솔루션피아

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

$$ \begin{cases} v_{A1}&=v_{A2}=40\ut{km/h}\\ v_{B}&=52\ut{km/h}\\ \Delta x_{A}&=60\ut{km} \end{cases} $$ $$ \begin{aligned} \Delta x_A&=v_{A1}t+v_{A2}t\\ t&=\frac{\Delta x_A}{v_{A1}+v_{A2}}\\ &=\frac{\Delta x_A}{2v_{A1}}\\ \Sigma \Delta x_B&=v_Bt\\ &=v_B\cdot \frac{\Delta x_A}{2v_{A1}}\\ &= \frac{\Delta x_A}{2}\cdot\frac{v_B}{v_{A1}}\\ &= \frac{60\ut{km}}{2}\cdot\frac{52\ut{km/h}}{40\ut{km/h}}\..

$$ \begin{cases} t_2&=\text{Start}\\ v_4&=0\\ t_8&=\text{Land}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} v&=v_0+a\Delta t,\\ 0&=v_0-g(2)\\ \end{aligned} $$ $$\therefore v_0=2g$$ $$\ab{a}$$ $$ \begin{aligned} \Delta y_{\max}&=\int_0^4v\dd t\\ &=\frac{1}{2}\(t_2+t_4\)v_0\\ &=\frac{1}{2}\(2+4\)2g\\ &=6g\\ &=\frac{588399}{10000}\ut{m}\\ &=58.8399\ut{m}\\ &\approx 6\times10\ut{m} \end{..

$$ \begin{cases} h_B&=h_A+15\ut{m}\\ t_{A0} &= 0\ut{s}\\ t_{B0} &= 1\ut{s}\\ \Delta y_3 &=0\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{cases} S_A&=v_{A0}t_A+\frac{1}{2}a_A{t_A}^2\\ S_B&=v_{B0}t_B+\frac{1}{2}a_B{t_B}^2\\ \end{cases} $$ $$ \begin{cases} -\Delta y_A&=(0)t_A+\frac{1}{2}(-g)(t_B+1)^2\\ -\Delta y_B&=v_{B0}{t_B}+\frac{1}{2}(-g){t_B}^2\\ \end{cases} $$ $..

$$ \begin{cases} h&=35.0\ut{m}\\ v&=1100\ut{km/h}=\dfrac{2750}{9}\ut{m/s}\\ \theta&=4.30\degree\\ \end{cases} $$ $$ \begin{aligned} \Delta x &= \frac{h}{\tan\theta}\\ vt&=\frac{h}{\tan\theta}\\ t&=\frac{h}{v\tan\theta}\\ &=\frac{35}{\(\dfrac{2750}{9}\)\tan4.3\degree}\ut{s}\\ &=\frac{63}{550}\cot4.3\degree\ut{s}\\ &\approx 1.523405760812364\ut{s}\\ &\approx 1.52\ut{s} \end{aligned} $$