10판/2. 직선운동

2-28 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 24. 22:36

{a=4.92[m/s2]v0=27.2[m/s]v=0 \begin{cases} a = -4.92 \ut{m/s^2} \\ v_0 = 27.2 \ut{m/s} \\ v = 0 \end{cases}
(a) t=? t =? v=v0+at=0(v=0) v = v_0+at = 0(\because v=0) t=v0a=27.2[m/s]4.92[m/s2]=680123[s]5.528455284552845[s]5.53[s] \begin{aligned} \\ t &= -\frac{v_0}{a} \\ &= -\frac{27.2 \ut{m/s}}{-4.92 \ut{m/s^2}} \\ &= \frac{680}{123}\ut{s} \\ &\approx 5.528455284552845\ut{s} \\ &\approx 5.53\ut{s} \end{aligned}
(b) S=? S = ? 2aS=v2v02, 2aS = v^2-v_0^2, S=v022a(v=0)=(27.2[m/s])22(4.92[m/s2])=9248123[m]75.1869918699187[m]75.2[m] \begin{aligned} S &= -\frac{v_0^2}{2a}(\because v=0) \\ &= -\frac{(27.2 \ut{m/s})^2}{2(-4.92 \ut{m/s^2})} \\ &= \frac{9248}{123}\ut{m} \\ &\approx 75.1869918699187\ut{m} \\ &\approx 75.2\ut{m} \end{aligned}
(c) x(t)=v0t+12at2=(27.2)t+12(4.92)t2=1365t12350t2 \begin{aligned} x(t) &= v_0t+\frac{1}{2}at^2 \\ &= (27.2)t+\frac{1}{2}(-4.92)t^2 \\ &= \frac{136}{5}t-\frac{123 }{50}t^2 \end{aligned} v(t)=x˙= ⁣dx ⁣dt= ⁣d ⁣dt(1365t12350t2)=136512325t \begin{aligned} v(t) &= \dot x = \dxt{x} \\ &= \dt(\frac{136}{5}t-\frac{123 }{50}t^2) \\ &= \frac{136}{5}-\frac{123}{25}t \end{aligned}