11판/2. 직선운동

2-10 할리데이 11판 솔루션 일반물리학

짱세디럭스 2023. 6. 25. 16:12
{vA0=60.0[m/s]vB0=3460.0[m/s]=45.0[m/s]ΔtA=5[s]ΔtB=4[s]vA5=0vB4=0ΔxA0B0=320[m] \begin{cases} v_{A0}&=60.0\ut{m/s}\\ v_{B0}&=\frac{3}{4}\cdot60.0\ut{m/s}=45.0\ut{m/s}\\ \Delta t_A&=5\ut{s}\\ \Delta t_B&=4\ut{s}\\ v_{A5}&=0\\ v_{B4}&=0\\ \Delta x_{A0\rarr B0}&=320\ut{m} \end{cases} S=12(v+v0)t,S=\frac{1}{2}(v+v_0)t, SA=12(vA5+vA0)ΔtA=12(0+60.0[m/s])(5[s])=150[m] \begin{aligned} S_A&=\frac{1}{2}(v_{A5}+v_{A0})\Delta t_A\\ &=\frac{1}{2}(0+60.0\ut{m/s})(5\ut{s})\\ &=150\ut{m} \end{aligned} SB=12(vB4+vB0)ΔtA=12(0+45.0[m/s])(4[s])=90.0[m] \begin{aligned} S_B&=\frac{1}{2}(v_{B4}+v_{B0})\Delta t_A\\ &=\frac{1}{2}(0+45.0\ut{m/s})(4\ut{s})\\ &=90.0\ut{m} \end{aligned} ΣS=150[m]+90[m]=240[m] \begin{aligned} \Sigma S&=150\ut{m}+90\ut{m}\\ &=240\ut{m} \end{aligned} Ans=320[m]240[m]=80.0[m] \begin{aligned} \Ans&=320\ut{m}-240\ut{m}\\ &=80.0\ut{m} \end{aligned}