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

The correct option is a/3b.

To find the time at which the acceleration of the particle becomes zero, we start with the given equation of position as a function of time:
$x=a{t}^2-b{t}^3$
where $x$ represents the position and $t$ represents the time.

First, we find the velocity ($v$) of the particle by taking the first derivative of position with respect to time:
$v=\frac{dx}{dt}=\frac{d}{dt}\left(a{t}^2-b{t}^3\right)$
Applying the power rule of differentiation:
$v=2at-3b{t}^2$

Next, we determine the acceleration ($A$) of the particle by taking the derivative of the velocity with respect to time:
$A=\frac{dv}{dt}=\frac{d}{dt}\left(2at-3b{t}^2\right)$
Differentiating term-by-term, we obtain:
$A=2a-6bt$

To find the time when the acceleration is zero, we set $A=0$:
$2a-6bt=0$
Solving this equation for $t$:
$6bt=2a$
$t=\frac{2a}{6b}$
Simplifying the fraction gives:
$t=\frac{a}{3b}$

Thus, the acceleration of the particle will be zero at time $t=\frac{a}{3b}$.