11판/4. 2차원 운동과 3차원 운동

4-2 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 25. 20:23
{A:AirP:Plane \begin{cases} A:\text{Air}\\ P:\text{Plane}\\ \end{cases} {vA=30.0[km/h]j^vP+A=vP+vAvP+A=vP+Ai^+0j^vP=60[km/h]vP=vPcosθi^+vPsinθj^ \begin{cases} \vec v_A&=-30.0\ut{km/h}\j\\ \vec v_{P+ A}&=\vec v_P+\vec v_A\\ \vec v_{P+ A}&=v_{P+ A}\i+0\j\\ v_P&=60\ut{km/h}\\ \vec v_P&=v_P\cos\theta\i+v_P\sin\theta\j\\ \end{cases} vP+A=vP+vA,vP+Ai^+0j^=(60cosθi^+60sinθj^)+(30j^)=60cosθi^+(60sinθ30)j^ \begin{aligned} \vec v_{P+ A}&=\vec v_P+\vec v_A,\\ v_{P+ A}\i+0\j&=\(60\cos\theta\i+60\sin\theta\j\)+(-30\j)\\ &=60\cos\theta\i+(60\sin\theta-30)\j\\ \end{aligned} {vP+Ai^=60cosθi^0j^=(60sinθ30)j^ \begin{cases} v_{P+ A}\i&=60\cos\theta\i \\ 0\j&=(60\sin\theta-30)\j\\ \end{cases} {60sinθ30=060cosθ>0 \begin{cases} 60\sin\theta-30&=0\\ 60\cos\theta&>0 \end{cases} θ=π6\therefore \theta=\frac{\pi}{6} vP+A=vP+Ai^+0j^,\vec v_{P+ A}=v_{P+ A}\i+0\j, vP+A=vP+A=60cosθ=303[km/h]51.96152422706631[km/h]52.0[km/h] \begin{aligned} \abs{\vec v_{P+ A}}&=v_{P+ A}\\ &=60\cos\theta\\ &=30\sqrt{3}\ut{km/h}\\ &\approx 51.96152422706631\ut{km/h}\\ &\approx 52.0\ut{km/h}\\ \end{aligned}