11판/4. 2차원 운동과 3차원 운동

4-26 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 2. 15:48
{r4=3.00i^+4.00j^[m]v4=2.00j^[m/s]a4=a4i^ \begin{cases} \vec r_4 &=3.00\i+4.00\j\ut{m}\\ \vec v_4 &=2.00\j\ut{m/s}\\ \vec a_4 &= a_4\i \end{cases} {v7=2.00i^[m/s]a7=a7j^ \begin{cases} \vec v_7 &=-2.00\i\ut{m/s}\\ \vec a_7 &= a_7\j \end{cases} t47=34TT=433[s]T=4[s] \begin{aligned} t_{4\rarr7}&=\frac{3}{4}T\\ T&=\frac{4}{3}\cdot3\ut{s}\\ T&=4\ut{s} \end{aligned} 2πR=vT,2\pi R=vT, R=vT2π=242π=4π[m] \begin{aligned} R&=\frac{vT}{2\pi}\\ &= \frac{2\cdot 4}{2\pi}\\ &=\frac{4}{\pi}\ut{m}\\ \end{aligned} O=r4+4πi^=3.00i^+4.00j^+4πi^=(3.00+4π)i^+4.00j^ \begin{aligned} \vec O &=\vec r_4 + \frac{4}{\pi}\i\\ &=3.00\i+4.00\j+ \frac{4}{\pi}\i\\ &=\(3.00+\frac{4}{\pi}\)\i+4.00\j\\ \end{aligned} (a)\ab{a} Ox=3.00+4π4.273239544735163[m]4.27[m] \begin{aligned} O_x &= 3.00+\frac{4}{\pi}\\ &\approx 4.273239544735163\ut{m}\\ &\approx 4.27\ut{m}\\ \end{aligned} (b)\ab{b} Oy=4.00[m] \begin{aligned} O_y &= 4.00\ut{m}\\ \end{aligned}