4-43 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec v_0 &= 8.00\times10^8\i\ut{cm/s}\\ \Delta x &= 2.00\ut{cm}\\ \vec a&=-6.70\times10^{16}\j\ut{cm/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} t&=\frac{\Delta x}{v_x}\\ &=\frac{2}{8\times10^8}\ut{s}\\ &=2.50\times10^{-9}\ut{s} \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \end{aligned}$$ $$\begin{aligned} \Delta y&=v_{y0}t+\frac{1}{2}a_yt^2\\ &=(0)t+\frac{1}{2}a_yt^2\\ &=\frac{1}{2}(-6.70\times10^{16})(2.5\times10^{-9})^2\\ &=-\frac{67}{320}\ut{cm}\\ &=-0.209375\ut{cm}\\ &\approx -0.209\ut{cm}\\ &\approx -2.09\ut{mm} \end{aligned}$$ $$\ab{c,d}$$ $$\begin{aligned} \vec v&=\vec v_0+\Delta \vec v\\ &=\vec v_0+\int_0^t\vec a \dd t\\ &=(8\times10^8\i)+\int_0^t(-6.70\times10^{16}\j) \dd t\\ &=8\times10^8\i-6.7\times10^{16}t\j \end{aligned}$$ $$\begin{aligned} \vec v(t)&=8\times10^8\i-6.7\times10^{16}t\j,\\ \end{aligned}$$ $$\begin{aligned} \Ans&=\vec v\(2.50\times10^{-9}\)\\ &=8\times10^8\i-1.675\times10^8\j\ut{cm/s}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \abs{v_x} =8\times10^8\ut{cm/s} \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \abs{v_y} &=1.675\times10^8\ut{cm/s}\\ &\approx 1.68\times10^8\ut{cm/s} \end{aligned}$$