2-51 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&:\text{Red Train}\\ G&:\text{Green Train}\\ \end{cases} $$ $$ \begin{cases} v_R&=72\ut{km/h}=20\ut{m/s}\\ v_G&=160\ut{km/h}=\frac{400}{9}\ut{m/s}\\ S&=950\ut{m}\\ a&=-1.0\ut{m/s^2}\\ \end{cases} $$ $$ 2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2(-1)S_{\text{Stop}}&=(0)^2-{v_0}^2\\ S_{\text{Stop}}&=\frac{{v_0}^2}{2}\\ \end{aligned} $$ $$ \begin{cases} S_{R\text{stop}}&=\cfrac{{v_{R0}}^2}{2}\\ S_{G\text{stop}}&=\cfrac{{v_{G0}}^2}{2}\\ \end{cases} $$ $$ \begin{cases} S_{R\text{stop}}&=\cfrac{20^2}{2}\\ S_{G\text{stop}}&=\cfrac{\(\frac{400}{9}\)^2}{2}\\ \end{cases} $$ $$ \begin{cases} S_{R\text{stop}}&=200\ut{m}\\ S_{G\text{stop}}&=\cfrac{80000}{81}\ut{m}\\ \end{cases} $$ $$ \begin{aligned} \Sigma S &= S_{R\text{stop}}+S_{G\text{stop}}\\ &=200+\frac{80000}{81}\\ &=\frac{96200}{81}\ut{m}\\ &\approx 1187.654320987654\ut{m}\\ \end{aligned} $$ $$\therefore \Sigma S>950\ut{m}$$ $$\title{Train Crash}$$

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$$ S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{cases} S_R&=v_{R0}t_A+\frac{1}{2}a_R{t_R}^2\\ S_G&=v_{G0}t_G+\frac{1}{2}a_G{t_G}^2\\ \end{cases} $$ $$ \begin{cases} S_R&=v_{R0}t+\frac{1}{2}a{t}^2\\ S_G&=v_{G0}t+\frac{1}{2}a{t}^2\\ \end{cases} $$ $$ \begin{aligned} \Sigma S &= S_R + S_G\\ &=t(v_{R0}+v_{G0})+2\cdot\frac{1}{2}at^2\\ 950&=\(20+\frac{400}{9}\)t+(1)t^2\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{5}{9} \(\sqrt{6442}-58\)\\ \end{aligned} $$ $$ v=v_0+at,$$ $$ \begin{cases} v_R&=v_{R0}+(-1)(t_R)\\ v_G&=v_{G0}+(-1)(t_G)\\ \end{cases} $$ $$ \begin{cases} v_R&=(20)-\bra{\dfrac{5}{9} \(\sqrt{6442}-58\)}\\ v_G&=\(\dfrac{400}{9}\)-\bra{\dfrac{5}{9} \(\sqrt{6442}-58\)}\\ \end{cases} $$ $$ \begin{cases} v_R&=\cfrac{5}{9} \left(94-\sqrt{6442}\right)\ut{m/s}\\ v_G&=\cfrac{5}{9} \left(138-\sqrt{6442}\right)\ut{m/s}\\ \end{cases} $$ $$ \begin{cases} v_R&\approx7.632182920397518\ut{m/s}\\ v_G&\approx32.07662736484196\ut{m/s}\\ \end{cases} $$ $$ \begin{cases} v_R&\approx7.6\ut{m/s}\\ v_G&\approx32\ut{m/s}\\ \end{cases} $$