4-9 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec r_A&=15\hat i-15\hat j\ut{m}\\ \vec r_B&=30\hat i-45\hat j\ut{m}\\ \vec r_C&=20\hat i-15\hat j\ut{m}\\ \vec r_D&=45\hat i+45\hat j\ut{m}\\ \end{cases} $$ $$\begin{cases} t_A&=0\\ t_B&=5.0\ut{min}\\ t_C&=10.0\ut{min}\\ t_D&=15.0\ut{min}\\ \end{cases} $$ $$ \begin{aligned} \bar v_{AB}=&\frac{\Delta \vec r_{AB}}{\Delta t_{AB}}=\frac{\vec r_B-\vec r_A}{t_B-t_A}\\ =&\frac{(30\hat i-45\hat j\ut{m})-(15\hat i-15\hat j\ut{m})}{5.0\ut{min}-0}\\ =&\frac{15\hat i-30\hat j\ut{m}}{5.0\ut{min}}\\ =&3\hat i-6\hat j\ut{m/min}\\ \end{aligned} $$ $$ \begin{aligned} \bar v_{AC}=&\frac{\Delta \vec r_{AC}}{\Delta t_{AC}}=\frac{\vec r_C-\vec r_A}{t_C-t_A}\\ =&\frac{(20\hat i-15\hat j\ut{m})-(15\hat i-15\hat j\ut{m})}{10.0\ut{min}-0}\\ =&\frac{5\hat i\ut{m}}{10.0\ut{min}}\\ =&\frac{1}{2}\hat i\ut{m/min}\\ \end{aligned} $$ $$ \begin{aligned} \bar v_{AD}=&\frac{\Delta \vec r_{AD}}{\Delta t_{AD}}=\frac{\vec r_D-\vec r_A}{t_D-t_A}\\ =&\frac{(45\hat i+45\hat j\ut{m})-(15\hat i-15\hat j\ut{m})}{15.0\ut{min}-0}\\ =&\frac{30\hat i+60\hat j\ut{m}}{15.0\ut{min}}\\ =&2\hat i+4\hat j\ut{m/min}\\ \end{aligned} $$ $$ \begin{cases} \abs{\bar v_{AB}}=&3\sqrt{5}\approx 6.7\ut{m/min} \\ \abs{\bar v_{AC}}=&0.5\ut{m/min}\\ \abs{\bar v_{AD}}=&2\sqrt{5}\approx 4.47\ut{m/min}\\ \end{cases} $$
(a) $$\abs{\bar v_{AC}}=0.5\ut{m/min}$$
(b) $$\theta_{\bar v_{AC}}=0$$
(c) $$\abs{\bar v_{AB}}=3\sqrt{5}\approx 6.7\ut{m/min}$$
(d) $$\theta_{\bar v_{AB}}=-\tan^{-1}(2)\approx -1.1\ut{rad}$$