\(v = 0 + 2 \times 10 = 20\,m/s\)

and distance travelled by it in \(10 \,sec\)

\({S_1} = \frac{1}{2} \times 2 \times {(10)^2} = 100\;m\)

then it moves with constant velocity (20 \,m/s)for \(30\;\sec \)

\(\;{S_2} = 20 \times 30 = 600\;m\)

After that due to retardation \((4\,m/{s^2})\) it stops

\({S_3} = \frac{{{v^2}}}{{2a}} = \frac{{{{(20)}^2}}}{{2 \times 4}} = 50\,m\)

Total distance travelled \({S_1} + {S_2} + {S_3} = 750\,m\)