솔루션피아

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

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

$$\begin{cases} t_0:\text{Drive}\\ t_1:\text{Brake Start}\\ t_2:\text{Stop}\\ \end{cases}$$ $$\begin{cases} v_{A0}&=80.5\ut{km/h}=\dfrac{805}{36}\ut{m/s}\\ \Sigma S_A&=63.4\ut{m}\\ v_{B0}&=48.3\ut{km/h}=\dfrac{161}{12}\ut{m/s}\\ \Sigma S_B&=28.9\ut{m}\\ \end{cases}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2aS_{1\rarr2}&={v_2}^2-{v_1}^2\\ S_{1\rarr2}&=\frac{{0}^2-{v_0}^2}{2a}\\ &=-\frac{{v_..

$$\begin{cases} a&=8.0t\ut{m/s^2}\\ v(2.0)&=17\ut{m/s} \end{cases}$$ $$\begin{aligned} v&=\int a\dd t\\ &=\int 8t\dd t\\ &=4t^2+C\\ \end{aligned}$$ $$v(2)=4(2)^2+C=17,\\ C=1$$ $$\therefore v(t)=4t^2+1$$ $$\begin{aligned} v(5)&= 4(5)^2+1\\ &=101\ut{m/s}\\ &\approx 1.0\times10^2\ut{m/s} \end{aligned}$$

$$\begin{cases} S_A&=-40.0\ut{m}\\ a_B&=1.50\ut{m/s^2}\\ v_2&=-3.00\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} S_A&=v_0t_A+\frac{1}{2}a_A{t_A}^2\\ -40&=(0)t_A+\frac{1}{2}(-g){t_A}^2\\ \end{aligned}$$ $$\therefore t_A=4\sqrt{\frac{5}{g}}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2(a_A)(S_A)&={v_1}^2-{v_0}^2\\ 2(-g)(-40)&={v_1}^2-{0}..

$$\begin{cases} v_A&=45.0\ut{km/h}\\ v_B&=70.0\ut{km/h}\\ \end{cases}$$ $$ \begin{aligned} \bar v&=\frac{\Sigma S}{\Sigma t}\\ &=\frac{S_A+S_B}{t_A+t_B}\\ &=\frac{2S}{\(\dfrac{S}{v_A}\)+\(\dfrac{S}{v_B}\)}\\ &=\frac{2}{\(\dfrac{1}{v_A}\)+\(\dfrac{1}{v_B}\)}\\ &=\frac{2}{\(\dfrac{1}{45}\)+\(\dfrac{1}{70}\)}\\ &=\frac{1260}{23}\ut{km/h}\\ &\approx 54.78260869565217\ut{km/h}\\ &\approx 54.8\ut{km..

$$\begin{aligned} S&=\frac{1}{2}(v+v_0)t,\\ &=\frac{1}{2}(15+30)5\\ &=\frac{225}{2}\ut{m}\\ &=112.5\ut{m}\\ &\approx1.1\times10^2\ut{m} \end{aligned}$$

$$\begin{cases} t_s&=7.3\ut{s}\\ t_{ns}&=24.0\ut{s}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \frac{v_s}{v_{ns}}&=\frac{\(\dfrac{S}{t_s}\)}{\(\dfrac{S}{t_{ns}}\)}\\ &=\frac{t_{ns}}{t_s}\\ &=\frac{24}{7.3}\\ &=\frac{240}{73}\\ \end{aligned}$$ $$\begin{aligned} v_s : v_{ns}=240:73 \end{aligned}$$ $$\ab{b}$$ $$v_s : v_{ns}=240:73,$$ $$ \begin{aligned} v_{ns}&=\frac{73}{240}v_s \end{aligned..