For first part of journey, $x=0, a=4 \mathrm{~ms}^{-2}$, time $=t_1$. Velocity attained by particle at end of time $t_1$ is $\quad v_1=u+a t_1=4 t_1$
$\therefore$ Displacement in first $t_1$ seconds is
$$
s_1=\frac{1}{2} a t_1^2=\frac{1}{2} \times 4 \times t_1^2=2 t_1^2
$$


For second part of journey, initial velocity, $v_1=4 t_1$, time $=t_2$ and acceleration $=0$.
$\therefore$ Displacement in next $t_2$ seconds is $s_2=v_1 t_2=4 t_1 t_2$ For third part of journey,
Initial velocity, $v_1=4 t_1$
Final velocity $=0$
Time interval $=t_1$
So, acceleration $=\frac{0-4 t_1}{t_1}=-4 \mathrm{~ms}^{-2}$
$\therefore$ Displacement in third part is
$$
s_3=\frac{-16 t_1^2}{-8}=2 t_1^2 \quad\left(\because v_2^2-v_1^2=2 a s_3\right)
$$
Now, $\quad v_{\text {avg }}=\frac{\text { Total displacement }}{\text { Total time }}$
$$
\begin{aligned}
& 5.1=\frac{s_1+s_2+s_3}{t_1+t_2+t_1} \\
& {\left[\therefore \text { third time interva] }=t_1\right]} \\
& 5.1=\frac{2 t_1^2+4 t_1 t_2+2 t_1^2}{10}
\end{aligned}
$$

$$
\begin{array}{lc}
\Rightarrow & 5 \mathrm{l}=4 t_1^2+4 t_1\left(10-2 t_1\right) \\
\Rightarrow & 5 \mathrm{l}=4 t_1^2+40 t_1-8 t_1^2 \\
\Rightarrow & 4 t_1^2-40 t_1+5 \mathrm{l}=0 \\
\Rightarrow & 4 t_1^2-34 t_1-6 t_1+5 \mathrm{l}=0
\end{array}
$$

$$
\begin{aligned}
& \Rightarrow \quad\left(2 t_1-3\right)\left(2 t_1-17\right)=0 \Rightarrow t_1=\frac{3}{2} \\
& \text { or } \\
& t_1=\frac{17}{2} \Rightarrow t_1=1.5 \mathrm{~s} \\
&
\end{aligned}
$$
(As $t_1=8.5 \mathrm{~s}$ is not possible. It exceeds total time.)