3-5 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a_1 = 3.66\hat j\ut{m}\\ \vec a_2:(1.83\ut{m},-\frac{\pi}{4})\\ \vec a_3:(0.91\ut{m},-\frac{3}{4}\pi)\\ \end{cases}$$ $$\begin{aligned} \vec a_2 &= 1.83\cos\(-\frac{\pi}{4}\)\hat i+1.83\sin\(-\frac{\pi}{4}\)\hat j\\ &= \frac{1.83}{\sqrt{2}}\hat i-\frac{1.83}{\sqrt{2}}\hat j\\ \end{aligned}$$ $$\begin{aligned} \vec a_3 &= 0.91\cos\(-\frac{3}{4}\pi\)\hat i+0.91\sin\(-\frac{3}{4}\pi\)\hat j\\ &= -\frac{0.91}{\sqrt{2}}\hat i-\frac{0.91}{\sqrt{2}}\hat j\\ \end{aligned}$$ $$\begin{aligned} \Sigma \vec a &= \vec a_1+\vec a_2+\vec a_3\\ &=\(\frac{1.83}{\sqrt{2}}-\frac{0.91}{\sqrt{2}}\)\hat i+\(3.66-\frac{1.83}{\sqrt{2}}-\frac{0.91}{\sqrt{2}}\)\hat j\\ &=\frac{23}{25\sqrt{2}}\hat i +\frac{366-137\sqrt{2}}{100} \hat j\ut{m}\\ \end{aligned}$$
(a)$\abs{\Sigma\vec a}=?$ $$\begin{aligned} \abs{\Sigma\vec a}&=\sqrt{\(\frac{23}{25\sqrt{2}}\)^2+\(\frac{366-137\sqrt{2}}{100}\)^2}\\ &=\frac{87863-50142 \sqrt{2}}{5000}\ut{m}\\ &\approx 3.39\ut{m} \end{aligned}$$
(b)$\theta_a=?$ $$\begin{aligned} \theta_a&=\tan^{-1}\(\frac{a_x}{a_y}\)\\ &=\tan^{-1}\(\frac{\frac{366-137\sqrt{2}}{100}}{\frac{23}{25\sqrt{2}}}\)\\ &=\tan^{-1}\(\frac{183\sqrt{2}-137}{46} \)\\ &\approx 1.21\ut{rad} \end{aligned}$$