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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: iorodov question
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ankitagg (285)

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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.

    
amitp91 (426)

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v = dy/dt = a -2bx

i am genius
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shinee (234)

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velocity along the negative y direction is a-2bx
velocity at the origin is a m/s

SHREYA
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ramyani (2539)

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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


it is not important where u stand, but in which direction u are moving
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karthik2007 (3380)

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The given equation is analogous to that of a projectile fired from the ground. Compare the given equation with  


Will nip in at times to solve problems :)
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rohanrony (17)

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it cant be solved unless value for dx/dt is given
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harry (17)

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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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karthik2007 (3380)

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My last reply was distorted. Here goes :


we have  tan\theta = c and \frac{a}{2u^2 cos^2 \theta} = b


Now you can solve for u.


Will nip in at times to solve problems :)
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sachinguptaiit (772)

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\text{y}=\text{x.tan}\theta-\frac{\text{ax}^2}{\text{2u}^2\text{cos}^2\theta}\\\\\text{y}\;=\;\text{cx}-\text{bx}^2\\\\\text{c=tan}\theta\\\\\text{b}=\frac{\text{a}}{\text{2u}^2\text{cos}^2\theta}\\\\\Rightarrow\;\text{u}=\sqrt{\frac{\text{a}.(\text{c}^2+1)}{\text{2b}}}


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