2-10 할리데이 10판 솔루션 일반물리학

(a) $$\begin{aligned} \\ v_{e1} &= \frac{v_1 + v_2}{2}= \frac{\frac{d}{t_1}+\frac{d}{t_2}}{2} \\ &= \frac{d}{2} \left( \frac{1}{t_1}+\frac{1}{t_2}\right) \\ &= \frac{d}{2} \frac{t_1+t_2}{t_1t_2} \\ &= \frac{d(t_1+t_2)}{2t_1t_2}\ne v_e \end{aligned}$$ $$\begin{aligned} \\ v_{e2} &= \frac{d}{\bar t}= \frac{d}{\frac{t_1+t_2}{2}} \\ &= \frac{2d}{t_1+t_2} \cdots({2-10-1} ) \\ &= \frac{d_1+d_2}{t_1+t_2} \\ &= \frac{\Sigma d}{\Sigma t} = v_e \end{aligned}$$
(b) $$\begin{cases} R&=0.0240 \\ v_c : v_w &= 1 : R \\ v_w &= Rv_c \\ v_1 &= v_c+v_w = v_c+Rv_c \\ v_2 &= v_c-v_w = v_c-Rv_c \end{cases}$$ $$\begin{aligned} \\ &({2-10-1}) \\ &\downarrow \\ v_{e2} &= \frac{2d}{t_1+t_2}= \frac{2d}{\frac{d}{v_1}+\frac{d}{v_2}} \\ &= \frac{2}{\frac{1}{v_1}+\frac{1}{v_2}} \\ &= \frac{2v_1v_2}{v_1+v_2} \end{aligned}$$ $$\begin{aligned} \\ \Ans &= \abs{v_{e1} - v_{e2}} \\ &= \abs{\frac{v_1 + v_2}{2}-\frac{2v_1v_2}{v_1+v_2}} \\ &= \abs{\frac{(v_1 + v_2)^2-4v_1v_2}{2(v_1+v_2)}} \\ &= \abs{\frac{(v_1 - v_2)^2}{2(v_1+v_2)}} \\ &= \abs{\frac{\left\{v_c+Rv_c - (v_c-Rv_c)\right\}^2}{2(v_c+Rv_c+v_c-Rv_c)}} \\ &= \abs{\frac{(2Rv_c)^2}{2(2v_c)}} \\ &= \abs{R^2v_c} = \abs{Rv_w} \\ &= 0.0240\abs{v_w} \end{aligned}$$