10판/4. 2차원운동과 3차원운동

4-14 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 7. 1. 22:48
{v0=4.0i^2.0j^+3.0k^[m/s]v4.0=2.0i^2.0j^+5.0k^[m/s]\begin{cases} \vec v_0 &= 4.0\i-2.0\j+3.0\k\ut{m/s}\\ \vec v_{4.0} &= -2.0\i-2.0\j+5.0\k\ut{m/s}\\ \end{cases}
(a) aavg=?\vec a_{avg}=? aavg=ΔvΔt=v4v04=(2.0i^2.0j^+5.0k^)(4.0i^2.0j^+3.0k^)4[m/s2]=1.5i^+0.5k^[m/s2]\begin{aligned} \vec a_{avg} &= \frac{\Delta \vec v}{\Delta t}\\ &=\frac{\vec v_4-\vec v_0}{4}\\ &=\frac{(-2.0\i-2.0\j+5.0\k)-(4.0\i-2.0\j+3.0\k)}{4}\ut{m/s^2}\\ &=-1.5\i+0.5\k\ut{m/s^2}\\ \end{aligned}
(b) aavg=?\abs{\vec a_{avg}}=? aavg=(1.5)2+(0.5)2[m/s2]=52[m/s2]1.58113883008[m/s2]1.6[m/s2]\begin{aligned} \abs{\vec a_{avg}} &= \sqrt{(-1.5)^2+(0.5)^2}\ut{m/s^2}\\ &=\sqrt{\frac{5}{2}}\ut{m/s^2}\\ &\approx1.58113883008\ut{m/s^2}\\ &\approx1.6\ut{m/s^2}\\ \end{aligned}
(c) θa=?\theta_{\vec a}=? θa=cos1(axa)=cos1(310)2.81984209919[rad]2.8[rad]\begin{aligned} \theta_{\vec a} &= \cos^{-1}\(\frac{a_x}{\abs{\vec a}}\)\\ &=\cos^{-1}\(-\frac{3}{\sqrt{10}}\)\\ &\approx2.81984209919\ut{rad}\\ &\approx2.8\ut{rad}\\ \end{aligned}