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  4. The position x of a particle varies with time, (t) as x = at2 - bt3. The acceleration will be zero at time t is equal to

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Q. The position x of a particle varies with time, (t) as $x = at^2 - bt^3.$ The acceleration will be zero at time t is equal to

AIPMTAIPMT 1997Motion in a Straight Line

Solution:

Distance $ (x) = at^2- bt^3.$
Therefore velocity $ (v) = \frac{dx}{dt} = \frac{d}{dt} (at^2- bt^3)$
$=2at -3bt^2$ and
acceleration $ = \frac{dv}{dt} = \frac{d}{dt} (2at- 3bt^2) = 2a - 6bt = 0$
or $ t = \frac{2a}{6b} = \frac{a}{3b}.$