2-42 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} t_0:\text{Drive}\\ t_1:\text{Brake Start}\\ t_2:\text{Stop}\\ \end{cases} $$ $$ \begin{cases} v_{A0}&=80.5\ut{km/h}=\dfrac{805}{36}\ut{m/s}\\ \Sigma S_A&=63.4\ut{m}\\ v_{B0}&=48.3\ut{km/h}=\dfrac{161}{12}\ut{m/s}\\ \Sigma S_B&=28.9\ut{m}\\ \end{cases} $$ $$ 2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2aS_{1\rarr2}&={v_2}^2-{v_1}^2\\ S_{1\rarr2}&=\frac{{0}^2-{v_0}^2}{2a}\\ &=-\frac{{v_0}^2}{2a}\\ \end{aligned} $$ $$ \begin{aligned} \Sigma S &= S_{0\rarr1}+S_{1\rarr2}\\ &=v_0t_{0\rarr1}-\frac{{v_0}^2}{2a} \end{aligned} $$ $$ \begin{cases} \Sigma S_A &= v_{A0}t_{0\rarr1}-\dfrac{{v_{A0}}^2}{2a}\\ \Sigma S_B &= v_{B0}t_{0\rarr1}-\dfrac{{v_{B0}}^2}{2a} \end{cases} $$ $$ \begin{cases} 63.4 &= \(\dfrac{805}{36}\)t_{0\rarr1}-\dfrac{\(\frac{805}{36}\)^2}{2a}\\ 28.9 &= \(\dfrac{161}{12}\)t_{0\rarr1}-\dfrac{\(\frac{161}{12}\)^2}{2a} \end{cases} $$ $$ \begin{cases} a&= -\cfrac{648025}{98712}\ut{m/s^2}\\ t_{0\rarr1}&= \cfrac{651}{575}\ut{s}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} t_{0\rarr1}&= \cfrac{651}{575}\ut{s}\\ &\approx 1.132173913043478\ut{s}\\ &\approx 1.13\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans &= \abs{a}\\ &= \abs{-\frac{648025}{98712}}\ut{m/s^2}\\ &\approx 6.564804684334225\ut{m/s^2}\\ &\approx 6.56\ut{m/s^2}\\ \end{aligned} $$