A $2.00‐kg$ box is moving to the right with speed $8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ on a horizontal, frictionless surface. At $t=0$ a horizontal force is applied to the box. The force is directed to the left and has magnitude $F\left(t\right)=\left(6.00 \raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)t^2$. What distance does the box move from its position at $t=0$ before its speed is reduced to zero(in m)?

Given data are :

$$\begin{gathered} m=2 kg \\ u=8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ F\left(t\right)=-\left(6 \raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)t^2 \end{gathered}$$

Here negative sign means the force is acting in the opposite direction to the direction of initial velocity $u$ (i.e. left side)

$$\begin{gathered} a=\frac{F\left(t\right)}{m}=\frac{-6t^2}{2}=-3t^2 \\ a=\frac{dv}{dt}=-3t^2 \\ {\int }_8^{0}dv={\int }_0^t-3t^2 dt \\ -8={\overline{)\frac{-3t^3}{3}}}_0^t \\ {t}^3=8 \\ t=2 \end{gathered}$$

Now,

$$\begin{gathered} \frac{dv}{dt}=-3t^2 \\ \int dv=\int -3t^2 dt \\ v=\frac{-3t^3}{3}+c \\ v=-t^3+c \cdots \cdots \cdots \cdots \cdots \left(1\right) \end{gathered}$$

When $t=0$, $v=8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ when we put these values in equation $\left(1\right)$ we will get

$c=8 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Where $c$ is some constant

$$\begin{gathered} v=-t^3+8 \\ v=\frac{ds}{dt}=-t^3+8 \\ {\int }_0^{s}ds{\int }_0^2\left(-t^3+8\right)dt \\ s={\overline{)\frac{-t^4}{4}+8t}}_0^2 \\ =-4+16 \\ s =12m \end{gathered}$$

Hence correct option is 1