Given that, At $t=0, x=0$ and $u=\mathbf{v}=-\mathbf{v}_0$
At any time $t$, velocity of particle,
$$
\mathbf{v}=\mathbf{v}_0\left(1-\frac{t}{t_0}\right)=\mathbf{v}_0-\frac{\mathbf{v}_0}{t_0} t
$$
Velocity will become zero at time $t_1$
$$
\begin{array}{ll}
\therefore & 0=\mathbf{v}_0-\frac{\mathbf{v}_0}{t_0} t_1 \\
\Rightarrow & t_1=\frac{\mathbf{v}_0}{\mathbf{v}_0} \times t_0=t_0=10 \mathrm{~s}
\end{array}
$$
$\therefore$ For time interval from 0 to $10 \mathrm{~s}$,
Acceleration, $\mathbf{a}_1=-\frac{\mathbf{v}_0}{t_0}=-\frac{10}{10}=-1 \mathrm{~m} / \mathrm{s}^2$
and for time interval from 10 to $20 \mathrm{~s}$
Acceleration, $\mathbf{a}_2=\frac{\mathbf{v}_0}{t_0}=1 \mathrm{~m} / \mathrm{s}^2$
Now, displacement in time 0 to $10 \mathrm{~s}$
$$
\begin{aligned}
\mathbf{s}_1=\mathbf{u} t_1+\frac{1}{2} \mathbf{a}_1 \Rightarrow t_1^2 & =\mathbf{v}_0(10)-\frac{1}{2} \times 1(10)^2 \\
& =10 \times 10-50=50 \mathrm{~m}
\end{aligned}
$$
and displacement for time $10 \mathrm{~s}$ to $20 \mathrm{~s}$,
$$
\begin{aligned}
t_2 & =20-10=10 \mathrm{~s} \\
\mathbf{s}_2=\mathbf{u} t+\frac{1}{2} \mathbf{a}_2 t_2^2= & 0+\frac{1}{2} \times 1 \times(10)^2 \\
& =50 \mathrm{~m}\left(\text { At } t_1=10 \mathrm{~s}, u_2=v=0\right)
\end{aligned}
$$
Total distance travelled, $d=\left|s_1\right|+\left|s_2\right|$ $=50+50=100 \mathrm{~m}$