5-10 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1(0) &= 1.50\ut{kg}\\ m_2 &= 3.00\ut{kg}\\ \frac{\Delta m_1}{\Delta t}&=-0.200\ut{kg/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} m_1(t)&=m_1(0)+\Delta m_1\\ &=1.50-0.200t \end{aligned} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma \vec F_1 &= m_1\vec a\\ \Sigma \vec F_2 &= m_2\vec a\\ \end{cases} $$ $$ \begin{cases} T-m_1g &= m_1 a\\ -T+m_2g &= m_2 a\\ \end{cases} $$ $$ \begin{aligned} a&=\frac{15+2t}{45-2t}g\\ \dxt{a}&=\dt\(\frac{15+2t}{45-2t}g\)\\ &=\frac{120 g}{(45-2 t)^2} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \dxt{a}_{t=0}&=\frac{120 g}{(45-2 (0))^2}\\ &=\frac{8g}{135}\\ &\approx 0.5811348148148148\ut{m/s^2}\\ &\approx 0.581\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \dxt{a}_{t=3}&=\frac{120 g}{(45-2 (3))^2}\\ &=\frac{40g}{507}\\ &\approx 0.7737001972386587\ut{m/s^2}\\ &\approx 0.774\ut{m/s^2}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} m_1(t)&=1.50-0.200t,\\ 0&\lt m_1\\ \end{aligned} $$ $$\therefore 0 \lt t \lt 7.5$$ $$ \begin{aligned} \dxt{a}&=\frac{120 g}{(45-2 t)^2}>0, \end{aligned} $$ $$ \begin{aligned} a_{\max}&=a(7.5)\\ \end{aligned} $$ $$\therefore 7.5\ut{s}$$