11판/4. 2차원 운동과 3차원 운동

4-68 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 14. 18:00
{Δt=3.15[h]Δr=11.2i^+24.5j^+2.88k^[km] \begin{cases} \Delta t &= 3.15\ut{h}\\ \Delta \vec r &= 11.2\i+24.5\j+2.88\k\ut{km} \end{cases} (a)\ab{a} vˉ=ΔrΔt=11.22+24.52+2.8823.15[km/h]=21834961315[km/h]8.600681416985678[km/h]8.60[km/h] \begin{aligned} \bar v &= \frac{\Delta r}{\Delta t}\\ &= \frac{\sqrt{11.2^2+24.5^2+2.88^2}}{3.15}\ut{km/h}\\ &=\frac{2\sqrt{1834961}}{315}\ut{km/h}\\ &\approx 8.600681416985678\ut{km/h}\\ &\approx 8.60\ut{km/h}\\ \end{aligned} (b)\ab{b} n=0i^+0j+1k^,π2θ=cos1rnrn=cos12.8811.22+24.52+2.882=cos11441834961θ=π2cos114418349610.1065051091644924[rad]0.107[rad] \begin{aligned} \vec n &= 0\i+0j+1\k,\\ \frac{\pi}{2}-\theta&=\cos^{-1}\frac{\vec r\cdot\vec n}{r n}\\ &=\cos^{-1}\frac{2.88}{\sqrt{11.2^2+24.5^2+2.88^2}}\\ &=\cos^{-1}\frac{144}{\sqrt{1834961}}\\ \theta &= \frac{\pi}{2}-\cos^{-1}\frac{144}{\sqrt{1834961}}\\ &\approx 0.1065051091644924\ut{rad}\\ &\approx 0.107\ut{rad}\\ \end{aligned}