2-9 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=0\\ v_1&=31\ut{m/s}\\ a_{0\rarr1}&=2.6\ut{m/s^2}\\ a_{1\rarr2}&=-1.4\ut{m/s^2}\\ v_2&=0 \end{cases} $$ $$\ab{a}$$ $$v=v_0+at,$$ $$ \begin{aligned} t_{0\rarr1}&=\frac{v_1-v_0}{a}\\ &=\frac{(31\ut{m/s})-0}{2.6\ut{m/s^2}}\\ &=\frac{155}{13}\ut{s} \end{aligned} $$ $$ \begin{aligned} t_{1\rarr2}&=\frac{v_2-v_1}{a}\\ &=\frac{(0)-(31\ut{m/s})}{-1.4\ut{m/s^2}}\\ &=\frac{155}{7}\ut{s} \end{aligned} $$ $$ \begin{aligned} \Sigma t&= \frac{155}{13}\ut{s}+\frac{155}{7}\ut{s}\\ &=\frac{3100}{91}\ut{s}\\ &\approx 34.06593406593407\ut{s}\\ &\approx 34\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} S_{0\rarr1}&=\frac{v_1^2-v_0^2}{2a_{0\rarr1}}\\ &=\frac{(31\ut{m/s})^2-0^2}{2(2.6\ut{m/s^2})}\\ &=\frac{4805}{26}\ut{m} \end{aligned} $$ $$ \begin{aligned} S_{1\rarr2}&=\frac{v_2^2-v_1^2}{2a_{1\rarr2}}\\ &=\frac{0^2-(31\ut{m/s})^2}{2(-1.4\ut{m/s^2})}\\ &=\frac{4805}{14}\ut{m} \end{aligned} $$ $$ \begin{aligned} \Sigma S&= \frac{4805}{26}\ut{m}+\frac{4805}{14}\ut{m}\\ &=\frac{48050}{91}\ut{m}\\ &\approx 528.021978021978\ut{m}\\ &\approx 5.3\times10^2\ut{m}\\ \end{aligned} $$