2-13 할리데이 10판 솔루션 일반물리학

(a) $$\begin{cases} v_1 &= 55\ut{km/h} \\ v_2 &= 90\ut{km/h} \\ t &= t_1 = t_2 \end{cases}$$ $$\begin{aligned} \Ans &= \bar v_a = \frac{\Sigma S}{\Sigma t} =\frac{S_1+S_2}{t_1+t_2} \\ &=\frac{S_1+S_2}{2t} \\ &=\frac{1}{2}\left(\frac{S_1}{t}+\frac{S_2}{t}\right) \\ &=\frac{v_1+v_2}{2} \\ &=\frac{55\ut{km/h}+90\ut{km/h}}{2} \\ &=72.5\ut{km/h} \\ &\approx 73\ut{km/h} \end{aligned}$$
(b) $$\begin{cases} v_3 &= 55\ut{km/h} \\ v_4 &= 90\ut{km/h} \\ S &= S_3 = S_4 \end{cases}$$ $$\begin{aligned} \Ans &= \bar v_b = \frac{\Sigma S}{\Sigma t} =\frac{S_3+S_4}{t_3+t_4} \\ &=\frac{2S}{\frac{S}{v_3}+\frac{S}{v_4}} \\ &=\frac{2}{\frac{1}{v_3}+\frac{1}{v_4}} \\ &=\frac{2v_3v_4}{v_3+v_4} \\ &=\frac{2\cdot55\ut{km/h}\cdot90\ut{km/h}}{55\ut{km/h}+90\ut{km/h}} \\ &=\frac{1980}{29}\ut{km/h} \\ &\approx 68.27586206896552\ut{km/h} \\ &\approx 68\ut{km/h} \end{aligned}$$
(c) $$\begin{aligned} \Ans&=\bar v=\frac{\Sigma S}{\Sigma t} \\ &=\frac{S_1+S_2+S_3+S_4}{t_1+t_2+t_3+t_4} \\ &=\frac{4S}{\frac{2S}{\bar v_a}+\frac{2S}{\bar v_b}} \\ &=\frac{2}{\frac{1}{\bar v_a}+\frac{1}{\bar v_b}} \\ &=\frac{2\bar v_a\bar v_b}{\bar v_a+\bar v_b} \\ &=\frac{2\cdot72.5\ut{km/h}\cdot\frac{1980}{29}\ut{km/h}}{72.5\ut{km/h}+\frac{1980}{29}\ut{km/h}} \\ &=\frac{114840}{1633}\ut{km/h} \\ &\approx 70.324556031843233313\ut{km/h} \\ &\approx 70\ut{km/h} \end{aligned}$$
(d)
변위가0이므로 평균속도는 0
(e)

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