2-62 할리데이 11판 솔루션 일반물리학

$$\begin{cases} a&=(5.0-1.2t)\ut{m/s^2}\\ v(0)&=2.7\ut{m/s}\\ x(0)&=7.3\ut{m}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \Delta v&=\int_0^t a\dd t\\ &=\int_0^t (5-1.2t)\dd t\\ &=5t-\frac{3}{5}t^2 \end{aligned}$$ $$\begin{aligned} \therefore v&=v(0)+\Delta v\\ &=2.7+5t-\frac{3}{5}t^2\\ \end{aligned}$$ $$\begin{aligned} a&=5-1.2t_1=0\\ t_1&=\frac{25}{6}\ut{s}\\ \end{aligned}$$ $$\begin{aligned} v_{\max}&=v(t_1)\\ &=\frac{787}{60}\ut{m/s}\\ &\approx 13.11666666666667\ut{m/s}\\ &\approx 13\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} v&=2.7+5t-\frac{3}{5}t^2=0\\ \end{aligned}$$ $$\begin{aligned} t&=\frac{1}{6} \left(25\pm\sqrt{787}\right)\ut{s}\\ \end{aligned}$$ $$\begin{cases} t_1&\approx -0.5089200463685124\ut{s}\\ t_2&\approx 8.842253379701845\ut{s} \end{cases}$$ $$0\lt t \lt6,\rarr v\gt 0$$ $$\begin{aligned} \Delta x_{0\rarr6} &= \int_0^6 v \dd t\\ &= \int_0^6 \(2.7+5t-\frac{3}{5}t^2\) \dd t\\ &=\[-\frac{t^3}{5}+\frac{5 t^2}{2}+\frac{27 t}{10}\]_0^6\\ &=63\ut{m} \end{aligned}$$