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Mechanics

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 Joined: 10 Jul 2008 Post: 598
15 Jul 2008 14:42:28 IST
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15
1655
problem from relative motion
Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Mechanics

Two ships A and B originally at a distance

d = 3 km from each other depart at the same time from a straight coastline. Ship A moves along a straight line perpendicular to the shore while ship B constantly heads for ship A, having at each moment the same speed as the later. After a suffciently large interval of time, ship B will obviously follow ship A at a certain distance. Find this distance.

Blazing goIITian

Joined: 22 Jul 2007
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15 Jul 2008 20:13:51 IST
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can u plz tell the initial direction of ship B

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15 Jul 2008 21:07:34 IST
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as is given, ship B always heads for A so initially the velocity of B will be along the line BA

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17 Jul 2008 09:53:59 IST
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any take on the problem guys? c'mon

Scorching goIITian

Joined: 7 Dec 2007
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17 Jul 2008 13:51:37 IST
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hmm i assumed after long time ship b will be in teh same vertical line as ship A then since they both have teh same speed always tehir distance won't change further assuming this i get 3d/2

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17 Jul 2008 14:16:04 IST
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well ... i am sorry but that is wrong..

Scorching goIITian

Joined: 18 Nov 2007
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17 Jul 2008 14:30:20 IST
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SHIP B ALWAYS HEAD TOWARDS SHIP A. NOW AS A IS MOVING FORWARD, DIRECTION OF B IS CONTINUOSLY CHANGING AND HENCE IT IS FOLLOWING THE CURVED PATH. YES, AFTER LONG TIME INTERVAL, THE CURVE OF B WILL INTERSECT THE STRAGHT LINE OF A. YOU CAN SOLVE THIS IF BY USING GRAPHICAL VIEW.

IF YOU CAN KNOW THE EQUATIONS OF THE CURVES, YOU CAN FIND THE DISTANCE. SEE THE GRAPH

Scorching goIITian

Joined: 7 Dec 2007
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17 Jul 2008 15:22:53 IST
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hmm  can u plz post the answer i think i got my calculation wrong somewhere

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17 Jul 2008 15:42:23 IST
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Scorching goIITian

Joined: 7 Dec 2007
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17 Jul 2008 16:09:37 IST
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hmm lemme post my solution as i said it is indeed a calculation mistake

consider at any time t the line joining A nd B makes an angle with the horizontal(the coast)

now let the distance between thema t any time t be given by x(t)

clearly x(0)=d

also

we assume the following

also consider the triangle formed by A B and the perpendicular from B to A's path

say it's foot is P the PAB is a right angle dtriangle with sides

so using pythagoras's theorem and the expression obtained for x(t)

we get the following equality

this follows from the equation for x(t)

now after sufficiently long time since B gets behind A and their distance(relative) remains constant we have

thus substituing this we get

done!!!

PS:i had wrongly taken before so the wrong answer

Blazing goIITian

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18 Jul 2008 12:45:25 IST
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excellent !!!

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18 Jul 2008 14:55:41 IST
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here is my own version

Scorching goIITian

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18 Jul 2008 16:44:35 IST
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nice one there i have always wondered how people use invariance :D

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18 Jul 2008 17:04:04 IST
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thanx... :)

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19 Jul 2008 21:04:13 IST
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Done??

Now see the ultimate  general method !!

I have already posted a similar but  a more general problem long ago and now I a going to use it. Let me post the whole thing again for conveneience :

The problem was :

Two particles A & B start from positions ( 0, 0 ) & ( 0, -d ) and move with constant speeds v & u respectively . A moves along x - axis and B moves such that its velocity is always aimed at A . Let r be the distance between them and  be the angle made by the velocity of B with X - axis , at some time t .

Prove that ,

So it is a more general case .  The r for a large time is found by letting  tends to zero ( as they now follow each other through a st. line ).

Now if we put u=v (as given here ) then taking the limit , it follows directly that r=d/2.

Now the derivation of the formula ( as done by me )is as follows:

see, the controlling differential eqns are clearly.........

dx/dt = u cos ..................... ( 1 )

dy/dt = usin  .......................( 2 )

y/( x - vt) = tan ....................... ( 3 )

y = - r sin  .................. ( 4 )

Now the task is to solve these coupled differential eqn .

From ( 3 ) we have

x - vt = y cot

dwrt 't'

dx/dt -v = dy/dt cot - y cosec^2() d/dt

combining  ( 1 ) & ( 2 )  with this eqn we have

u cos   - v = u sin  cos /sin  + r cosec d/dt

or ,                 d/dt = -v/r sin  .................. ( A )

differentiating ( 4 ) wrt 't ' , we get

u sin   = - d/d  ( r sin ) d/dt

combining with ( A ) we get

u/v  =  cos   + sin  1/r  dr/d

or ,         u/v cosec  - cot  = 1/r dr/d

Now integrating it and putting the boundary condition that r= d at = /2

we get

What do u say about that , Kayamant?

Blazing goIITian

Joined: 12 Jun 2007
Posts: 1078
20 Jul 2008 09:02:36 IST
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@pardesi, feynamann and Anant sir.....

awesome solutions!!!!

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