솔루션피아

$$\vec r = (3.00t^3-6.00t)\hat i +(7.00-8.00t^4)\hat j\ut{m}$$ (a) $$\begin{aligned} \vec r_3 &= (3.00\cdot 3^3-6.00\cdot 3)\hat i +(7.00-8.00\cdot 3^4)\hat j\ut{m}\\ &=63\hat i -641\hat j\ut{m} \end{aligned}$$ (b) $$\begin{aligned} \vec v =&\dxt{\vec r}\\ =& (9 t^2-6)\hat i -32 t^3\hat j\ut{m/s} \end{aligned}$$ $$\begin{aligned} \vec v_3 =&(9 \cdot 3^2-6)\hat i -32 \cdot 3^3\hat j\ut{m/s}\\ =& ..

$$\vec r = 5.00t\ut{m}\hat i +(et+ft^2)\ut{m}\hat j$$ $$\begin{cases} \theta_0 &= 35\degree\\ \theta_{12} &= 0\\ \end{cases} $$ $$\begin{aligned} \vec v =&\dxt{\vec r}\\ =& 5.00\ut{m/s}\hat i +(e+2ft)\ut{m/s}\hat j \end{aligned}$$ (a) $$\begin{aligned} \vec v_0 =&5.00\ut{m/s}\hat i +(e+2f\cdot 0)\ut{m/s}\hat j\\ =&5.00\ut{m/s}\hat i +e\ut{m/s}\hat j\\ \end{aligned}$$ $$\begin{aligned} \theta_0 =..

$$\begin{cases} \vec r_A&=15\hat i-15\hat j\ut{m}\\ \vec r_B&=30\hat i-45\hat j\ut{m}\\ \vec r_C&=20\hat i-15\hat j\ut{m}\\ \vec r_D&=45\hat i+45\hat j\ut{m}\\ \end{cases} $$ $$\begin{cases} t_A&=0\\ t_B&=5.0\ut{min}\\ t_C&=10.0\ut{min}\\ t_D&=15.0\ut{min}\\ \end{cases} $$ $$ \begin{aligned} \bar v_{AB}=&\frac{\Delta \vec r_{AB}}{\Delta t_{AB}}=\frac{\vec r_B-\vec r_A}{t_B-t_A}\\ =&\frac{(30\hat..

$$\begin{cases} t_0=\text{City A}\\ t_1=\text{City B}\\ t_2=\text{City C}\\ \end{cases} $$ $$\begin{cases} \vec r_{0\to1} = 483\i\ut{km}\\ t_{0\to1}=48.0\ut{min}\\ \vec r_{1\to2} = -966\j\ut{km}\\ t_{1\to2}=1.50\ut{h}\\ \end{cases} $$ (a)$\abs{\Sigma \vec r}=?$ $$ \begin{aligned} \abs{\Sigma \vec r}=&\abs{\vec r_{0\to1}+\vec r_{1\to2}}\\ =&\abs{483\i-966\j}\ut{km}\\ =&\sqrt{483^2+(-966)^2}\ut{km..

$$\begin{cases} \vec r_1 = 6.0\i - 7.0\j + 3.0\k\ut{m}\\ \vec r_2 = 3.0\i + 9.0\j - 3.0\k\ut{m}\\ t_{1\to2}=10\ut{s} \end{cases} $$ $$ \begin{aligned} \bar v=&\frac{\Delta\vec r}{\Delta t}\\ =&\frac{\vec r_2-\vec r_1}{\Delta t}\\ =&\frac{(6.0\i - 7.0\j + 3.0\k)-(3.0\i + 9.0\j - 3.0\k)}{10}\\ =&0.3\i-1.0\j+0.6\k\ut{m/s} \end{aligned} $$

(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{..

$$\begin{cases} t_{0\to1}=40.0\ut{min}\\ t_{1\to2}=20.0\ut{min}\\ t_{2\to3}=50.0\ut{min}\\ \end{cases} $$ $$\begin{cases} \vec v_{0\to1}=60.0\ut{km/h}\i\\ \vec v_{1\to2}:(60.0\ut{km/h},90^\circ-50^\circ)\\ \vec v_{2\to3}=-60.0\ut{km/h}\i=-\vec v_{0\to1}\\ \end{cases} $$ $$ \begin{aligned} \vec r_{0\to1}=&v_{0\to1}t_{0\to1}\\ =&60.0\ut{km/h}\i\cdot40.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&4..

$$\begin{cases} t_0=15\ut{min}\\ t_1=30\ut{min}\\ t_2=t_1+30\ut{min}\\ t_3=t_2+60\ut{min}\\ \end{cases} $$ (a)(b)$\Delta \vec r_{0\to1}=?$ $$ \begin{aligned} \Delta\vec r_{0\to1} =& \vec r_1-\vec r_0\\ =&(12,\pi)-(12,\frac{\pi}{2})\\ =&(-12\j)-(12\i)\\ =&-12\i-12\j\\ \end{aligned} $$ (a)$\abs{\Delta r_{0\to1}}=?$ $$ \begin{aligned} \abs{\Delta r_{0\to1}}=&\sqrt{(-12)^2+(-12)^2}\\ =&12\sqrt2\\ \a..

$$ \vec r_0 =?$$ $$\vec r_0 + \Delta\vec r = \vec r$$ $$ \begin{aligned} \vec r_0 =& \vec r - \Delta\vec r\\ =&(4.0\j — 5.0\k) - (2.0\i - 4.0\j + 8.0\k) \\ =&-2.0\i+8.0\j-13\k\\ \end{aligned} $$

$$\begin{cases} p_x = -5.0\ut{m}\\ p_y = 8.0\ut{m}\\ p_z = 0.0\ut{m}\\ \end{cases} $$ (a)$\vec p=?$ $$ \vec p = (-5.0\ut{m})\i+8.0\ut{m}\j$$ (b)$p=?$ $$ \begin{aligned} p =& \sqrt{(-5)^2+8^2}\\ =& \sqrt{89}\\ \approx&9.4\ut{m} \end{aligned} $$ (c)$\phi_{p|\i}=?$ $$ \begin{aligned} \phi_{\vec p|\i}=&\cos^{-1}\(\frac{\vec p \cdot \i}{\abs{p}\abs{\i}}\)\\ =&\cos^{-1}\(\frac{-5}{\sqrt{89}}\)\\ \appr..