2-40 할리데이 11판 솔루션 일반물리학

$$\begin{cases} S_A&=-40.0\ut{m}\\ a_B&=1.50\ut{m/s^2}\\ v_2&=-3.00\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} S_A&=v_0t_A+\frac{1}{2}a_A{t_A}^2\\ -40&=(0)t_A+\frac{1}{2}(-g){t_A}^2\\ \end{aligned}$$ $$\therefore t_A=4\sqrt{\frac{5}{g}}$$ $$2aS=v^2-{v_0}^2,$$ $$\begin{aligned} 2(a_A)(S_A)&={v_1}^2-{v_0}^2\\ 2(-g)(-40)&={v_1}^2-{0}^2\\ \end{aligned}$$ $$\therefore v_1=-4\sqrt{5g}$$ $$v=v_0+at,$$ $$\begin{aligned} v_2&=v_1+(a_B)t_B,\\ -3&=(-4\sqrt{5g})+(1.5)t_B,\\ \end{aligned}$$ $$\therefore t_B=\frac{8 }{3}\sqrt{5g}-2$$ $$\begin{aligned} \Ans&=\Sigma t\\ &=t_A+t_B\\ &=4\sqrt{\frac{5}{g}}+\frac{8 }{3}\sqrt{5g}-2\\ &=\(\frac{226133 }{15}\sqrt{\frac{2}{980665}}-2\)\ut{s}\\ &\approx 19.52917288841768\ut{s}\\ &\approx 19.5\ut{s} \end{aligned}$$ $$\ab{b}$$ $$S=vt-\frac{1}{2}at^2,$$ $$\begin{aligned} S_B&=v_2t_B-\frac{1}{2}a_B{t_B}^2\\ &=(-3)\(\frac{8 }{3}\sqrt{5g}-2\)-\frac{1}{2}(1.5)\(\frac{8 }{3}\sqrt{5g}-2\)^2\\ &=3-\frac{80}{3}g\\ &=-\frac{193883}{750}\ut{m}\\ \end{aligned}$$ $$\begin{aligned} \Ans&=\abs{S_A+S_B}\\ &=\abs{(-40)+\(-\frac{193883}{750}\)}\ut{m}\\ &=\frac{223883}{750}\ut{m}\\ &\approx 298.5106666666667\ut{m}\\ &\approx 299\ut{m} \end{aligned}$$