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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Jul 2008 19:45:12 IST
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A particle moves in x-y plane with constant accelaration a directed along negative y- axis. the equation of motion of the particle has the form y=cx - bx^2, where c and b are constants. find the velocity of the particles at the origin of cordinates.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Jul 2008 19:52:26 IST
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v = dy/dt = a -2bx
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i am genius |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Jul 2008 20:28:41 IST
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velocity along the negative y direction is a-2bx velocity at the origin is a m/s
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SHREYA |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Jul 2008 02:22:35 IST
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the method is there and look at the comment
http://www.goiit.com/posts/list/community-shelf-a-problem-of-irodov-component-method-39256.htm
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it is not important where u stand, but in which direction u are moving |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Jul 2008 15:52:54 IST
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The given equation is analogous to that of a projectile fired from the ground. Compare the given equation with 
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Jul 2008 17:06:29 IST
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it cant be solved unless value for dx/dt is given
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Jul 2008 21:49:58 IST
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y = cx - bx2 differating it u get vy = (c-2bx)vx-------1 again differatating u get a = 2bvx ^2------------2 solve 1,2 fr vx , vy so with da values of vx , vy u can find velocity at any point
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Jul 2008 18:08:31 IST
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My last reply was distorted. Here goes :
we have and 
Now you can solve for u.
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Will nip in at times to solve problems :)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Jul 2008 00:51:41 IST
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"Before you start some work, always ask yourself three questions - Why am I doing it, What the results might be and Will I be successful. Only when you think deeply and find satisfactory answers to these questions, go ahead."
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