2-58 할리데이 11판 솔루션 일반물리학

$$\begin{cases} t_0&:\text{Start}\\ t_1&:\text{Max Speed}\\ t_2&:\text{Stop}\\ t_A&:t_{0\rarr1}\\ t_B&:t_{1\rarr2}\\ \end{cases}$$ $$\begin{cases} S_A&=2.1\ut{m}\\ v_1&=6.0\ut{m/s}\\ a_B&=-3.0\ut{m/s^2}\\ \end{cases}$$ $$\title{Time A}$$ $$$$ $$\begin{aligned} S&=\frac{1}{2}(v+v_0)t,\\ S_A&=\frac{1}{2}(v_1+v_0)t_A\\ 2.1&=\frac{1}{2}(6+0)t_A\\ \end{aligned}$$ $$\begin{aligned} t_A&=\frac{7}{10}\ut{s}\\ &=0.7\ut{s}\\ \end{aligned}$$ $$\title{Time B}$$ $$\begin{aligned} v&=v_0+at,\\ 0&=(6)+(-3)t_B\\ \end{aligned}$$ $$\begin{aligned} t_B&=2\ut{s} \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} \Sigma t&=2+0.7\\ &=2.7\ut{s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-3)S_B&=(0)^2-{6}^2\\ \end{aligned}$$ $$\begin{aligned} S_B&=6\ut{m}\\ \Sigma S &= S_A+S_B\\ &= (2.1)+(6)\\ &= 8.1\ut{m}\\ \end{aligned}$$