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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The position of a particle as a function of time is given by equation: $x =2t^3-6t^2+12t+6$. The acceleration of the body is zero at

$\begin{array}{1 1} t=1\;s \\ t=2\;s \\ t=3\;s \\t=0.5\;s \end{array} $

Can you answer this question?

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Answer :$ t=1\;s $
$v= \large\frac{d}{dt} $$(x) =\large\frac{d}{dt}$$(2t^3)-\large\frac{d}{dt}$$(6t^2)+\large\frac{d}{dt}$$(12 t) $$+\large\frac{d}{dt} $$(6)$
$\quad= 6t^2-12 t+12$
$a= \large\frac{d}{dt}$$(v)= \large\frac{d}{dt}$$(6t^2)$$-\large\frac{d}{dt}$$(12 t) +\frac{d}{dt}$$(12)$
or $ a = 12 t -12$
When $a=0, 12t-12=0$ or $t=1\;s$
answered Aug 13, 2014 by meena.p
edited Aug 13, 2014 by meena.p

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