The correct option is D The acceleration of the particle is zero at t=α3β.
Differentiate x=αt2−βt3 w.r.t time to get velocity
v=dxdt=2αt−3βt2.....(1)
Differentiate v w.r.t time to get acceleration
a=dvdt=2α−6βt....(2)
The particle will return to its initial position when x=0
i.e., 0=αt2−βt3
which gives t=αβ
The particle will come to rest when v=0
i.e., 0=2αt−3βt2
which gives t=2α3β
Using equation (1) and (2), we say that initial velocity of the particle is zero but initial acceleration is a=2α
a=0⇒2α−6βt=0, which gives t=α3β
Thus, acceleration is zero at t=α3β