2-18 할리데이 10판 솔루션 일반물리학

$$x = 12t^2-2t^3$$ $$\begin{aligned} v &= \dot x = \dxt{x} \\ &= \dt(12t^2-2t^3) \\ &= 24t-6t^2 \end{aligned}$$ $$\begin{aligned} a &= \dot v = \dxt{v}=\ddot x = \dxtt{x} \\ &= \dt(24t-6t^2) \\ &= 24-12t \end{aligned}$$ $$\therefore \begin{cases} x &= 12t^2-2t^3 \\ v &= 24t-6t^2 \\ a &= 24-12t \end{cases}\cdots({2-18-1})$$
(a) $$\Ans = x(3.5) =?$$ $$(2-18-1),$$ $$\begin{aligned} x(3.5) &= 12(3.5)^2-2(3.5)t^3 \\ &= \frac{245}{4}\ut{m} \\ &= 61.25\ut{m} \\ &\approx 61\ut{m} \end{aligned}$$
(b) $$\Ans = v(3.5) =?$$ $$(2-18-1),$$ $$\begin{aligned} v(3.5) &= 24(3.5)-6(3.5)^2 \\&= \frac{21}{2}\ut{m/s} \\&= 10.5\ut{m/s} \\&\approx 11\ut{m/s} \end{aligned}$$
(c) $$\Ans = a(3.5) =?$$ $$(2-18-1),$$ $$\begin{aligned} a &= 24-12t \\ a(3.5) &= 24-12(3.5) \\ &= -18\ut{m/s^2} \end{aligned}$$
(d) $$\max x = ?$$ $$(2-18-1),$$ $$v = 24t_d-6t_d^2 =0$$ $$t_d=0,4$$ $$a(t_d) = (0,24), (4,-24)\cdots({2-18-2})$$ $$\therefore \begin{cases} x_{LocalMin}&=x(0) \\ x_{LocalMax}&=x(4) \end{cases}$$ $$\begin{aligned} \\ \max x &= x(4) \ (\because t>0)\cdots({2-18-3}) \\ x(4) &= 12(4)^2-2(4)^3 \\ &=64\ut{m} \end{aligned}$$
(e) $$(2-18-3),$$ $$t_e =4\ut{s}$$
(f) $$\max v = ?$$ $$(2-18-1),$$ $$a = 24-12t_f =0$$ $$t_f=2\cdots(2-18-4)$$ $$\dot a = -12$$ $$\therefore v_{LocalMax}=v(2)$$ $$\therefore \max v = v(2)$$ $$\begin{aligned} v(2) &= 24(2)-6(2)^2 \\ &=24\ut{m/s} \end{aligned}$$
(g) $$(2-18-4),$$ $$t_g =2\ut{s}$$
(h) $$\Ans = a_{v=0}$$ $$(2-18-2),$$ $$a_{v=0} = -24\ut{m/s^2}$$
(i) $$\begin{aligned} \\ \Ans &= \bar x=\frac{x(3)-x(0)}{3-0} \\ &= \frac{\left\{12(3)^2-2(3)^3 \right\}-\left\{12(0)^2-2(0)^3 \right\}}{3-0} \\ &= 18\ut{m/s} \end{aligned}$$