솔루션피아

$$ \begin{cases} t_0&:\text{Start}\\ t_1&:\text{Max Speed}\\ t_2&:\text{End}\\ \end{cases} $$ $$ \begin{cases} v_{1}&=11.00\ut{m/s}\\ S_{0\rarr1}&=12.0\ut{m}\\ S_{0\rarr2}&=100\ut{m} \end{cases} $$ $$\ab{a}$$ $$ S=\frac{1}{2}(v+v_0)t,$$ $$ \begin{aligned} S_{0\rarr1}&=\frac{1}{2}(v_1+v_0)t_{0\rarr1}\\ &=\frac{1}{2}(11+0)t_{0\rarr1}\\ t_{0\rarr1}&=\frac{2S_{0\rarr1}}{11}\\ \end{aligned} $$ $$ \be..

$$ \begin{cases} R&:\text{Red Train}\\ G&:\text{Green Train}\\ \end{cases} $$ $$ \begin{cases} v_R&=72\ut{km/h}=20\ut{m/s}\\ v_G&=160\ut{km/h}=\frac{400}{9}\ut{m/s}\\ S&=950\ut{m}\\ a&=-1.0\ut{m/s^2}\\ \end{cases} $$ $$ 2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2(-1)S_{\text{Stop}}&=(0)^2-{v_0}^2\\ S_{\text{Stop}}&=\frac{{v_0}^2}{2}\\ \end{aligned} $$ $$ \begin{cases} S_{R\text{stop}}&=\cfrac{{v_{R0..

$$ S=vt-\frac{1}{2}at^2, $$ $$ \begin{cases} S_{1.5}=v_{1.5}t_{1.5}-\frac{1}{2}a_{1.5}{t_{1.5}}^2\\ S_1=v_1t_1-\frac{1}{2}a_1{t_1}^2 \end{cases} $$ $$ \begin{cases} H_{1.5}=(0)t_{1.5}-\frac{1}{2}(-g){t_{1.5}}^2\\ H_1=(0)t_1-\frac{1}{2}(-g){t_1}^2 \end{cases} $$ $$ \begin{cases} H_{1.5}=\frac{1}{2}g{t_{1.5}}^2\\ H_1=\frac{1}{2}g{t_1}^2 \end{cases} $$ $$ \begin{aligned} \frac{H_{1.5}}{H_1}&=\frac{..

$$ \begin{cases} H&=45.0\ut{m}\\ t&=2.00\ut{s}\\ h_1&=11.8\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ S=vt-\frac{1}{2}at^2, $$ $$ \begin{aligned} h_1-H&=v(2)-\frac{1}{2}(-g)(2)^2\\ (11.8)-(45)&=v(2)-\frac{1}{2}(-g)(2)^2\\ \end{aligned} $$ $$ \begin{aligned} \abs{v}&=\abs{-\(\frac{83}{5}+g\)}\\ &=\frac{528133}{20000}\ut{m/s}\\ &=26.40665\ut{m/s}\\ &\approx 26.4\ut{m/s}\\ \end{aligned} $$

$$ \begin{cases} a&=50.0\ut{m/s^2}\\ v&=80.0\ut{km/h} \end{cases} $$ $$ \begin{aligned} 80\ut{km/h}&=80\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}}\\ &=\frac{200}{9}\ut{m/s} \end{aligned} $$ $$ v=v_0+at,$$ $$ \begin{aligned} \frac{200}{9}\ut{m/s}&=(0)+(50\ut{m/s^2})t\\ t&=\frac{4}{9}\ut{s}\\ &\approx 0.4444444444444444\ut{s}\\ &\approx 0.444\ut{s}\\ &\approx 444\ut{ms..

$$ \begin{cases} v_1&=65\ut{mi/h}\\ v_2&=70\ut{mi/h}\\ S &= 700\ut{km}\\ \end{cases} $$ $$ \begin{cases} 1\ut{mile}&=5280\ut{ft}\\ 1\ut{ft}&=12\ut{in}\\ 1\ut{in}&=2.54\ut{cm}\\ \therefore 1\ut{mile}&=1.609344\ut{km} \end{cases} $$ $$ \begin{aligned} \Delta t &= t_1-t_2\\ &=\frac{S}{v_1}-\frac{S}{v_2}\\ &=S\(\frac{1}{v_1}-\frac{1}{v_2}\)\\ &=(700\ut{km})\(\frac{1}{65\ut{mi/h}}-\frac{1}{70\ut{mi/h..

$$ \begin{cases} t&=2\ut{h}+41\ut{min}\\ \Delta x &= 3.90\ut{m}\\ \end{cases} $$ $$ \begin{aligned} t&=2\ut{h}+41\ut{min}\\ &=120\ut{min}+41\ut{min}\\ &=161\ut{min} \end{aligned} $$ $$ \begin{aligned} \bar v&=\frac{\Delta x}{t}\\ &=\frac{390\ut{cm}}{161\ut{min}}\\ &=\frac{390}{161}\ut{cm/min}\\ &\approx 2.422360248447205\ut{cm/min}\\ &\approx 2.42\ut{cm/min}\\ \end{aligned} $$

$$x=50t+10t^2$$ $$\ab{a}$$ $$ \begin{aligned} \bar v_{0\rarr3}&=\frac{\Sigma \Delta x_{0\rarr3}}{\Sigma \Delta t_{0\rarr3}}\\ &=\frac{x(3)-x(0)}{3}\\ &=\frac{(240)-(0)}{3}\ut{m/s}\\ &=80\ut{m/s}\\ &=80.0\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=\dxt{x}\\ &=\dt\(50t+10t^2\)\\ &=50+20t \end{aligned} $$ $$ \begin{aligned} a&=\dxt{v}\\ &=\dt\(50+20t\)\\ &=20\\ \end{aligned} $$ $$ ..

$$y=(4.0\ut{cm})\sin\frac{\pi t}{4}$$ $$\ab{a}$$ $$ \begin{aligned} \bar v_{0\rarr2}&=\frac{\Sigma \Delta y_{0\rarr2}}{\Sigma t_{0\rarr2}}\\ &=\frac{y(2)-y(0)}{2}\\ &=\frac{\(4\sin\dfrac{\pi\cdot2}{4}\)-\(4\sin\dfrac{\pi\cdot0 }{4}\)}{2}\\ &=\frac{\(4\)-\(0\)}{2}\ut{cm/s}\\ &=2\ut{cm/s}\\ &= 2.0\ut{cm/s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=\dxt{y}\\ &=\dt\(4\sin\frac{\pi t}{4}\)\\ ..

풀이자 주 : 자유낙하로 가정하지 않으면 문제는 초기속도에 종속된 답이 나오므로, 자유낙하로 가정하고 풀었습니다. $$ \begin{cases} A&:\text{First Half}\\ B&:\text{Second Half}\\ \end{cases} $$ $$ \begin{cases} v_0&=0\\ H&=120\ut{m}\\ h_1&=60\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} -h_1&=(0)t+\frac{1}{2}(-g){t_A}^2\\ h_1&=\frac{1}{2}g{t_A}^2\\ 60&=\frac{1}{2}g{t_A}^2\\ \end{aligned..