4-20 할리데이 10판 솔루션 일반물리학

$$\begin{cases} A_y&=30\ut{m}\\ \vec v_A&=3.0\i\ut{m/s}\\ \vec v_{B0}&=0\\ \abs{\vec a_{B}}&=0.40\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} \vec r_A &=\Delta \vec r_A + \vec r_{A0}\\ &=\int \vec v_A\dd t+\vec r_{A0}\\ &=\int 3\i\dd t+30\j\ut{m}\\ &=3t\i+30j\ut{m}\\ \end{aligned} $$ $$ \vec a_B = \frac{2}{5}\sin\theta\i+\frac{2}{5}\cos\theta\j\ut{m/s^2}$$ $$ \begin{aligned} \vec v_B&=\vec v_{B0}+\vec a_B t\\ &=\frac{2}{5}t\sin\theta\i+\frac{2}{5}t\cos\theta\j\ut{m/s} \end{aligned} $$ $$ \begin{aligned} \vec r_B&=\Delta \vec r_B+\vec r_{B0}\\ &=\int \vec v_B \dd t\\ &=\int \frac{2}{5}t\sin\theta\i+\frac{2}{5}t\cos\theta\j \dd t\\ &=\frac{1}{5}t^2\sin\theta\i+\frac{1}{5}t^2\cos\theta\j\ut{m}\\ \end{aligned} $$ $$\title{A, B Meet on y}$$ $$30=\frac{1}{5}t^2\cos\theta$$ $$t=5\sqrt{6\sec\theta}\ut{sec}$$ $$\title{A, B Meet on x}$$ $$ \begin{aligned} 3t&=\frac{1}{5}t^2\sin\theta\\ 3&=\frac{1}{5}t\sin\theta\\ &=\frac{1}{5}\(5\sqrt{6\sec\theta}\)\sin\theta\\ \end{aligned} $$ $$ \begin{aligned} \theta&=\frac{\pi}{3}\ut{rad}\\ &\approx1.047197551196598\ut{rad}\\ &\approx1.0\ut{rad}\\ \end{aligned} $$