4-23 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec v_0&=5.00\i\ut{m/s}\\ \vec a&=\(-0.600\i-0.405\j\)\ut{m/s^2}\\ t_1&:x_{\max}\text{ Time} \end{cases}$$ $$\begin{aligned} \vec v&=\vec v_{0}+\Delta \vec v\\ &=\vec v_0+\int_0^t\vec a\dd t\\ &=5\i+\int_0^t\(-0.600\i-0.405\j\)\dd t\\ &=5\i-0.6t\i-0.405t\j\\ &=(5-0.6t)\i-0.405t\j\\ \end{aligned}$$ $$\begin{aligned} v_x&=0\\ &=5-0.6t_1\\ \end{aligned}$$ $$t_1=\frac{25}{3}\ut{s}$$ $$\ab{a}$$ $$\begin{aligned} \vec v(t)&=(5-0.6t)\i-0.405t\j\\ \vec v_1&=\vec v\(\frac{25}{3}\)\\ &=\bra{5-0.6\(\frac{25}{3}\)}\i-0.405\(\frac{25}{3}\)\j\\ &=-\frac{27}{8}\j\ut{m/s}\\ &=-3.375\j\ut{m/s}\\ &\approx -3.38\j\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \vec r &= \vec r_0+\Delta \vec r\\ &= (0)+\int_0^t\vec v\dd t\\ \vec r(t)&= \int_0^t\bra{(5-0.6t)\i-0.405t\j}\dd t\\ &=\(5 t-\frac{3}{10} t^2\)\i-\frac{81 }{400}t^2\j\\ \end{aligned}$$ $$\begin{aligned} \vec r_1&=\vec r\(\frac{25}{3}\)\\ &=\bra{5 \(\frac{25}{3}\)-\frac{3 }{10}\(\frac{25}{3}\)^2}\i-\frac{81 }{400}\(\frac{25}{3}\)^2\j\\ &=\(\frac{125}{6}\i-\frac{225}{16}\j\)\ut{m}\\ &\approx \(20.83333333333333\i-14.0625\j\)\ut{m}\\ &\approx \(20.8\i-14.1\j\)\ut{m}\\ \end{aligned}$$