4-8 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec A &= 0\\ \vec B &= -90\ut{km}\i\\ \end{cases} $$ $$ \begin{cases} r_1&= 75\ut{km}, \theta_1=37\degree,t_1=55\ut{h}\\ \vec r_2&=-65\ut{km}\j, t_2=45\ut{h}\\ t_3&=5.0\ut{h}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Sigma \vec r&=\vec r_1 + \vec r_2\\ &=75\cos37\degree\i+(75\sin37\degree-65)\j\\ \end{aligned} $$ $$ \begin{aligned} r&=\sqrt{\(75\cos37\degree\)^2+(75\sin37\degree-65)^2}\\ &=\sqrt{9850-9750 \sin37\degree}\ut{km}\\ &\approx 63.10549519865548\ut{km}\\ &\approx 63\ut{km}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_r&=\tan^{-1}\(\frac{75\sin37\degree-65}{75\cos37\degree}\)\\ &\approx -0.3202169402769146\ut{rad}\\ &\approx -0.32\ut{rad}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \bar v&=\frac{\vec r}{\Sigma t}\\ &=\frac{\sqrt{9850-9750 \sin37\degree}\ut{km}}{55\ut{h}+45\ut{h}+5.0\ut{h}}\\ &\approx 0.6010047161776713\ut{km/h}\\ &\approx 0.60\ut{km/h} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \theta_{\bar v}&=\theta_r\\ &\approx -0.32\ut{rad}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \bar v&=\frac{\Sigma S}{\Sigma t}\\ &=\frac{75\ut{km}+65\ut{km}}{55\ut{h}+45\ut{h}+5.0\ut{h}}\\ &=\frac{4}{3}\ut{km/h}\\ &\approx 1.333333333333333\ut{km/h}\\ &\approx 1.3\ut{km/h}\\ \end{aligned} $$ $$\ab{f,g}$$ $$ \begin{aligned} t&=135\ut{h}-(55\ut{h}+45\ut{h}+5.0\ut{h})\\ &=30\ut{h} \end{aligned} $$ $$ \begin{aligned} \vec r_{B} &= -\Sigma \vec r + \overrightarrow{AB} \\ &=-\bra{75\cos37\degree\i+(75\sin37\degree-65)\j}-90\ut{km}\i\\ &=\(-75\cos37\degree-90\)\i+(-75\sin37\degree+65)\j\\ \end{aligned} $$ $$\ab{f}$$ $$ \begin{aligned} \bar v&=\sqrt{(65-75 \sin37\degree)^2+(-75 \cos37\degree-90)^2}\\ &=5 \sqrt{-390 \sin37\degree+540 \cos37\degree+718}\ut{km/h}\\ &\approx 151.2080781899763\ut{km/h}\\ &\approx 1.5\times10^2\ut{km/h}\\ \end{aligned} $$ $$\ab{g}$$ $$ \begin{aligned} \theta_v&=\frac{-75\sin37\degree+65}{-75\cos37\degree-90}\\ &\approx -0.1325162302897097\ut{rad}\\ &\approx -0.13\ut{rad}\\ \end{aligned} $$