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)