2-20 할리데이 10판 솔루션 일반물리학

$$ x = 25t-6.0t^3 $$ $$ \begin{aligned} v &= \dot x = \frac{dx}{dt} \\ &= \frac{d}{dt}(25t-6.0t^3) \\ &= 25-18 t^2 \end{aligned} $$ $$ \begin{aligned} a &= \dot v = \frac{dv}{dt} \\ &= \frac{d}{dt}(25-18 t^2) \\ &= -36 t \end{aligned} $$ $$ \therefore \begin{cases} x &= 25t-6.0t^3 \\ v &= 25-18 t^2 \\ a &= -36 t \end{cases}\cdots(2-20-1) $$
(a) $$\Ans =t_{v=0} = ? $$ $$ (2-20-1), $$ $$v = 25-18 t_a^2 =0 $$ $$t_a = \pm \frac{5}{3 \sqrt{2}}\ut{s} $$
(b) $$ \Ans =t_{a=0} = ? $$ $$ (2-20-1), $$ $$ a = -36 t_b =0 $$ $$ t_b = 0\ut{s} $$
(c) $$ \Ans =t_{a<0} $$ $$(2-20-1), $$ $$a = -36 t_c <0 $$ $$ t_c >0\ut{s} $$
(d) $$ \Ans =t_{a>0} $$ $$ (2-20-1), $$ $$ a = -36 t_d >0 $$ $$ t_d <0\ut{s} $$
(e)