Ta thấy đồ thị qua gốc tọa độ \(O\left(0;0\right)\)
\(\Rightarrow s\left(0\right)=0\Leftrightarrow d=0\)
\(\Rightarrow s\left(t\right)=at^3+bt^2+cx\)
\(s\left(2\right)=12\Leftrightarrow a.2^3+b.2^2+c.2=12\Leftrightarrow8a+4b+2c=12\)
\(\Leftrightarrow4a+2b+c=6\left(1\right)\)
\(s\left(4\right)=24\Leftrightarrow a.4^3+b.4^2+c.4=12\Leftrightarrow64a+16b+4c=24\)
\(\Leftrightarrow16a+4b+c=6\left(2\right)\)
\(v=s'\left(t\right)=3at^2+2bt+c\)
\(s'\left(t\right)=0\Leftrightarrow s'\left(4\right)=0\Leftrightarrow3a.4^2+2b.4+c=0\) (\(s\left(t=4\right)_{max}\))
\(\Leftrightarrow48a+8b+c=0\left(3\right)\)
\(\left(1\right);\left(2\right);\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}4a+2b+c=6\\16a+4b+c=6\\48a+8b+c=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{4}\\b=\dfrac{9}{2}\\c=0\end{matrix}\right.\)
\(\Rightarrow s\left(t\right)=-\dfrac{3}{4}t^3+\dfrac{9}{2}t^2\)
\(\Rightarrow v\left(t\right)=s'\left(t\right)=-\dfrac{9}{4}t^2+9t\)
\(v'\left(t\right)=-\dfrac{9}{2}t+9\)
\(v'\left(t\right)=0\Leftrightarrow-\dfrac{9}{2}t+9=0\Leftrightarrow t=2\)
\(\Rightarrow v\left(2\right)=v_{max}=-\dfrac{9}{4}.2^2+9.2=9\left(km/h\right)\)
Ta thấy đồ thị qua gốc tọa độ \(O\left(0;0\right)\)
\(\Rightarrow s\left(0\right)=0\Leftrightarrow d=0\)
\(\Rightarrow s\left(t\right)=at^3+bt^2+cx\)
\(s\left(2\right)=12\Leftrightarrow a.2^3+b.2^2+c.2=12\Leftrightarrow8a+4b+2c=12\)
\(\Leftrightarrow4a+2b+c=6\left(1\right)\)
\(s\left(4\right)=24\Leftrightarrow a.4^3+b.4^2+c.4=12\Leftrightarrow64a+16b+4c=24\)
\(\Leftrightarrow16a+4b+c=6\left(2\right)\)
\(v=s'\left(t\right)=3at^2+2bt+c\)
\(s'\left(t\right)=0\Leftrightarrow s'\left(4\right)=0\Leftrightarrow3a.4^2+2b.4+c=0\) (\(s\left(t=4\right)_{max}\))
\(\Leftrightarrow48a+8b+c=0\left(3\right)\)
\(\left(1\right);\left(2\right);\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}4a+2b+c=6\\16a+4b+c=6\\48a+8b+c=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{4}\\b=\dfrac{9}{2}\\c=0\end{matrix}\right.\)
\(\Rightarrow s\left(t\right)=-\dfrac{3}{4}t^3+\dfrac{9}{2}t^2\)
\(\Rightarrow v\left(t\right)=s'\left(t\right)=-\dfrac{9}{4}t^2+9t\)
\(v'\left(t\right)=-\dfrac{9}{2}t+9\)
\(v'\left(t\right)=0\Leftrightarrow-\dfrac{9}{2}t+9=0\Leftrightarrow t=2\)
\(\Rightarrow v\left(2\right)=v_{max}=-\dfrac{9}{4}.2^2+9.2=9\left(km/h\right)\)
