10판/2. 직선운동

2-9 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 22. 19:36

($L_1=1\ut{km}$ 로 추정함.) $$ \begin{cases} S &= 1\ut{km} \\ t_1 &= 2\ut{min}+27.95\ut{s} = 147.95\ut{s} \\ t_2 &= 2\ut{min}+28.15\ut{s} = 148.15\ut{s} \\ L_2 &= L_1 + \Delta L \\ v_1 &> v_2 \end{cases} $$ $$ \begin{aligned} v_1 &> v_2 \\ \frac{L_1}{t_1} &> \frac{L_2}{t_2} \\ \frac{L_1}{t_1} &> \frac{L_1 + \Delta L}{t_2} \\ \frac{L_1}{t_1} &> \frac{L_1}{t_2}+\frac{\Delta L}{t_2} \\ \\ \frac{\Delta L}{t_2} &< \frac{L_1}{t_1} - \frac{L_1}{t_2} \\ \Delta L &< t_2\(\frac{L_1}{t_1} - \frac{L_1}{t_2}\) \\ \Delta L &< t_2L_1\(\frac{1}{t_1} - \frac{1}{t_2}\) \\ \Delta L &< L_1t_2\frac{t_2-t_1}{t_1t_2} \\ \Delta L &< L_1\frac{t_2-t_1}{t_1} \\ \Delta L &< L_1\(\frac{t_2}{t_1}-1\) \\ \\ \Delta L &< 1\ut{km}\cdot \(\frac{148.15\ut{s}}{147.95\ut{s}}-1\)\cdot \frac{1000\ut{m}}{1\ut{km}} \\ \Delta L &< \frac{4000}{2959}\ut{m} \approx 1.352\ut{m} \end{aligned} $$