x=9t2−t3.....(I)
We know that,
v=dxdt
The maximum speed is
v=18t−3t2
v maximum, at dvdt=0
dvdt=18−6t
The time will be
0=18−6t
t=3 sec
The position of the particle at 3 sec
Put the value of t in equation (I)
x=9×9−27
x=81−27
x=54 m
The position of the particle will be 54 m.
Hence, Option A is correct