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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Feb 2008 21:17:43 IST
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A point moves in plane so dat its at =a and normal acc. an =bt2,a,b r const. n at t=0,pt. is at rest.The variation of radius of curvature R with total distance covered S is???
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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5 Feb 2008 22:00:25 IST
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Hello anyone der???
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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5 Feb 2008 22:30:31 IST
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it shud be independent of S
an =bt2=v2/r
dv/dt=a
v=at
v2/r=bt2
a2t2/r = bt2
r=a2/b.......................(a2 is a square ,b2 is b square) am i correct???????//
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5 Feb 2008 22:51:22 IST
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As at=a dv/dt=a integrate both the sides v=at...........(1) Now an=bt2 v2/r=bt2 r is the radius v2=brt2 put it in eqn (1) we get r=a2/b hence it is constant
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5 Feb 2008 22:56:20 IST
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ans. given is a3/(2*b*s)....even i'm gettin de same ans. as u guyz...
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MaNuTd RoXxXx..MaNuTd 2 WiN PrEmIeR LeAgUe ThIs SeAsOn ToO AlOnG WiTh ChAmPiOnS LeAgUe.....HaiL RoNaLdO ...HaiL LaMpArD... |
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