4-6 할리데이 10판 솔루션 일반물리학

(풀이자주:(c),(d)의 경우 모든t에 대한 v를 지칭하는지 t=3일때를 지칭하는지 애매합니다. 하여, 두가지 모두를 풀었습니다.) $$\vec r = 3.00t\i - 4.00t^2\j + 2.00\k$$
(a)$\vec v(t)=?$ $$ \begin{aligned} \vec v(t)=& \dt\vec r\\ =&\dt\(3.00t\i - 4.00t^2\j + 2.00\k\)\\ =&\dt\(3.00t\)\i+\dt\(-4.00t^2\)\j+\dt\(2.00\)\k\\ =&3.00\i-8.00t\j \end{aligned} $$
(b)$\vec v(3.00)=?$ $$ \begin{aligned} \vec v(3.00)=&3.00\i-8.00(3.00)\j\\ =&3.00\i-24.0\j \end{aligned} $$
(c)$\abs{\vec v(t)}=?$ $$ \begin{aligned} \abs{\vec v(t)}=&\sqrt{3.00^2+(-8.00t)^2}\\ =&\sqrt{64.0t^2+9.00}\ut{m/s} \end{aligned} $$ $\abs{\vec v(3.00)}=?$ $$ \begin{aligned} \vec v(3.00)=&\sqrt{64.0(3.00)^2+9.00}\\ =&3\sqrt{65}\ut{m/s}\\ \approx&24.2\ut{m/s} \end{aligned} $$
(d)$\phi_{\vec v\i}=?$ $$ \begin{aligned} \phi_{\vec v\i}=&\cos^{-1}\(\frac{\vec v \cdot \i}{\abs{v}\abs{\i}}\)\\ =&\cos^{-1}\(\frac{(3\i-8t\j) \cdot \i}{\sqrt{64t^2+9}}\)\\ =&\cos^{-1}\(\frac{3}{\sqrt{64t^2+9}}\)\\ \end{aligned} $$ $\phi_{\vec v(3)\i}=?$ $$ \begin{aligned} \phi_{\vec v(3)\i}=&\cos^{-1}\(\frac{3}{\sqrt{64(3.00)^2+9}}\)\\ =&\sec ^{-1}\left(\sqrt{65}\right)\ut{rad}\\ \approx&1.45\ut{rad} \end{aligned} $$