2-52 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} t_0&:\text{Start}\\ t_1&:\text{Max Speed}\\ t_2&:\text{End}\\ \end{cases} $$ $$ \begin{cases} v_{1}&=11.00\ut{m/s}\\ S_{0\rarr1}&=12.0\ut{m}\\ S_{0\rarr2}&=100\ut{m} \end{cases} $$ $$\ab{a}$$ $$ S=\frac{1}{2}(v+v_0)t,$$ $$ \begin{aligned} S_{0\rarr1}&=\frac{1}{2}(v_1+v_0)t_{0\rarr1}\\ &=\frac{1}{2}(11+0)t_{0\rarr1}\\ t_{0\rarr1}&=\frac{2S_{0\rarr1}}{11}\\ \end{aligned} $$ $$ \begin{aligned} S_{1\rarr2}&=v_1t_{1\rarr2}\\ 100-S_{0\rarr1}&=11t_{1\rarr2}\\ t_{1\rarr2}&=\frac{100-S_{0\rarr1}}{11}\\ \end{aligned} $$ $$ \begin{aligned} \Sigma t&=t_{0\rarr1}+t_{1\rarr2}\\ &=\(\frac{2S_{0\rarr1}}{11}+\frac{100-S_{0\rarr1}}{11}\)\\ &=\frac{S_{0\rarr1}+100}{11}\\ &=\frac{(12)+100}{11}\\ &=\frac{112}{11}\ut{s}\\ &\approx 10.18181818181818\ut{s}\\ &\approx 10.2\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma t&=\frac{S_{0\rarr1}+100}{11}\\ 9.9&=\frac{S_{0\rarr1}+100}{11}\\ S_{0\rarr1}&=\frac{89}{10}\ut{m}\\ &=8.9\ut{m}\\ \end{aligned} $$