4-8 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_0=\text{City A}\\ t_1=\text{City B}\\ t_2=\text{City C}\\ \end{cases} $$ $$\begin{cases} \vec r_{0\to1} = 483\i\ut{km}\\ t_{0\to1}=48.0\ut{min}\\ \vec r_{1\to2} = -966\j\ut{km}\\ t_{1\to2}=1.50\ut{h}\\ \end{cases} $$
(a)$\abs{\Sigma \vec r}=?$ $$ \begin{aligned} \abs{\Sigma \vec r}=&\abs{\vec r_{0\to1}+\vec r_{1\to2}}\\ =&\abs{483\i-966\j}\ut{km}\\ =&\sqrt{483^2+(-966)^2}\ut{km}\\ =&483\sqrt5\ut{km}\\ \approx&1080\ut{km} \end{aligned} $$
(b)$\theta_{\Sigma \vec r}=?$ $$ \begin{aligned} \theta_{\Sigma \vec r}=&\tan^{-1}\(\frac{-966}{483}\)\\ =&-\tan^{-1}2\\ \approx&-1.11\ut{rad} \end{aligned} $$
(c)$\abs{\bar v}=?$ $$ \begin{aligned} \abs{\bar v}=&\abs{\frac{\Sigma \vec r}{\Sigma t}}\\ =&\abs{\frac{\vec r_1+\vec r_2}{t_{0\to2}}}\\ =&\abs{\frac{483\i-966\j\ut{km}}{138\ut{min}}}\\ =&\abs{\frac{7}{2}\i-7\j}\ut{km/min}\\ =&\sqrt{\(\frac{7}{2}\)^2+7^2}\ut{km/min}\\ =&\frac{7}{2}\sqrt5\ut{km/min}\\ \approx&7.83\ut{km/min}\\ \end{aligned} $$
(d)$\theta_{\bar v}=?$ $$ \begin{aligned} \theta_{\bar v}=&\tan^{-1}\(\frac{-7}{\frac{7}{2}}\)\\ =&-\tan^{-1}2\\ \approx&-1.11\ut{rad} \end{aligned} $$
(e)$\frac{\Sigma \abs{\vec r}}{\Sigma t}=?$ $$ \begin{aligned} \frac{\Sigma \abs{\vec r}}{\Sigma t}=& \frac{\abs{r_{0\to1}}+\abs{r_{1\to2}}}{t_{0\to2}}\\ =& \frac{\abs{483\i\ut{km}}+\abs{-966\j\ut{km}}}{138\ut{min}}\\ =&\frac{21}{2}\ut{km/min}\\ =&10.5\ut{km/min} \end{aligned} $$