST

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We have to prove that the maximum steepness can never be less than 4.

So, let us look for a curve that could possibly violate this condition

Intuitively we can eliminate curves with more turns as these produce more steepness. So, we are looking for a curve with just one point at which f'(t) changes sign

Now, we can have the curve to be convex or concave and our choice is obvious that the candidate curve must be concave (sine curve like). Again, symmetry with x = 1/2 is a preference.

For such a curve, the maximum acceleration takes place at the end-points.

Since the net displacement is 1, we have

$1 = \int_0^1 f(t) dt$

By integration by parts, we get

$\int_0^1 f(t) dt = |t f(t)|_0^1 - \int_0^1 t f$

Since, f(1) = f(0) = 0 (given), we have

$\int_0^1 f(t) dt = - \int_0^1 t f$

So, now $<br/>\frac{|a_{max}|}{2} = 1-\int_0^1 \frac{t^2}{2} f"(t) dt$

Since the curve is concave, $f"(t) \le 0$

Hence, for a^_maxto be least, we must have $f"(t) \rightarrow 0$ . This means the concavity must approach that of a straight line. This function is nothing but a triangle.

This leads us to the situation of uniform acceleration which has been done by Anchit where the maximum acceleration is obtained as 4. (notice the same modulus of acceleration is present at the end-points i.e. t =0 and t = 1)

Hence, the maximum acceleration for any other situation is going to be atleast 4 which is what we set out to prove

ravi teja

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30 Jul 2008 06:27:17 IST

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when a body starts from rest nd come to rest mean that it had 2 acc of opp signs..so option A is ri8
bt if we consider option D we cannot say A is ri8 as other options cannot be assertioned so
in option C
if we jus checkwith some assumed values of time we get it ri8 but
if we check with B it will not be true bcoz it has to acc nd retard so Z may be higher so A nd C
rate me if i my ans is satisfactory

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