10판/2. 직선운동

2-37 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 25. 22:56

{x0=2[m]x1=0x2=6.0[m]tx=x[s]a=Constant \begin{cases} x_0 &= -2\ut{m} \\ x_1 &= 0 \\ x_2 &= 6.0\ut{m} \\ t_x &= x\ut{s} \\ a &= \Cons \end{cases} a=?\vec{a} =? (Make System of Equations)\text{(Make System of Equations)} Δx=v0xt+12ayt2, \Delta x = \overset{x}{v_0}t+\frac{1}{2}\overset{y}{a}t^2, (OverSet x,y Means Unknown Value)\text{(OverSet x,y Means Unknown Value)} {Δx01=v0xt01+12ayt012Δx02=v0xt02+12ayt022 \begin{cases} \Delta x_{0 \to 1} &= \overset{x}{v_0}t_{0 \to 1}+\frac{1}{2}\overset{y}{a}t_{0 \to 1}^2 \\ \Delta x_{0 \to 2} &= \overset{x}{v_0}t_{0 \to 2}+\frac{1}{2}\overset{y}{a}t_{0 \to 2}^2 \end{cases} {0(2)=v0x(1)+12ay(1)26(2)=v0x(2)+12ay(2)2 \begin{cases} 0 - (-2) &= \overset{x}{v_0}(1)+\frac{1}{2}\overset{y}{a}(1)^2 \\ 6 - (-2) &= \overset{x}{v_0}(2)+\frac{1}{2}\overset{y}{a}(2)^2 \end{cases} a=+4[m/s2] a = +4\ut{m/s^2}
(a) a=4[m/s2] |a| = 4\ut{m/s^2}
(b) a=+xDirection a = +x \, \text{Direction}