Answer
The distance $d$ the train moved is given by
$$
d=\int_{0}^{t} v\left(t^{\prime}\right) d t^{\prime}$$$$=\int_{0}^{360}\left(100+\frac{3}{2} t\right)^{1 / 2} d t$$$$=\left.\frac{4}{9}\left(100+\frac{3}{2} t\right)^{3 / 2}\right|_{0} ^{360}$$$$=6.7 \times 10^{3} \mathrm{m} .
$$
Work Step by Step
The distance $d$ the train moved is given by
$$
d=\int_{0}^{t} v\left(t^{\prime}\right) d t^{\prime}$$$$=\int_{0}^{360}\left(100+\frac{3}{2} t\right)^{1 / 2} d t$$$$=\left.\frac{4}{9}\left(100+\frac{3}{2} t\right)^{3 / 2}\right|_{0} ^{360}$$$$=6.7 \times 10^{3} \mathrm{m} .
$$
