Giải bởi Vietjack
\(\int {\left( {{t^2} + 4t} \right)} dt = \frac{{{t^3}}}{3} + 2{t^2} + C \Rightarrow v(t) = \frac{{{t^3}}}{3} + 2{t^2} + C.\)
\({\rm{v}}(0) = 15 \Rightarrow {\rm{v}}({\rm{t}}) = \frac{{{{\rm{t}}^3}}}{3} + 2{{\rm{t}}^2} + 15.\)
\({\rm{s}} = \int {{0^3}} \left( {15 + \frac{{{{\rm{t}}^3}}}{3} + 2{{\rm{t}}^2}} \right){\rm{dt}} = \left. {\left( {15{\rm{t}} + \frac{{{{\rm{t}}^4}}}{{12}} + \frac{{2{{\rm{t}}^3}}}{3}} \right)} \right|_0^3 = 69,75(\;{\rm{m}}).\) Chọn C.
