4-4 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} P:\text{Plane}\\ A:\text{Air}\\ t_{\alpha}:\text{go out}\\ t_{\beta}:\text{come in}\\ \end{cases} $$ $$ \begin{cases} \Delta x &= 4000\ut{km}\\ v_{P\larr A}&= 1000\ut{km/h}\\ \Delta t_{\alpha\rarr\beta} &= 100.0\ut{min}=\cfrac{5}{3}\ut{h}\\ \end{cases} $$ $$ \begin{cases} v_{P\alpha} &= v_{P\larr A}+v_A\\ v_{P\beta} &= v_{P\larr A}-v_A \end{cases} $$ $$ \begin{cases} t_\alpha &=\cfrac{\Delta x}{v_{P\alpha}}\\ t_\beta &=\cfrac{\Delta x}{v_{P\beta}}\\ \end{cases} $$ $$ \begin{aligned} t_\beta-t_\alpha&=\cfrac{\Delta x}{v_{P\beta}}-\cfrac{\Delta x}{v_{P\alpha}}\\ &=\Delta x\(\cfrac{1}{ v_{P\larr A}-v_A}-\cfrac{1}{v_{P\larr A}+v_A}\) \end{aligned} $$ $$ \begin{aligned} v_A&=\frac{\sqrt{(v_{P\larr A})^2 \(\Delta t_{\alpha\rarr\beta}\)^2+(\Delta x)^2}-\Delta x}{\Delta t_{\alpha\rarr\beta}}\\ &=\frac{\sqrt{\(1000\ut{km/h}\)^2 \(\frac{5}{3}\ut{h}\)^2+\(4000\ut{km}\)^2}-4000\ut{km}}{\frac{5}{3}\ut{h}}\\ &=200\ut{km/h} \end{aligned} $$