2-16 할리데이 10판 솔루션 일반물리학

$$x = 4.0- 6.0t^2 \cdots({2-16-1})$$
(a) $$\Ans = t_{\dot x=0} =?$$ $$\begin{aligned} \dot x &= \dxt{x} \\&= \dt(4.0- 6.0t^2) \\&= -12t=0 \end{aligned}$$ $$t=0\ut{s}$$
(b) $$\begin{aligned} \Ans &= x(0) =? \\ &= 4.0- 6.0(0)^2 \\ &= 4.0\ut{m} \end{aligned}$$
(c) $$\Ans = t_{x=0,t<0} =?$$ $$\begin{aligned} x &= 4.0- 6.0t^2 = 0 \\ t &= \pm \sqrt{\frac{2}{3}} \cdots({2-16-2}) \end{aligned}$$ $$\therefore t = -\sqrt{\frac{2}{3}}$$
(d) $$\Ans = t_{x=0,t>0} =?$$ $$(2-16-2),$$ $$t = \sqrt{\frac{2}{3}}$$
(e)

(f)
+20t
(그래프참조)
(g)
증가시킴
(그래프참조)