4-5 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_{0\to1}=40.0\ut{min}\\ t_{1\to2}=20.0\ut{min}\\ t_{2\to3}=50.0\ut{min}\\ \end{cases} $$ $$\begin{cases} \vec v_{0\to1}=60.0\ut{km/h}\i\\ \vec v_{1\to2}:(60.0\ut{km/h},90^\circ-50^\circ)\\ \vec v_{2\to3}=-60.0\ut{km/h}\i=-\vec v_{0\to1}\\ \end{cases} $$ $$ \begin{aligned} \vec r_{0\to1}=&v_{0\to1}t_{0\to1}\\ =&60.0\ut{km/h}\i\cdot40.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&40.0\ut{km}\i \end{aligned} $$ $$ \begin{aligned} \vec r_{1\to2}=&v_{1\to2}t_{1\to2}\\ =&(60.0\ut{km/h},40^\circ)\cdot20.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&(20.0\ut{km/h},40^\circ)\\ =&20.0\cos40^\circ\ut{km}\i+20.0\sin40^\circ\ut{km}\j \end{aligned} $$ $$ \begin{aligned} \vec r_{2\to3}=&v_{2\to3}t_{2\to3}\\ =&-60.0\ut{km/h}\i\cdot50.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&-50.0\ut{km}\i \end{aligned} $$ $$ \begin{aligned} \bar v =& \frac{\Sigma\Delta\vec r}{\Sigma\Delta t}\\ =&\frac{\vec r_{0\to1}+\vec r_{1\to2}+\vec r_{2\to3}}{t_{0\to3}}\\ =&\frac{\vec r_{0\to1}}{t_{0\to3}}+\frac{\vec r_{1\to2}}{t_{0\to3}}+\frac{\vec r_{2\to3}}{t_{0\to3}}\\ =&\frac{40.0\ut{km}\i}{110\ut{min}}\\ &+\frac{20.0\cos40^\circ\ut{km}\i+20.0\sin40^\circ\ut{km}\j}{110\ut{min}}\\ &+\frac{-50.0\ut{km}\i}{110\ut{min}}\\ =&\frac{2\cos40^\circ-1}{11}\ut{km/min}\i+\frac{2\sin40^\circ}{11}\ut{km/min}\\ \end{aligned} $$
(a)$\bar v=?$ $$ \begin{aligned} \bar v=&\sqrt{\(\frac{2\cos40^\circ-1}{11}\)^2+\(\frac{2\sin40^\circ}{11}\)^2}\\ =&\frac{1}{11} \sqrt{5-4 \cos (40{}^{\circ})}\ut{km/min}\\ =&\frac{50}{33} \sqrt{5-4 \cos (40{}^{\circ})}\ut{m/s}\\ \approx&2.11\ut{m/s} \end{aligned} $$
(b)$\theta \bar v=?$ $$ \begin{aligned} \theta \bar v=&\tan^{-1}\(\frac{\frac{2\sin40^\circ}{11}}{\frac{2\cos40^\circ-1}{11}}\)\\ =&\tan ^{-1}\left(\frac{2 \sin (40{}^{\circ})}{2 \cos (40{}^{\circ})-1}\right)\\ \approx&1.18\ut{rad} \end{aligned} $$