10판/4. 2차원운동과 3차원운동

4-7 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 14. 18:31

{r1=6.0i^7.0j^+3.0k^[m]r2=3.0i^+9.0j^3.0k^[m]t12=10[s]\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} vˉ=ΔrΔt=r2r1Δt=(6.0i^7.0j^+3.0k^)(3.0i^+9.0j^3.0k^)10=0.3i^1.0j^+0.6k^[m/s] \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}