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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Apr 2008 09:57:44 IST
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A particle moves with deceleration along a circle of radius so that at any moment its tangential and normal acc. are equal in magnitude.At initial moment t=0 ,the speed of the particle equal to v
Q1.Speed of the particle as a function of time t is given by v(t)=v/k1+k2*v*t/k3*R
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22 Apr 2008 12:42:48 IST
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Will nip in at times to solve problems :)
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22 Apr 2008 13:30:52 IST
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^^ one correction
it's decelaration , so -dv/dt = v^2/r
solved here: http://www.goiit.com/posts/list/mechanics-please-solve-54679.htm
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