5-14 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=5.00\ut{kg}\\ \vec v &= 8.00t\i+3.00t^2\j\\ \Sigma F&=45.0\ut{N} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \dxt{\vec v}&=\dt\(8.00t\i+3.00t^2\j\)\\ \vec a&=8\i+6t\j\\ \end{aligned} $$ $$ \begin{aligned} a&=\frac{F}{m}\\ &=\frac{45}{5}\\ 9&=\sqrt{8^2+(6t)^2}\\ t&=\frac{\sqrt{17}}{6} \end{aligned} $$ $$ \begin{aligned} \vec F &= m\vec a\\ &=5\bra{8\i+6\(\frac{\sqrt{17}}{6}\)\j}\\ &=40\i+5\sqrt{17}\j\\ \end{aligned} $$ $$ \begin{aligned} \theta_{\vec F}&=\tan^{-1}\frac{\sqrt{17}}{8}\\ &\approx 0.4758822496604165\ut{rad}\\ &\approx 0.476\ut{rad}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec v &=8.00\(\frac{\sqrt{17}}{6}\)\i+3.00\(\frac{\sqrt{17}}{6}\)^2\j\\ &=\frac{4}{3}\sqrt{17}\i+\frac{17}{12}\j\\ \end{aligned} $$ $$ \begin{aligned} \theta_{\vec v}&=\tan^{-1}\frac{\frac{17}{12}}{\frac{4}{3}\sqrt{17}}\\ &=\tan^{-1}\frac{\sqrt{17}}{16}\\ &\approx 0.2522069584584807\ut{rad}\\ &\approx 0.252\ut{rad}\\ \end{aligned} $$