10판/4. 2차원운동과 3차원운동

4-4 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 13. 22:49

{t0=15[min]t1=30[min]t2=t1+30[min]t3=t2+60[min]\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)Δr01=?\Delta \vec r_{0\to1}=? Δr01=r1r0=(12,π)(12,π2)=(12j^)(12i^)=12i^12j^ \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)Δr01=?\abs{\Delta r_{0\to1}}=? Δr01=(12)2+(12)2=12217[cm] \begin{aligned} \abs{\Delta r_{0\to1}}=&\sqrt{(-12)^2+(-12)^2}\\ =&12\sqrt2\\ \approx&17\ut{cm} \end{aligned}
(b)ϕ01=?\phi_{0\to1}=? ϕ01=cos1(r0r1r0r1)=π2[rad] \begin{aligned} \phi_{0\to1}=&\cos^{-1}\(\frac{\vec r_0 \cdot \vec r_1}{\abs{r_0}\abs{r_1}}\)\\ =&\frac{\pi}{2}\ut{rad} \end{aligned}
(c)(d)Δr12=?\Delta \vec r_{1\to2}=? Δr12=r2r1=(12,32π)(12,π2)=(12i^)(12i^)=24i^ \begin{aligned} \Delta\vec r_{1\to2} =& \vec r_2-\vec r_1\\ =&(12,\frac{3}{2}\pi)-(12,\frac{\pi}{2})\\ =&(-12\i)-(12\i)\\ =&-24\i\\ \end{aligned}
(c)Δr12=?\abs{\Delta r_{1\to2}}=? Δr12=24\abs{\Delta r_{1\to2}}=24
(d)ϕ12=?\phi_{1\to2}=? ϕ12=π\phi_{1\to2}=\pi
(e)(f)Δr23=?\Delta \vec r_{2\to3}=? Δr23=r3r2=(12,32π+2π)(12,32π)=(12,32π)(12,32π)=0 \begin{aligned} \Delta\vec r_{2\to3} =& \vec r_3-\vec r_2\\ =&(12,\frac{3}{2}\pi+2\pi)-(12,\frac{3}{2}\pi)\\ =&(12,\frac{3}{2}\pi)-(12,\frac{3}{2}\pi)\\ =&0 \end{aligned}
(e)Δr23=?\abs{\Delta r_{2\to3}}=? Δr23=0 \abs{\Delta r_{2\to3}} = 0
(f)ϕ23=?\phi_{2\to3}=? ϕ23=0\phi_{2\to3}=0