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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

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

$x=at^2-bt^3$
$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

edited Jan 24, 2014 by meena.p

+1 vote