The correct option is C x(t)=t33−2t2
Given, a(t)=2t−4
As we know that a=dvdt⇒ ∫v0dv=∫t0a(t)dt
and also v=dxdt⇒ ∫x0dx=∫t0v(t)dt
Thus, v=∫t0a(t)dt=∫t0(2t−4)dt
⇒ v(t)=t2−4t
also, x=∫t0v(t)dt=∫t0(t2−4t)dt
⇒ x(t)=t33−2t2
Hence, displacement of the particle as a function of time is x(t)=t33−2t2