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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 May 2007 20:24:13 IST
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soo nw the soln. of ur ques...........
v=-t^2 + 3t +1
dv/dt= -2t + 3
nw for function to be max or min. dv/dt=0
so we get t=3/2 sec
now d^2 v/d t^2 = -2 so dat means our function is max at t = 3/2 sec
hence tym is t=3/2 sec
cheers
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