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the relation between time t and positon x for a particle moving on x axis is given by t =p(x)square +qx find relation between velocity and acceleration??
note: dx/dt = v dv/dx=a and (dv/dx)*(dx/dt)=a t=px2+qx on differentiating wrt t , we have 1=2pxv+qv .....(1) on differentiating wrt x , we have 0=2pv+2px.dv/dx +q.dv/dx 0=2pv+2px.(a/v) +q.(a/v) 0=2pv2+2pxa +qa ....(2) from (1) and (2) ..we have 2pv3+a=0 -- regards Ramesh
note: dx/dt = v dv/dx=a and (dv/dx)*(dx/dt)=a
t=px2+qx
on differentiating wrt t , we have
1=2pxv+qv .....(1)
on differentiating wrt x , we have
0=2pv+2px.dv/dx +q.dv/dx
0=2pv+2px.(a/v) +q.(a/v)
0=2pv2+2pxa +qa ....(2)
from (1) and (2) ..we have
2pv3+a=0
--
regards
Ramesh
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