(d)

$x=2+t-3 t^2, v=\frac{d x}{d t}=1-6 t$, velocity will become

zero at time, $0=1-6 t_0$ or $t_0=\frac{1}{6}\,s$,

Since, the given time $t=1 s$ is greater than $t_0=\frac{1}{6}\,s$,

distance $ > \mid$ displacement $\mid$

$\text { Displacement } s=x_f-x_i$

$=(2+1-3)-(2+0-0)=-2\,m$

$\text { Distance } d=\left|s_{0-t_0}\right|+\left|s_{t-t_0}\right|$

$=\frac{u^2}{2|a|}+\frac{1}{2}|a|\left(t-t_0\right)^2$

$\text { Comparing } v=1-6 t \text { with } v=u+u t \text {, we have }$

$u=1\,m / s \text { and } a=-6\,m / s ^2$

$\therefore \quad \text { Distance }=\frac{(1)^2}{2 \times 6}+\frac{1}{2} \times 6 \times\left(1-\frac{1}{6}\right)^2$

$=2.1\,m$