Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 64

\[\boxed{\mathbf{64.}}\]

\[\textbf{а)}\ s(t) = 5t^{2} - 10t + 1;\]

\[t = 0;\ \ t = 2;\ \ t = 4.\]

\[v(t) = s^{'}(t) = 10t - 10;\]

\[a(t) = s"(t) = 10.\]

\[v(0) = - 10;\]

\[v(2) = 20 - 10 = 10;\]

\[v(4) = 40 - 10 = 30.\ \]

\[a(t) = 10\ в\ любой\ момент\ \]

\[времени.\]

\[v(t) = 0:\]

\[10t - 10 = 0\]

\[10t = 10\]

\[t = 1.\]

\[\textbf{б)}\ s(t) = 2t^{2} - 8t + 1;\]

\[t = 0;\ \ t = 1;\ \ t = 5.\]

\[v(t) = s^{'}(t) = 4t - 8;\]

\[a(t) = s"(t) = 4.\]

\[v(0) = - 8;\]

\[v(1) = 4 - 8 = - 4;\]

\[v(5) = 20 - 8 = 12.\]

\[a(t) = 4\ в\ любой\ момент\]

\[\ времени.\]

\[v(t) = 0:\]

\[4t - 8 = 0\]

\[4t = 8\]

\[t = 2.\]

\[\textbf{в)}\ s(t) = t^{3} - 2t^{2} + t + 1;\]

\[t = 0;\ \ t = 2;\ \ t = 3.\]

\[v(t) = s^{'}(t) = 3t^{2} - 4t + 1;\]

\[a(t) = s"(t) = 6t - 4.\ \]

\[v(0) = 1;\]

\[v(2) = 3 \cdot 4 - 8 + 1 = 5;\]

\[v(3) = 3 \cdot 9 - 12 + 1 = 16.\]

\[a(0) = - 4;\]

\[a(2) = 12 - 4 = 8;\]

\[a(3) = 18 - 4 = 14.\]

\[v(t) = 0:\]

\[3t^{2} - 4t + 1 = 0\]

\[D_{1} = 4 - 3 = 1\]

\[t_{1} = \frac{2 + 1}{3} = 1;\ \ \]

\[t_{2} = \frac{2 - 1}{3} = \frac{1}{3}.\]