10판/4. 2차원운동과 3차원운동

4-5 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 14. 16:31

{t01=40.0[min]t12=20.0[min]t23=50.0[min]\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} {v01=60.0[km/h]i^v12:(60.0[km/h],9050)v23=60.0[km/h]i^=v01\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} r01=v01t01=60.0[km/h]i^40.0[min]1[h]60[min]=40.0[km]i^ \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}}\\ =&40.0\ut{km}\i \end{aligned} r12=v12t12=(60.0[km/h],40)20.0[min]1[h]60[min]=(20.0[km/h],40)=20.0cos40[km]i^+20.0sin40[km]j^ \begin{aligned} \vec r_{1\to2}=&v_{1\to2}t_{1\to2}\\ =&(60.0\ut{km/h},40^\circ)\cdot20.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&(20.0\ut{km/h},40^\circ)\\ =&20.0\cos40^\circ\ut{km}\i+20.0\sin40^\circ\ut{km}\j \end{aligned} r23=v23t23=60.0[km/h]i^50.0[min]1[h]60[min]=50.0[km]i^ \begin{aligned} \vec r_{2\to3}=&v_{2\to3}t_{2\to3}\\ =&-60.0\ut{km/h}\i\cdot50.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&-50.0\ut{km}\i \end{aligned} vˉ=ΣΔrΣΔt=r01+r12+r23t03=r01t03+r12t03+r23t03=40.0[km]i^110[min]+20.0cos40[km]i^+20.0sin40[km]j^110[min]+50.0[km]i^110[min]=2cos40111[km/min]i^+2sin4011[km/min] \begin{aligned} \bar v =& \frac{\Sigma\Delta\vec r}{\Sigma\Delta t}\\ =&\frac{\vec r_{0\to1}+\vec r_{1\to2}+\vec r_{2\to3}}{t_{0\to3}}\\ =&\frac{\vec r_{0\to1}}{t_{0\to3}}+\frac{\vec r_{1\to2}}{t_{0\to3}}+\frac{\vec r_{2\to3}}{t_{0\to3}}\\ =&\frac{40.0\ut{km}\i}{110\ut{min}}\\ &+\frac{20.0\cos40^\circ\ut{km}\i+20.0\sin40^\circ\ut{km}\j}{110\ut{min}}\\ &+\frac{-50.0\ut{km}\i}{110\ut{min}}\\ =&\frac{2\cos40^\circ-1}{11}\ut{km/min}\i+\frac{2\sin40^\circ}{11}\ut{km/min}\\ \end{aligned}
(a)vˉ=?\bar v=? vˉ=(2cos40111)2+(2sin4011)2=11154cos(40)[km/min]=503354cos(40)[m/s]2.11[m/s] \begin{aligned} \bar v=&\sqrt{\(\frac{2\cos40^\circ-1}{11}\)^2+\(\frac{2\sin40^\circ}{11}\)^2}\\ =&\frac{1}{11} \sqrt{5-4 \cos (40{}^{\circ})}\ut{km/min}\\ =&\frac{50}{33} \sqrt{5-4 \cos (40{}^{\circ})}\ut{m/s}\\ \approx&2.11\ut{m/s} \end{aligned}
(b)θvˉ=?\theta \bar v=? θvˉ=tan1(2sin40112cos40111)=tan1(2sin(40)2cos(40)1)1.18[rad] \begin{aligned} \theta \bar v=&\tan^{-1}\(\frac{\frac{2\sin40^\circ}{11}}{\frac{2\cos40^\circ-1}{11}}\)\\ =&\tan ^{-1}\left(\frac{2 \sin (40{}^{\circ})}{2 \cos (40{}^{\circ})-1}\right)\\ \approx&1.18\ut{rad} \end{aligned}