4-47 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R_t&=(x,y)\ut{m}\\ R_0&=(0,2)\ut{m}\\ R_{0.5}&=(5,6)\ut{m}\\ R_{1}&=(10,8)\ut{m}\\ R_{1.5}&=(15,8)\ut{m}\\ R_{2}&=(20,6)\ut{m}\\ R_{2.5}&=(25,2)\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_x &=\frac{\Delta x}{\Delta t}\\ &=\frac{10}{1}\ut{m}\\ &=10\ut{m}\\ \end{aligned} $$ $$ \begin{aligned} v_y&=\frac{\Delta y}{\Delta t}\\ v_{0\rarr0.5}&=\frac{4}{0.5}=8\ut{m/s}\\ v_{0.5\rarr1}&=\frac{2}{0.5}=4\ut{m/s}\\ v_{1\rarr1.5}&=\frac{0}{0.5}=0\ut{m/s}\\ v_{1.5\rarr2}&=\frac{-2}{0.5}=-4\ut{m/s}\\ v_{2\rarr2.5}&=\frac{-4}{0.5}=-8\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} a_y&=\frac{\Delta v_y}{\Delta t}\\ a_{0.25\rarr0.75}&=\frac{-4}{0.5}=-8\ut{m/s^2}\\ a_{0.75\rarr1.25}&=\frac{-4}{0.5}=-8\ut{m/s^2}\\ a_{1.25\rarr1.75}&=\frac{-4}{0.5}=-8\ut{m/s^2}\\ a_{1.75\rarr2.25}&=\frac{-4}{0.5}=-8\ut{m/s^2}\\ \end{aligned} $$ $$ \begin{aligned} \therefore a_y&=-8\ut{m/s^2} \end{aligned} $$ $$ \begin{aligned} v_y&=v_{y0}+\Delta v_y\\ &=v_{y0}+\int_0^ta_y\dd t\\ &=v_{y0}+\int_0^t(-8)\dd t\\ &=v_{y0}-8t\\ \end{aligned} $$ $$ \begin{aligned} y&=y_0+\Delta y\\ &=2+\int_0^tv_y\dd t\\ &=2+\int_0^t(v_{y0}-8t)\dd t\\ &=2+v_{y0}t-4t^2\\ y(1)&=2+v_{y0}(1)-4(1)^2=8\\ \end{aligned} $$ $$ \begin{aligned} \therefore v_{y0}=10\ut{m/s} \end{aligned} $$ $$\therefore \vec v_0=10\i+10\j\ut{m/s}$$ $$\ab{b}$$ $$ \begin{aligned} a_y&=8\ut{m/s^2} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_{y0}t+\frac{1}{2}a_yt^2\\ -2&=(10)t+\frac{1}{2}(-8)t^2\\ t&=\frac{1}{4} \(5+\sqrt{33}\)\\ &\approx 2.686140661634507\ut{s}\\ &\approx 2.69\ut{s}\\ \end{aligned} $$ $$\ab{d}$$ $$g=9.80665\ut{m/s^2},$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_{y0}t+\frac{1}{2}(-g)t^2\\ -2&=(10)t-\frac{1}{2}gt^2\\ t&=\frac{2 \left(\sqrt{g+25}+5\right)}{g}\\ &\approx 2.222923443757904\ut{s}\\ &\approx 2.22\ut{s}\\ \end{aligned} $$