A particle starts from rest and has an acceleration of 2 ~m / s^2...

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, $\begin{gathered}s_1=u t_1+\frac{1}{2}a_1 t_1^2=(0 \times10)+\frac{1}{2} \times2 \times(10)^2 \\ =100 \mathrm{~m} \end{gathered}$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 $\mathrm{~m}$