2-17 할리데이 10판 솔루션 일반물리학

$$ x = 9.75 + 1.50t^3 $$
(a) $$ \begin{aligned} \Ans &= \bar v = \frac{x(t_3)-x(t_2)}{t_3-t_2} \\ &= \frac{x(3.00)-x(2.00)}{3.00-2.00} \\ &= \frac{9.75 + 1.50(3.00)^3-\bra{9.75 + 1.50(2.00)^3}}{3.00-2.00} \\ &= \frac{57}{2}\ut{cm/s} \\ &= 28.5\ut{cm/s} \\ \end{aligned} $$
(b) $$\Ans = \dot x(2.00) =? $$ $$ \begin{aligned} \\ \dot x &= \dxt{x} \\ &= \dt(9.75 + 1.50t^3) \\ &= \frac{9}{2}t^2 \cdots({2-17-1}) \end{aligned} $$ $$ \begin{aligned} \dot x(2.00) &= \frac{9}{2}(2.00)^2 \\ &= 18.0\ut{cm/s} \end{aligned} $$
(c) $$ \Ans = \dot x(3.00) =? $$ $$ (2-17-1), $$ $$ \begin{aligned} \dot x(3.00) &= \frac{9}{2}(3.00)^2 \\ &= \frac{81}{2}\ut{cm/s} \\ &= 40.5\ut{cm/s} \end{aligned} $$
(d) $$\Ans = \dot x(2.50) =? $$ $$ (2-17-1), $$ $$ \begin{aligned} \dot x(2.50) &= \frac{9}{2}(2.50)^2 \\ &= \frac{225}{8}\ut{cm/s} \\ &= 28.125\ut{cm/s} \\ &\approx 28.1\ut{cm/s} \end{aligned} $$
(e) $$ \Ans = \dot x(t_e) =? $$ $$ \begin{aligned} x_e &= \frac{x(2.00)+x(3.00)}{2} \\ &= \frac{9.75 + 1.50(2.00)^3 + 9.75 + 1.50(3.00)^3}{2} \\ &= 36\ut{cm} \end{aligned} $$ $$ x = 9.75 + 1.50t_e^3 = 36\ut{cm} $$ $$t_e = \sqrt[3]{\frac{35}{2}}\ut{s} $$ $$ (2-17-1), $$ $$ \begin{aligned} \dot x(t_e) &= \frac{9}{2}t_e^2 \\ &= \frac{9}{2}\sqrt[3]{\frac{35}{2}}^2\ut{cm/s} \\ &\approx 30.33224437247831\ut{cm/s} \\ &\approx 30.3\ut{cm/s} \end{aligned} $$
(f)