Given that, $a=$ deceleration $=-3 v^{2 / 3} \mathrm{~m} / \mathrm{s}^2$
At initial point $(t=0)$ velocity $=8 \mathrm{~m} / \mathrm{s}$
Distance travelled by the object before it stop is We know that,

$$
\begin{aligned}
a & =\frac{d v}{d t} \Rightarrow a=\frac{d v}{d s} \cdot \frac{d s}{d t} \\
a & =v d v / d s \\
a d s & =v d v \\
-3 v^{2 / 3} d s & =v d s \\
d \quad \quad d s & =-\frac{1}{3} v^{1 / 3} d v
\end{aligned}
$$

Integrate both sides, we get
$$
\begin{aligned}
& \int d s=\frac{-1}{3} \int v^{1 / 3} d v \\
& \Rightarrow \quad s=\frac{-1}{4} v^{4 / 3}+C \\
& \Rightarrow \text { At } \quad t=0, u=8 \mathrm{~m} / \mathrm{s}, s=0 \\
& C=\frac{1}{4}(8)^{4 / 3}=\frac{16}{4}=4 \\
&
\end{aligned}
$$
So,
Therefore, from Eq. (i)
$$
s=\frac{-1}{4} v^{4 / 3}+4
$$

Body stops, when $v=0$.
So,
$$
s=4 \mathrm{~m}
$$