4-51 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} A:\text{Ship}\\ B:\text{Water} \end{cases} $$ $$ \begin{cases} \Delta x &= 8.0\ut{km}\\ \vec v_{B}&=2.6\ut{km/h}\j\\ v_{A\larr B}&=8.0\ut{km/h}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec v_{A\larr B}&=8\i\ut{km/h},\\ \vec v_A &= \vec v_{A\larr B}+\vec v_B\\ &= 8\i+2.6\j\\ \end{aligned} $$ $$ \begin{aligned} \theta_A&=\tan^{-1}\frac{2.6}{8}\\ &\approx 0.3142318990843383\ut{rad}\\ &\approx 0.31\ut{rad}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} t&=\frac{\Delta x}{v_{Ax}}\\ &=\frac{8\ut{km}}{8\ut{km/h}}\\ &=1\ut{h}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} t_1&=\frac{S_1}{v_1}\\ &=\frac{4}{2.6+8}\ut{h}\\ &=\frac{20}{53}\ut{h} \end{aligned} $$ $$ \begin{aligned} t_2&=\frac{S_2}{v_2}\\ &=\frac{-4}{2.6-8}\ut{h}\\ &=\frac{20}{27}\ut{h} \end{aligned} $$ $$ \begin{aligned} \Ans &= t_1 + t_2\\ &=\frac{20}{53}+\frac{20}{27}\\ &=\frac{1600}{1431}\ut{h}\approx 1.1\ut{h}\\ &=\frac{32000}{477}\ut{min}\approx 67\ut{min} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} t_1&=\frac{S_1}{v_1}\\ &=\frac{-4}{2.6-8}\ut{h}\\ &=\frac{20}{27}\ut{h} \end{aligned} $$ $$ \begin{aligned} t_2&=\frac{S_2}{v_2}\\ &=\frac{4}{2.6+8}\ut{h}\\ &=\frac{20}{53}\ut{h} \end{aligned} $$ $$ \begin{aligned} \Ans &= t_1 + t_2\\ &=\frac{20}{27}+\frac{20}{53}\\ &=\frac{1600}{1431}\ut{h}\approx 1.1\ut{h}\\ &=\frac{32000}{477}\ut{min}\approx 67\ut{min} \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \vec v_A &= \vec v_{A\larr B}+\vec v_B\\ &= 8\cos\theta\i+8\sin\theta\j+2.6\ut{km/h}\j\\ \end{aligned} $$ $$ \begin{aligned} v_{Ax}&=8\cos\theta\\ \end{aligned} $$ $$\theta=0$$ $$\ab{f}$$ $$ \begin{aligned} t&=\frac{\Delta x}{v_{Ax}}\\ &=\frac{8\ut{km}}{8\ut{km/h}}\\ &=1\ut{h}\\ \end{aligned} $$