4-12 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec r_A&=0 &\\ t_{A\rarr B}&=45.0\ut{min},&\vec r_B=500\i\ut{km}\\ t_{B\rarr C}&=1.50\ut{h},&\vec r_C=-1100\j\ut{km}\\ \end{cases}$$ $$\Sigma \vec r = \(500\i-1100\j\)\ut{km}$$ $$\ab{a}$$ $$\begin{aligned} r&= \sqrt{500^2+1100^2}\\ &=100\sqrt{146}\ut{km}\\ &\approx 1208.304597359457\ut{km}\\ &\approx 1.21\times10^3\ut{km}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta_r &= \tan^{-1}\(\frac{-1100}{500}\)\\ &\approx -1.144168833668021\ut{rad}\\ &\approx -1.14\ut{rad}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \vec v&=\frac{(500\i-1100\j)\ut{km}}{45\ut{min}+1.5\ut{h}}\\ &=\(\frac{2000}{9}\i-\frac{4400}{9}\j\)\ut{km/h}\\ v&=\sqrt{\(\frac{2000}{9}\)^2+\(\frac{4400}{9}\)^2}\\ &=\frac{400 \sqrt{146}}{9}\ut{km/h}\\ &\approx 537.0242654930921\ut{km/h}\\ &\approx 537\ut{km/h}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \theta_{\vec v}&=\theta_r\\ &= \tan^{-1}\(\frac{-1100}{500}\)\\ &\approx -1.144168833668021\ut{rad}\\ &\approx -1.14\ut{rad}\\ \end{aligned}$$ $$\ab{e}$$ $$\begin{aligned} v_L&=\frac{\Sigma L}{\Sigma t}\\ &=\frac{(500+1100)\ut{km}}{45\ut{min}+1.5\ut{h}}\\ &=\frac{6400}{9}\ut{km/h}\\ &\approx 711.1111111111111\ut{km/h}\\ &\approx 711\ut{km/h}\\ \end{aligned}$$