4-19 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a&=3t\i+4t\j\ut{m/s^2}\\ \vec r_0&=20.0\i+40.0\j\ut{m}\\ \vec v_0&=5.00\i+2.00\j\ut{m/s}\\ \end{cases} $$ $$ \begin{aligned} \vec v&=\Delta \vec v+\vec v_0\\ &=\int_0^t \vec a\dd t+\vec v_0\\ &=\int_0^t 3t\i+4t\j\dd t+\vec v_0\\ &=\(\frac{3}{2}t^2+5\)\i+\(2 t^2+2\)\j\\ \end{aligned} $$ $$ \begin{aligned} \vec r&=\Delta \vec r+\vec r_0\\ &=\int_0^t \vec v\dd t+\vec r_0\\ &=\int_0^t \(\frac{3}{2}t^2+5\)\i+\(2 t^2+2\)\j\dd t+\vec r_0\\ &=\frac{1}{8} \left(t^4+20 t^2+160\right)\i+\(\frac{t^4}{6}+t^2+40\)\j\\ \end{aligned} $$
(a) $\vec r_{t=4.00}=?$ $$ \begin{aligned} \vec r_{4}&=\frac{1}{8} \left(4^4+20\cdot 4^2+160\right)\i+\(\frac{4^4}{6}+4^2+40\)\j\\ &=92\i+\frac{296}{3}\j\ut{m}\\ &\approx 92.0\i+98.66666666666667\j\ut{m}\\ &\approx 92.0\i+98.7\j\ut{m}\\ \end{aligned} $$
(b) $\theta_{\vec v_{t=4.00}}=?$ $$ \begin{aligned} \vec v_{4}&=\(\frac{3}{2}4^2+5\)\i+\(2\cdot4^2+2\)\j\\ &=29\i+34\j\\ \end{aligned} $$ $$ \begin{aligned} \theta_{\vec v_4}&=\tan^{-1}\frac{34}{29}\\ &\approx0.8645972343668997\ut{rad}\\ &\approx0.865\ut{rad}\\ \end{aligned} $$