1-23 할리데이 11판 솔루션 일반물리학

$$ \text{put}\begin{cases} t_A=&a_A\cdot t_B+b_A\\ t_B=&\text{ref}\\ t_C=&a_C\cdot t_B+b_C\\ \end{cases} $$ $$t_A=a_A\cdot t_B+b_A$$ $$ \begin{cases} 312=&a_A\cdot 125&+b_A\\ 512=&a_A\cdot 290&+b_A\\ \end{cases} $$ $$ \begin{cases} a_A=&\cfrac{40}{33}\\ b_A=&\cfrac{5296}{33}\\ \end{cases} $$ $$t_A=\cfrac{40}{33}\cdot t_B+\cfrac{5296}{33}~\cdots~(1)$$ $$ \begin{cases} 92=&a_C\cdot 25&+b_C\\ 142=&a_C\cdot 200&+b_C\\ \end{cases} $$ $$ \begin{cases} a_C=&\cfrac{2}{7}\\ b_C=&\cfrac{594}{7}\\ \end{cases} $$ $$t_C=\cfrac{2}{7}\cdot t_B+\cfrac{594}{7}~\cdots~(2)$$ $$ \begin{cases} t_A=&\cfrac{40}{33}\cdot t_B+\cfrac{5296}{33}\\ t_B=&\text{ref}\\ t_C=&\cfrac{2}{7}\cdot t_B+\cfrac{594}{7}\\ \end{cases} $$ $$ \Delta t_A=350\ut{s},$$ $$\ab{a}$$ $$ \begin{aligned} 350\ut{s}=&\Delta t_A\\ =&\Delta\(\frac{40}{33}\cdot t_B+\cfrac{5296}{33}\)\\ =&\frac{40}{33}\Delta t_B\\ \end{aligned} $$ $$ \begin{aligned} \Delta t_B=&\frac{33}{40}\cdot350\ut{s}\\ =&\frac{1155}{4}\ut{s}\\ =&288.75\ut{s}\\ \approx&289\ut{s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} 350\ut{s}=&\Delta t_A\\ =&\frac{40}{33}\Delta t_B\\ =&\frac{40}{33}\Delta \bra{\frac{7}{2}\(t_C-\cfrac{594}{7}\)}\\ =&\frac{40}{33}\cdot\frac{7}{2}\Delta \(t_C-\cfrac{594}{7}\)\\ =&\frac{40}{33}\cdot\frac{7}{2}\Delta t_C\\\ \end{aligned} $$ $$ \begin{aligned} \Delta t_C=&\frac{33}{40}\cdot\frac{2}{7}\cdot350\ut{s}\\ =&\frac{165}{2}\ut{s}\\ =&82.5\ut{s} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} 350\ut{s}=&t_A\\ =&\cfrac{40}{33}\cdot t_B+\cfrac{5296}{33}\\ \end{aligned} $$ $$ \begin{aligned} t_B=&\frac{3127}{20}\ut{s}\\ =&156.35\ut{s}\\ \approx&156\ut{s} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} 30.0\ut{s}=&t_C\\ =&\cfrac{2}{7}\cdot t_B+\cfrac{594}{7} \end{aligned} $$ $$t_B=-192\ut{s}$$