Initial velocity $(u)=0$,
Acceleration $\left(a_1\right)=2 \mathrm{~m} / \mathrm{s}^2$ and time during acceleration $\left(t_1\right)=10 \mathrm{sec}$.
Time during constant velocity $\left(t_2\right)=30 \mathrm{sec}$ and retardation $\left(a_2\right)=-4 \mathrm{~m} / \mathrm{s}^2$ (-ve sign due to retardation). Distance covered by the particle during acceleration,
$$
\begin{gathered}
s_1=u t_1+\frac{1}{2} a_1 t_1^2=(0 \times 10)+\frac{1}{2} \times 2 \times(10)^2 \\
=100 \mathrm{~m}
\end{gathered}
$$
And velocity of the particle at the end of acceleration,
$$
v=u+a_1 t_1=0+(2 \times 10)=20 \mathrm{~m} / \mathrm{s} .
$$
Therefore distance covered by the particle during constant velocity $\left(s_2\right)=v \times t_2$
$$
=20 \times 30=600 \mathrm{~m}
$$
Relation for the distance covered by the particle during retardation $\left(s_3\right)$ is $v^2=u^2+2 a_2 s_3$
or, $\quad(0)^2=(20)^2+2 \times(-4) \times s_3=400-8 s_3$
or, $s_3=400 / 8=50 \mathrm{~m}$
Therefore total distance covered by the particle


$$
s=s_1+s_2+s_3=100+600+50=750 \mathrm{~m}
$$