솔루션피아

$$\ab{a}$$ $$ \begin{aligned} v&=v_0+at,\\ 3.0\times10^8\cdot\frac{0.1}{100}&=0+(9.8)t\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{1500000}{49}\ut{s}\\ &=\frac{25000}{49}\ut{min}\\ &=\frac{1250}{147}\ut{h}\\ &=8\ut{h}+30\ut{min}+\frac{600}{49}\ut{s}\\ &\approx8\ut{h}+30\ut{min}+12.24489795918367\ut{s}\\ &\approx8\ut{h}+30\ut{min}+12\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} S&=v_..

$$ \begin{aligned} \Ans&=\Delta x\\ &=\int_0^{16}v\dd t\\ &=\frac{1}{2}\cdot2\cdot12+8\cdot12+\frac{1}{2}\cdot(6+12)\cdot2+4\cdot6\\ &=150\ut{m} \end{aligned} $$

$$ \begin{cases} v&=18.0\ut{m/s}\\ H&=90.0\ut{m} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -H&=(18)t+\frac{1}{2}(-g)t^2\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{600 }{196133}\left(600+\sqrt{2321330}\right)\ut{s}\\ &\approx 6.496380983846016 \ut{s}\\ &\approx 6.50 \ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-g)(-H)&=v^2-(18.0)^2\..

$$ \begin{cases} d&=20.0\ut{m}\\ v_p&=14.0\ut{m/s}\\ D_{23}&=200\ut{m}\\ D_{12}&=300\ut{m} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Delta t &= \frac{d}{v}\\ &=\frac{20.0\ut{m}}{14.0\ut{m/s}}\\ &=\frac{10}{7}\ut{s}\\ &\approx1.428571428571429\ut{s}\\ &\approx1.43\ut{s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} t_r&=0.500\ut{s}\\ a&=5.00\ut{m/s^2}\\ v_0&=0\\ t_\Ans &=t_r+\Delta t\\ S&=D_{1..

$$ \begin{cases} \text{A : in Air}\\ \text{B : in Water}\\ \end{cases} $$ $$ \begin{cases} H_A&=7.00\ut{m}\\ t_B&=4.00\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-g)(-H)&=v^2-(0)^2 \end{aligned} $$ $$ \begin{aligned} v_B&=\sqrt{2gH}\\ &=\sqrt{14g} \end{aligned} $$ $$ \begin{aligned} H_B&=v_B\cdot t\\ &=\(\sqrt{14g}\)\cdot (4\ut{s})\\ &=\frac{7}{..

12-23 할리데이 11판 솔루션 일반물리학$$ \begin{cases} \text{A : Car}\\ \text{B : Truck} \end{cases} $$ $$ \begin{cases} a_A&=1.80\ut{m/s^2}\\ v_{A0}&=0\\ v_B&=7.30\ut{m/s} \end{cases} $$ $$\ab{a}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2\\ v_Bt&=(0)t+\frac{1}{2}(a_A)t^2\\ \(7.30\ut{m/s}\)&=\frac{1}{2}\(1.80\ut{m/s^2}\)t\\ \end{aligned} $$ $$ t=\frac{73}{9}\ut{s} $$ $$ \begin{al..

$$ \begin{cases} v_A&=v\j\\ v_B&=\frac{1}{2}v\j\\ y_B&=y_A+4\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} 2aS&=v^2-v_0^2\\ 2(-g)(4\ut{m})&=\(\frac{1}{2}v\)^2-(v)^2\\ \end{aligned} $$ $$ \begin{aligned} v&=\sqrt{\frac{32g}{3}}\\ &=\frac{1}{25}\sqrt{\frac{196133}{3}}\ut{m/s}\\ &\approx 10.22762272801782\ut{m/s}\\ &\approx 1\times10\ut{m/s}\\ \end{ali..

$$ \begin{cases} v&=303\ut{km/h}\\ t&=120\ut{ms}\\ \end{cases} $$ $$ \begin{aligned} \Delta x &= vt\\ &=303\ut{km/h}\cdot120\ut{ms}\\ &=303\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}}\cdot120\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &=\frac{101}{10}\ut{m}\\ &=10.1\ut{m}\\ &\approx 1.0\times10\ut{m} \end{aligned} $$

$$ \begin{cases} t_1 &= 1.12\ut{s} &=\text{Time at }h_{\max}\\ t_2 &= 6.00\ut{s} &=\text{Time at Ground}\\ a&=-g&=-9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$v=v_0+at,$$ $$ \begin{aligned} v_0&=v-at\\ &=0-(-g)t_1\\ &=9.80665\ut{m/s^2}\cdot1.12\ut{s}\\ &=\frac{1372931}{125000}\ut{m/s}\\ &=10.983448\ut{m/s}\\ &\approx11.0\ut{m/s} \end{aligned} $$ $$\ab{b}$$ $$S=vt-\frac{1}{2}at^2,$$ $$ \begin{a..

$$ \begin{cases} a&=-g=-9.80665\ut{m/s^s}\\ v_0&=0\\ t_1&=50\ut{ms}=0.05\ut{s}\\ t_2&=100\ut{ms}=2t_1\\ t_3&=150\ut{ms}=3t_1\\ t_4&=200\ut{ms}=4t_1\\ t_5&=250\ut{ms}=5t_1 \end{cases} $$ $$$$ $$ \begin{aligned} S_k&=v_0t+\frac{1}{2}at^2,\\ &=\frac{1}{2}g(t_k)^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} S_{1}&=\frac{1}{2}gt_1^2\\ &=\frac{1}{2}(9.80665\ut{m/s^2})(0.05\ut{s})^2\\ &=\frac{1961..