11판/4. 2차원 운동과 3차원 운동

4-39 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 7. 19:20

{v0=4.50i^[m/s]a=1.2j^[m/s2]t=4.70[s] \begin{cases} \vec v_0&=4.50\i\ut{m/s}\\ \vec a&=1.2\j\ut{m/s^2}\\ t&=4.70\ut{s} \end{cases} v(t)=v0+Δv=v0+0ta ⁣dt=4.50i^+0t1.2j^ ⁣dt=4.50i^+1.2tj^ \begin{aligned} \vec v(t)&=\vec v_0+\Delta \vec v\\ &=\vec v_0+\int_0^{t}\vec a\dd t\\ &=4.50\i+\int_0^{t}1.2\j\dd t\\ &=4.50\i+1.2t\j\\ \end{aligned} (a)\ab{a} v(4.70)=4.50i^+1.2(4.7)j^[m/s]=4.50i^+5.64j^[m/s]4.5i^+5.6j^[m/s] \begin{aligned} \vec v(4.70)&=4.50\i+1.2(4.7)\j\ut{m/s}\\ &=4.50\i+5.64\j\ut{m/s}\\ &\approx 4.5\i+5.6\j\ut{m/s}\\ \end{aligned} (b)\ab{b} r=r0+Δr=0+0tv ⁣dt=0t(4.50i^+1.2tj^) ⁣dt=4.5ti^+0.6t2j^ \begin{aligned} \vec r&=\vec r_0+\Delta r\\ &=0+\int_0^{t}\vec v\dd t\\ &=\int_0^{t}\(4.50\i+1.2t\j\)\dd t\\ &=4.5t\i+0.6t^2\j \end{aligned} r(4.70)=21.15i^+13.254j^[m]21i^+13j^[m] \begin{aligned} \vec r(4.70)&=21.15\i+13.254\j\ut{m}\\ &\approx 21\i+13\j\ut{m}\\ \end{aligned}