솔루션피아

$$ \begin{aligned} \\ S &= 1.0\ut{mm}-5.0\ut{\mu m} \\ &= 1.0\ut{mm}\frac{1\ut{m}}{10^3\ut{mm}}-5.0\ut{\mu m}\frac{1\ut{m}}{10^6\ut{\mu m}} \\ &= \frac{199}{200000}\ut{m} \\ &= 9.95\times 10^{-4}\ut{m} \end{aligned} $$ $$ \begin{cases} v_0 &= 1.6\ut{m/s} \\ S &= 9.95\times 10^{-4}\ut{m} \\ v &= 0 \\ a &= Const. \\ g &= 9.80665 \ut{m/s^2} \end{cases} $$ (a) t=? $$S = \frac{1}{2}(v+v_0)t ,$$ $$ \b..

$$ \begin{cases} v_0 &= 0 \\ a &= \Cons \\ t_A &= \frac{1}{5}\ut{s} \\ t_B &= 5.00\ut{s} \end{cases} $$ $$ \Ans = S_A : S_B =?$$ $$\begin{aligned} \\ S &= v_0t+\frac{1}{2}at^2 \\ &= \frac{1}{2}at^2 (\because v_0=0) \end{aligned} $$ $$ \begin{aligned} \Ans &= S_A : S_B \\ &= \frac{1}{2}at_A^2 : \frac{1}{2}at_B^2 \\ &= t_A^2 : t_B^2 \\ &= \left(\frac{1}{5}\ut{s}\right)^2 : (5.00\ut{s})^2 \\ &= 1:6..

$$x= ct^2 - bt^3$$ (a) $$ \Ans = c =? $$ $$ \put k_n = \text{AnyNumber} $$ $$\begin{aligned} x &= ct^2 - bt^3=k_1\ut{m} \\ &\Rightarrow c(t\ut{s})^2 - b(t\ut{s})^3 = k_1\ut{m} \cdots(2-22-1) \\ &\Rightarrow c(t\ut{s})^2 = k_2\ut{m} \\ &\Rightarrow c(t^2\ut{s^2}) = k_2\ut{m} \\ &\Rightarrow c = \frac{k_2\ut{m}}{t^2\ut{s^2}} \\ &\Rightarrow c = \frac{k_2}{t^2}\ut{m/s^2} \end{aligned} $$ (b) $$ \An..

$$ \begin{aligned} v_0 &= 130\ut{km/h} \\&= 130\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}} \\ &= \frac{325}{9}\ut{m/s} \end{aligned} $$ $$ \begin{cases} v_0 &= \frac{325}{9}\ut{m/s} \\ v &= 0 \\ S &= 210\ut{m} \\ a &= \Cons \end{cases} $$ (a) $$\Ans = \abs{a} =? $$ $$2aS = v^2-v_0^2, $$ $$ \begin{aligned} \Ans &= \abs{a} = \abs{\frac{v^2-v_0^2}{2S}} \\ &= \abs{\frac{..

$$ x = 25t-6.0t^3 $$ $$ \begin{aligned} v &= \dot x = \frac{dx}{dt} \\ &= \frac{d}{dt}(25t-6.0t^3) \\ &= 25-18 t^2 \end{aligned} $$ $$ \begin{aligned} a &= \dot v = \frac{dv}{dt} \\ &= \frac{d}{dt}(25-18 t^2) \\ &= -36 t \end{aligned} $$ $$ \therefore \begin{cases} x &= 25t-6.0t^3 \\ v &= 25-18 t^2 \\ a &= -36 t \end{cases}\cdots(2-20-1) $$ (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) $$ \begin{aligned} \Ans &= \frac{x(t_2)-x(t_1)}{t_2-t_1} =? \\ &= \frac{x(3.0)-x(2.0)}{3.0-2.0} \\ &= \bra{18(3.0) + 5.0(3.0)^2}-\bra{18(2.0) + 5.0(2.0)^2} \\ &= 43\ut{m/s} \end{aligned} $$