2-5 할리데이 11판 솔루션 일반물리학

풀이자주:(b)문제가 묻는것이 불분명하여 임의로 위치를 구했습니다. $$\begin{cases} x\ut{cm}&=5.00(t\ut{sec})-0.500(t\ut{sec})^3\\ &0\lt x\lt15\ut{cm} \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} v&=\dxt{x}=\dt\(5t-0.5t^3\)\\ &=5-1.5t^2=0 \end{aligned}$$ $$\begin{aligned} t&=\sqrt{\frac{5}{1.5}}\ut{sec}\\ &=\sqrt{\frac{10}{3}}\ut{sec}\\ &\approx1.8257418583505538\ut{sec}\\ &\approx1.83\ut{sec}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} x&=5t-\frac{1}{2}t^3\\ &=5\(\sqrt{\frac{10}{3}}\)-\frac{1}{2}\(\sqrt{\frac{10}{3}}\)^3\\ &=\frac{10}{3}\sqrt{\frac{10}{3}}\ut{cm}\\ &\approx6.0858061945018465\ut{cm}\\ &\approx6.09\ut{cm}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} a&=\dxt{v}=\dt\(5-1.5t^2\)\\ &=-3t\\ &=-3\(\sqrt{\frac{10}{3}}\)\ut{cm/s^2}\\ &=-\sqrt{30}\ut{cm/s^2} \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} v&=5-1.5t^2,\\ \end{aligned}$$ $$\begin{cases} t\lt\sqrt{\cfrac{10}{3}}~&\rarr~v>0\\ t=\sqrt{\cfrac{10}{3}}~&\rarr~v=0\\ t>\sqrt{\cfrac{10}{3}}~&\rarr~v\lt0\\ \end{cases}$$ $$\text{Before stop, forward}$$ $$\ab{e}$$ $$\text{After stop, backward}$$ $$\ab{f}$$ $$\begin{aligned} x&=5t-\frac{1}{2}t^3\\ &(t>0,)\\ x_{\text{Max}}&\lt15\\ \therefore x&=0\\ 5t-\frac{1}{2}t^3&=0\\ t&=0,\pm\sqrt{10}\\ \therefore t&=\sqrt{10} \end{aligned}$$