4-12 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec r_0 &= (30.0,0)\ut{m}\\ \vec v_0 &= (0,-10.0)\ut{m/s}\\ \vec r_{30.0} &= (0,40.0)\ut{m}\\ \vec v_{30.0} &= (10.0,0)\ut{m/s}\\ \end{cases}$$
(a),(b)$\bar r_{0\rarr30}=?$ $$\begin{aligned} \bar r_{0\rarr 30} &= r_{30.0} - r_0\\ &=(0,40.0)-(30.0,0)\\ &=(-30.0,40.0)\ut{m}\\ \end{aligned}$$
(a) $$\begin{aligned} \abs{\bar r_{0\rarr 30}} &= \sqrt{(-30.0)^2+40.0^2}\\ &=50.0\ut{m}\\ \end{aligned}$$
(b) $$\begin{aligned} \theta_{\bar r_{0\rarr 30}} &= \pi-\tan^{-1}\frac{4}{3}\\ &\approx 2.21429743559\ut{rad}\\ &\approx 2.21\ut{rad}\\ \end{aligned}$$
(c),(d)$\bar v_{0\rarr30}=?$ $$\begin{aligned} \bar v_{0\rarr 30.0} &= \frac{r_{0\rarr 30.0}}{t_{0\rarr30.0}}\\ &=\frac{(-30.0,40.0)}{30.0}\\ &=\(-1,\frac{4}{3}\)\ut{m/s}\\ \end{aligned}$$
(c) $$\begin{aligned} \abs{\bar v_{0\rarr 30.0}} &= \sqrt{(-1)^2+\(\frac{4}{3}\)^2}\\ &=\frac{5}{3}\ut{m/s}\\ &\approx1.66666666667\ut{m/s}\\ &\approx1.67\ut{m/s}\\ \end{aligned}$$
(d) $$\begin{aligned} \theta_{\bar v_{0\rarr 30.0}} &= \pi-\tan^{-1}\frac{4}{3}\\ &\approx 2.21429743559\ut{rad}\\ &\approx 2.21\ut{rad}\\ \end{aligned}$$
(e),(f)$\bar a_{0\rarr30}=?$ $$\begin{aligned} \bar a_{0\rarr 30.0} &= \frac{v_{0\rarr 30.0}}{t_{0\rarr30.0}}\\ &=\frac{\(-1,\frac{4}{3}\)}{30.0}\\ &=\(-\frac{1}{30},\frac{2}{45}\)\ut{m/s^2}\\ \end{aligned}$$
(e) $$\begin{aligned} \abs{\bar a_{0\rarr 30.0}} &= \sqrt{\(-\frac{1}{30}\)^2+\(\frac{2}{45}\)^2}\\ &=\frac{1}{18}\ut{m/s^2}\\ &\approx0.0555555555556\ut{m/s^2}\\ &\approx5.56\times10^{-2}\ut{m/s^2}\\ \end{aligned}$$
(f) $$\begin{aligned} \theta_{\bar v_{0\rarr 30.0}} &= \pi-\tan^{-1}\frac{4}{3}\\ &\approx 2.21429743559\ut{rad}\\ &\approx 2.21\ut{rad}\\ \end{aligned}$$