4-40 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec v_0&=10\j\ut{m/s}\\ \vec a&=6.0\i+2.0\j\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} \vec v &= \vec v_0+\Delta \vec v\\ &= 10\j+\int_0^t\vec a\dd t\\ &= 10\j+\int_0^t6\i+2\j\dd t\\ &= 6t\i+\(10+2t\)\j\\ \end{aligned} $$ $$ \begin{aligned} \vec r &= \vec r_0+\Delta \vec r\\ &= 0+\int_0^t\vec v\dd t\\ &= \int_0^t6t\i+\(10+2t\)\j\dd t\\ &=3t^2\i+\(10t+t^2\)\j \end{aligned} $$ $$ \begin{aligned} r_x &= 40\\ &=3t^2\\ t&=2\sqrt{\frac{10}{3}}\ut{s}\\ \end{aligned} $$ $$\ab{a}$$ $$r_y(t)=10t+t^2,$$ $$ \begin{aligned} \Ans&=r_y\(2\sqrt{\frac{10}{3}}\)\\ &=\frac{40}{3}+20 \sqrt{\frac{10}{3}}\ut{m}\\ &\approx 49.84817050034441\ut{m}\\ &\approx 50\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v(t)&=\sqrt{(6t)^2+\(10+2t\)^2}\\ &=\sqrt{40 t^2+40 t+100}\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=v\(2\sqrt{\frac{10}{3}}\)\\ &=2 \sqrt{\frac{5}{3} \(95+4 \sqrt{30}\)}\ut{m/s}\\ &\approx 27.91760523399845\ut{m/s}\\ &\approx 28\ut{m/s}\\ \end{aligned} $$