4-15 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec v_0 &= 7.0\i\ut{m/s}\\ \vec a &= -9.0\i+3.0\j\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v &= \vec v_0+\vec at\\ &= 7.0\i+(-9.0\i+3.0\j)t\ut{m/s}\\ &= (7.0-9.0t)\i+3.0t\j\ut{m/s}\\ \end{aligned}$$ $$\vec v_x = 7.0-9.0t=0,$$ $$t_{r_{xM}} = \frac{7}{9}\ut{s}$$
(a) $v_{r_{xM}}=?$ $$\begin{aligned} v_{r_{xM}} &= v_{\frac{7}{9}}\\ &=\(7.0-9.0\cdot\frac{7}{9}\)\i+3.0\cdot\frac{7}{9}\j\ut{m/s}\\ &=\frac{7}{3}\j\ut{m/s}\\ &\approx2.333333333333334\j\ut{m/s}\\ &\approx2.33\j\ut{m/s}\\ \end{aligned}$$
(b) $r_{xM}=?$ $$\begin{aligned} \vec r&=\int_0^{\frac{7}{9}} \vec v\dd t\\ &=\int_0^{\frac{7}{9}} \{(7.0-9.0t)\i+3.0t\j\}\dd t\\ &=\[\(7t-\frac{9}{2}t^2\)\i+\frac{3}{2}t^2\j\]_0^{\frac{7}{9}} \\ &=\frac{49}{18}\i+\frac{49}{54}\j\ut{m}\\ &\approx2.72222222222\i+0.907407407407\j\ut{m}\\ &\approx2.7\i+0.91\j\ut{m}\\ \end{aligned}$$