11판/4. 2차원 운동과 3차원 운동

4-40 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 7. 19:37
{v0=10j^[m/s]a=6.0i^+2.0j^[m/s2] \begin{cases} \vec v_0&=10\j\ut{m/s}\\ \vec a&=6.0\i+2.0\j\ut{m/s^2}\\ \end{cases} v=v0+Δv=10j^+0ta ⁣dt=10j^+0t6i^+2j^ ⁣dt=6ti^+(10+2t)j^ \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} r=r0+Δr=0+0tv ⁣dt=0t6ti^+(10+2t)j^ ⁣dt=3t2i^+(10t+t2)j^ \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} rx=40=3t2t=2103[s] \begin{aligned} r_x &= 40\\ &=3t^2\\ t&=2\sqrt{\frac{10}{3}}\ut{s}\\ \end{aligned} (a)\ab{a} ry(t)=10t+t2,r_y(t)=10t+t^2, Ans=ry(2103)=403+20103[m]49.84817050034441[m]50[m] \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} (b)\ab{b} v(t)=(6t)2+(10+2t)2=40t2+40t+100 \begin{aligned} v(t)&=\sqrt{(6t)^2+\(10+2t\)^2}\\ &=\sqrt{40 t^2+40 t+100}\\ \end{aligned} Ans=v(2103)=253(95+430)[m/s]27.91760523399845[m/s]28[m/s] \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}