Объяснение:

[tex]a=12t^2+6t \ \ \ \ \ \ t=1 \ c\ \ \ \ \ v=8\ m/c\ \ \ \ \ S=6 \ m.\\v(t)=\int(12t^2+6t)dt=\frac{12t^3}{3}+\frac{6t^2}{2}+C=4t^3 +3t^2+C.\\ v(1)=4*1^3+3*1^2+C=8\\4+3+C=8\\7+C=8\\C=1.\ \ \ \ \Rightarrow\\v(t)=4t^3+3t^2+1.\\s(t)=\int(4t^3+3t^2+1)dt=\frac{4t^4}{4} +\frac{3t^3}{3}+t+C_1=t^4+t^3+t+C_1=6.\\ s(1)=1^4+1^3+1+C_1=6\\1+1+1+C_1=6\\3+C_1=6\\C_1=3.\ \ \ \ \Rightarrow\\s(t)=t^4+t^3+t+3.[/tex]

Ответ: s(t)=t⁴+t³+t+3.