솔루션피아

$$ x = 25t-6.0t^3 $$ $$ \begin{aligned} v &= \dot x = \frac{dx}{dt} \\ &= \frac{d}{dt}(25t-6.0t^3) \\ &= 25-18 t^2 \end{aligned} $$ $$ \begin{aligned} a &= \dot v = \frac{dv}{dt} \\ &= \frac{d}{dt}(25-18 t^2) \\ &= -36 t \end{aligned} $$ $$ \therefore \begin{cases} x &= 25t-6.0t^3 \\ v &= 25-18 t^2 \\ a &= -36 t \end{cases}\cdots(2-20-1) $$ (a) $$\Ans =t_{v=0} = ? $$ $$ (2-20-1), $$ $$v = 25-18 ..

$$ \begin{aligned} \Ans &= \bar a = \frac{v(2.4) - v(0)}{t_{2.4}-t_0} \\ &=\frac{-30\ut{m/s} - 18\ut{m/s}}{2.4\ut{s}} \\ &=-20\ut{m/s^2} \end{aligned} $$

$$ x = 12t^2-2t^3 $$ $$ \begin{aligned} v &= \dot x = \dxt{x} \\ &= \dt(12t^2-2t^3) \\ &= 24t-6t^2 \end{aligned} $$ $$ \begin{aligned} a &= \dot v = \dxt{v}=\ddot x = \dxtt{x} \\ &= \dt(24t-6t^2) \\ &= 24-12t \end{aligned} $$ $$ \therefore \begin{cases} x &= 12t^2-2t^3 \\ v &= 24t-6t^2 \\ a &= 24-12t \end{cases}\cdots({2-18-1}) $$ (a) $$ \Ans = x(3.5) =? $$ $$(2-18-1), $$ $$ \begin{aligned} x(3...

$$ x = 9.75 + 1.50t^3 $$ (a) $$ \begin{aligned} \Ans &= \bar v = \frac{x(t_3)-x(t_2)}{t_3-t_2} \\ &= \frac{x(3.00)-x(2.00)}{3.00-2.00} \\ &= \frac{9.75 + 1.50(3.00)^3-\bra{9.75 + 1.50(2.00)^3}}{3.00-2.00} \\ &= \frac{57}{2}\ut{cm/s} \\ &= 28.5\ut{cm/s} \\ \end{aligned} $$ (b) $$\Ans = \dot x(2.00) =? $$ $$ \begin{aligned} \\ \dot x &= \dxt{x} \\ &= \dt(9.75 + 1.50t^3) \\ &= \frac{9}{2}t^2 \cdots..

$$ x = 4.0- 6.0t^2 \cdots({2-16-1})$$ (a) $$ \Ans = t_{\dot x=0} =? $$ $$ \begin{aligned} \dot x &= \dxt{x} \\&= \dt(4.0- 6.0t^2) \\&= -12t=0 \end{aligned} $$ $$ t=0\ut{s} $$ (b) $$ \begin{aligned} \Ans &= x(0) =? \\ &= 4.0- 6.0(0)^2 \\ &= 4.0\ut{m} \end{aligned} $$ (c) $$ \Ans = t_{x=0,t0} =? $$ $$ (2-16-2), $$ $$ t = \sqrt{\frac{2}{3}} $$ (e) (f) +20t (그래프참조) (g) 증가시킴 (그래프참조)

$$x = 18t + 5.0t^2 $$ (a) $$ \Ans = \dot x(2) =? $$ $$ \begin{aligned} \dot x &= \dxt{x} \\&= \dt(18t + 5.0t^2) \\&= 18 + 10 t \\ \dot x(2)&=18 + 10 (2) \\ &= 38\ut{m/s} \end{aligned} $$ (b) $$ \begin{aligned} \Ans &= \frac{x(t_2)-x(t_1)}{t_2-t_1} =? \\ &= \frac{x(3.0)-x(2.0)}{3.0-2.0} \\ &= \bra{18(3.0) + 5.0(3.0)^2}-\bra{18(2.0) + 5.0(2.0)^2} \\ &= 43\ut{m/s} \end{aligned} $$

$$ x = 16te^{-t} $$ $$ \Ans = \abs{x(t_{\dot x=0})} =? $$ $$ \begin{aligned} \dot x &= \dxt{x} \\&= \dt(16te^{-t}) \\&= 16 e^{-t}-16 te^{-t} \\&= -16 e^{-t} (t-1) = 0 \end{aligned} $$ $$ t=1\ut{s} $$ $$ \begin{aligned} \\\abs{x(1)} &= \abs{16(1)e^{-(1)}} \\ &= \frac{16}{e}\ut{m} \\&\approx 5.886071058743077\ut{m} \end{aligned} $$

(a) $$ \begin{cases} v_1 &= 55\ut{km/h} \\ v_2 &= 90\ut{km/h} \\ t &= t_1 = t_2 \end{cases} $$ $$ \begin{aligned} \Ans &= \bar v_a = \frac{\Sigma S}{\Sigma t} =\frac{S_1+S_2}{t_1+t_2} \\ &=\frac{S_1+S_2}{2t} \\ &=\frac{1}{2}\left(\frac{S_1}{t}+\frac{S_2}{t}\right) \\ &=\frac{v_1+v_2}{2} \\ &=\frac{55\ut{km/h}+90\ut{km/h}}{2} \\ &=72.5\ut{km/h} \\ &\approx 73\ut{km/h} \end{aligned} $$ (b) $$ \beg..

$$\begin{cases} v_s &= 5.00\ut{m/s} \\ v &= 25.0\ut{m/s} \\ L &= 12.0\ut{m} \end{cases} $$ (a) $$ \put \begin{cases} x &= \text{Wave End Point} \\ L_a &= \text{Added Wave Length} \\ n &= \text{Number Of Added Cars} \\ t_a &= \text{Time Of Add 1 Car} \end{cases} $$ $$ \begin{aligned} t_a &= \frac{d}{v-v_s} \\ t &= nt_a \\ n &= \frac{t}{t_a} = t\frac{v-v_s}{d} \end{aligned} $$ $$ \begin{aligned} x..

$$ \begin{cases} v_c &= 15.00\ut{m/s} \\ t &= 150.0\ut{s} \\ x_{c0}-x_{a0} &= 1.500\ut{km} \end{cases} $$ $$ \begin{aligned} \Ans &= v_a = ? \\ x_{c0}+\Delta x_c &= x_{a0}+\Delta x_a \\ x_{c0}-x_{a0} &= \Delta x_a-\Delta x_c \\ &= v_at-v_ct \\ &= (v_a-v_c)t \\ v_a-v_c&= \frac{x_{c0}-x_{a0}}{t} \\ v_a&= \frac{x_{c0}-x_{a0}}{t}+v_c \\ &= \frac{1.500\ut{km}}{150.0\ut{s}}\cdot\frac{1000\ut{m}}{1\ut{..