Distance travelled by the particle between 0 to $10 \mathrm{~s}=$ Area of $\triangle \mathrm{OAB}=\frac{1}{2} \times 10 \times 12=60 \mathrm{~m}$.
Average speed $=60 / 10=6 \mathrm{~m} / \mathrm{s}$
Let $s_1$ and $s_2$ be the distances covered by the particle in the time interval $2 \mathrm{~s}$ to $5 \mathrm{~s}$ and $5 \mathrm{~s}$ to $6 \mathrm{~s}$, then the total distance covered in the given time interval is $\mathrm{s}_1+\mathrm{s}_2$
To find $s_1:$ Let, $u_1=$ velocity of particle after $2 \mathrm{~s}$ and $a_1$ $=$ acceleration of the particle during the time interval zero to $5 \mathrm{~s}$.
Then $u=0, v=12 \mathrm{~m} / \mathrm{s}, a=a_1$ and $t=5 \mathrm{~s}$.
We have, $a_1=(v-u) / t=(12-0) / 5=2.4 \mathrm{~ms}^{-2}$
$$
\therefore u_1=u+a_1 t=0+2.4 \times 2.4=4.8 \mathrm{~ms}^{-2}
$$
Thus for the distance travelled by particle in $3 \mathrm{~s}$ (i.e. time interval $2 \mathrm{~s}$ to $5 \mathrm{~s}$ ), we get,
$$
\begin{aligned}
u_1 &=4.8 \mathrm{~ms}^{-1}, t_1=3 \mathrm{~s}, a_1=2.4 \mathrm{~ms}^{-2} \\
\mathrm{~s}_1=u_1 \mathrm{t}_1+\frac{1}{2} a_1 \mathrm{t}_1{ }^2=4.8 \times 3+\frac{1}{2} \times 2.4 \times 32 \\
=25.2 \mathrm{~m}
\end{aligned}
$$
To find $s_2$ : Let, $a_2=$ acceleration of the particle during the motion, $t=5 \mathrm{~s}$ to $t=10 \mathrm{~s}$.
We have, $a_1=(0-12) /(10-5)=-2.4 \mathrm{~ms}^{-2}$
Taking motion of the particle during the motion, $t=5 \mathrm{~s}$ to $t=10 \mathrm{~s}$, we get,
$$
u_2=12 \mathrm{~ms}^{-1}, t_2=1 \mathrm{~s}, a_2=-2.4 \mathrm{~ms}^{-2}
$$
$$
\begin{aligned}
\mathrm{s}_2=u_2 t_2+\frac{1}{2} a_2 t_2{ }^2 &=12 \times 1+\frac{1}{2} \times(-2.4) \times 12 \\
&=10.8 \mathrm{~m}
\end{aligned}
$$
$\therefore \quad$ Total distance travelled, $\mathrm{S}=25.2+10.8$
$=36 \mathrm{~m}$
Average speed $=36 /(6-2)=36 / 4=9 \mathrm{~ms}^{-1}$