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If $s=r^{3}-4t^{2}+7,$ the velocity when the acceleration is zero is

\[\begin{array}{1 1}(1)\frac{32}{3}m/sec&(2)\frac{-16}{3}m/sec\\(3)\frac{16}{3}m/sec&(4)\frac{-32}{3}m/sec\end{array}\]

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1 Answer

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$s=t^3-4t^2+7$
$\large\frac{ds}{dt}$$=3t^2-8t$
=> $\large\frac{d^2s}{dt^2}$$=6t-8$
Given acceleration =0
$6t-8=0$
$t =\large\frac{8}{6} =\frac{4}{3}$
Velocity $\large\frac{ds}{dt} $$=3 \times \large\frac{16}{9} $$ -8 \times \large\frac{4}{3}$
$\qquad= \large\frac{16}{3} -\frac{32}{3} =\frac{-16}{3} $$ m/sec$
Hence 2 is the correct answer
answered May 19, 2014 by meena.p
 

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