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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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The position x of a particle varies with time as $x=at^2-bt^3$. The acceleration of the particle will be zero at time t equal to

\[(a)\;\frac{a}{b}\quad (b)\;\frac{2a}{3b} \quad (c)\;\frac{a}{3b} \quad (d)\;zero\]

Can you answer this question?
 
 

1 Answer

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$x=at^2-bt^3$
$v=\large\frac{dx}{dt}$$=2at-3bt^2$
$a=\large\frac{dv}{dt}$$=2a-6bt=0$
$=>t=\large\frac{a}{3b}$
Hence c is the correct answer

 

answered Jun 21, 2013 by meena.p
edited Jan 24, 2014 by meena.p
 

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