The position x of a particle varies with time t as x=at2−bt3. The acceleration of the particle will be zero at time t equal to

A
$\frac{a}{b}$
B
$\frac{2 a}{3 b}$
C
$\frac{a}{3 b}$
D
zero

Solution:

$\begin{array}{l} given Position of particle x = {at}^{2} - {bt}^{3} \\ d i f f e r e n t i a t e w . r . t . t i m e t t o g e t v e l o c i t y \\ velocity = \frac{d x}{d t} = \frac{d (a t^{2} - b t^{3})}{d t} \\ \Rightarrow v = 2 a t - 3 b t^{2} \\ a g a i n d i f f e r e n t i a t e w . r . t . t i m e t t o g e t a c c e l e r a t i o n \\ Acceleration = \frac{dv}{dt} \\ a = \frac{d (2 a t - 3 b t^{2})}{d t} \Rightarrow 2 a - 6 b t \\ Acceleration will become zero \\ O = 2 a - 6 bt \\ \Rightarrow 2 a = 6 bt \\ \Rightarrow t = \frac{a}{3 b} = t i m e i n w h i c h a c c e l e r a t i o n b e c o m e s z e r o \end{array}$

The position x of a particle varies with time t as $x = a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time t equal to

Solution:

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