솔루션피아

(풀이자주) 이 문제의 (b)는 정보가 부족하여 풀 수 없습니다. 주어진 정보만으로는 총 활강시간이 3초 이하일 수도 있는, 성립할 수 없는 문제입니다. 수직성분 기준으로 총 활강이 3초이상 걸렸을 것으로 추정할 수 있을 뿐 아무런 정보가 없습니다. $$\begin{cases} v_{x0} = 29.4\ut{m/s}\\ t_{1\to2} = 3.00\ut{s}\\ \end{cases}$$ (a)$v_{x1}=?$ $$ a_x = 0, $$ $$ v_x = \Cons $$ $$ v_{x1} = v_{x0} = 29.4\ut{m/s}$$

$$\begin{cases} x_0 = 80\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = 12\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ (a)$t_{0\to1}=?$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ -x_0 &= v_0t_1-\frac{1}{2}gt_1^2\\ \end{aligned}$$ $$\begin{aligned} t_1 &= \frac{v_0\pm\sqrt{2 g x_0+v_0^2}}{g}\\ &= \frac{12\ut{m/s}\pm\sqrt{2 (9.80665..

$$\begin{cases} x_0 = 30.0\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = -15.0\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ (a)$t_{0\to1}=?$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ -x_0 &= v_0t_1-\frac{1}{2}gt_1^2\\ \end{aligned}$$ $$\begin{aligned} t_1 &= \frac{v_0\pm\sqrt{2 g x_0+v_0^2}}{g}\\ &= \frac{-15\ut{m/s}\pm\sqrt{2 (9..

$$ \begin{cases} x_0 = 98\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = 14\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases} $$ (a)$\abs{v_1}=?$ $$ 2a\Delta x = v^2-v_0^2, $$ $$ v^2 = 2a\Delta x +v_0^2 $$ $$ \begin{aligned} \abs{v_1} &= \sqrt{2(-g)(-x_0) +v_0^2}\\ &= \sqrt{2gx_0+v_0^2}\\ &= \sqrt{2(9.80665\ut{m/s^2})(98\ut{m})+\(14\ut{m/s}\)^2}\\ &= \frac{7 }{50}\sqrt{\frac{216133}{2}}\ut{m/s}\\ &\approx 46.0..

$$\begin{cases} x_0 = 1800\ut{m}\\ x_1 = 0\\ v_0 = 0\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ (a)$\abs{v_1}=?$ $$2a\Delta x = v^2-v_0^2,$$ $$v^2 = 2a\Delta x +v_0^2$$ $$\begin{aligned} \abs{v_1} &= \sqrt{2(-g)(-x_0) +0^2}\\ &= \sqrt{2gx_0}\\ &= \sqrt{2(9.80665\ut{m/s^2})(1800\ut{m})}\\ &= \sqrt{\frac{1765197}{50}}\ut{m/s}\\ &\approx 187.8934272400182\ut{m/s}\\ &\approx 187.9\ut{m/s} \end{al..

$$\begin{cases} x_0 = H\\ x_1 =\frac{9}{25}H\\ x_2 = 0\\ t_{1\to2} = 1\ut{s}\\ v_0 = 0\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$H=?$$ $$\begin{cases} \Delta x = v_0t+\frac{1}{2}at^2\\ 2a\Delta x = v^2-v_0^2\\ \end{cases}$$ $$\begin{aligned} x_2-x_1 &= v_1t_{1\to2}+\frac{1}{2}(-g)t_{1\to2}^2 \\ 0-\frac{9}{25}H &= v_1(1)+\frac{1}{2}(-g)(1)^2 \\ \frac{9}{25}H &= \frac{1}{2}g-v_1 \taag{2-45-1..

$$\begin{cases} t_2 = 0.200\ut{s}\\ x_2 = 0.588\ut{m}\\ x_1 = 0.544\ut{m}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $\abs{v_0}=?$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} \abs{v_0}&= \frac{x_2-\frac{1}{2}(-g)t_2^2}{t_2}\\ &= \frac{x_2}{t_2}+\frac{gt_2}{2}\\ &= \frac{0.588\ut{m}}{0.200\ut{s}}+\frac{9.80665\ut{m/s^2}(0.200\ut{s})}{2}\\ &= \frac{254933}{20000}\ut{m/s}\\ &= 12...

$$\begin{cases} A = \text{Speed Train} \\ B = \text{Slow Train} \\ t_1 = \text{A,B Minimum Distance Time} \\ \end{cases}$$ $$\begin{aligned} v_{A0} &= 161\ut{km/h} \\ &= 161\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{805}{18}\ut{m/s} \\ \end{aligned}$$ $$\begin{aligned} v_B &= 29.0\ut{km/h} \\ &= 29.0\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\u..

$$\begin{cases} D = \text{Driver} \\ C = \text{Cop} \\ t_1 = \text{Driver Look Cop Speed Down}\\ t_2 = \text{Driver Brake Start}\\ t_3 = \text{Car Crash}\\ \end{cases}$$ $$\begin{aligned} v_0 &= v_{D0}= v_{C0}= 120\ut{km/h} \\ &= 120\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{100}{3}\ut{m/s} \\ \end{aligned}$$ $$\begin{cases} x_{D0} = 0 \\ x_{C0} = 25\ut{..

$$v_0 = 64\ut{m/s}$$ (a)$a=?$ $$\begin{cases} a = \Cons \\ t = 3.0\ut{s} \\ v = 0 \\ \end{cases}$$ $$v = v_0+at,$$ $$\begin{aligned} a &= \frac{v-v_0}{t}\\ &= \frac{0-64\ut{m/s}}{3.0\ut{s}}\\ &= -\frac{64}{3}\ut{m/s^2}\\ &\approx -21.33333333333333\ut{m/s^2}\\ &\approx -21\ut{m/s^2}\\ \end{aligned}$$ (a)$x_i=0, \Delta x=?$ $$\begin{aligned} \Delta x &= \frac{1}{2}(v+v_0)t,\\ &= \frac{1}{2}(0+64\..