2-39 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_A&=45.0\ut{km/h}\\ v_B&=70.0\ut{km/h}\\ \end{cases}$$ $$\begin{aligned} \bar v&=\frac{\Sigma S}{\Sigma t}\\ &=\frac{S_A+S_B}{t_A+t_B}\\ &=\frac{2S}{\(\dfrac{S}{v_A}\)+\(\dfrac{S}{v_B}\)}\\ &=\frac{2}{\(\dfrac{1}{v_A}\)+\(\dfrac{1}{v_B}\)}\\ &=\frac{2}{\(\dfrac{1}{45}\)+\(\dfrac{1}{70}\)}\\ &=\frac{1260}{23}\ut{km/h}\\ &\approx 54.78260869565217\ut{km/h}\\ &\approx 54.8\ut{km/h}\\ \end{aligned}$$