10판/4. 2차원운동과 3차원운동

4-16 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 7. 3. 02:33
v=(6.0t4.0t2)i^+8.0j^[m/s]\vec v = (6.0t-4.0t^2)\i+8.0\j\ut{m/s}
(a) a2.5=?\vec a_{2.5}=? a= ⁣dv ⁣dt= ⁣d ⁣dt{(6.0t4.0t2)i^+8.0j^}=(68t)i^[m/s2]\begin{aligned} \vec a &= \dxt{\vec v}\\ &= \dt\{(6.0t-4.0t^2)\i+8.0\j\}\\ &= (6-8t)\i\ut{m/s^2}\\ \end{aligned} at=2.5=(682.5)i^[m/s2]=14i^[m/s2] \begin{aligned} \vec a_{t=2.5}&=(6-8\cdot2.5)\i\ut{m/s^2}\\ &=-14\i\ut{m/s^2} \end{aligned}
(b) ta=0=?t_{a=0}=? a=(68t)i^=0\vec a = (6-8t)\i=0 t=34[s]=0.75[s] \begin{aligned} t&=\frac{3}{4}\ut{s}\\ &= 0.75\ut{s} \end{aligned}
(c) tv=0=?t_{v=0}=? v=(6.0t4.0t2)2+8.02=0v = \sqrt{(6.0t-4.0t^2)^2+8.0^2}=0 [Impossible] \title{Impossible}
(d) tv=10=?t_{v=10}=? v=(6.0t4.0t2)2+8.02=10v = \sqrt{(6.0t-4.0t^2)^2+8.0^2}=10 t=14(3+33)[s](t>0)2.186140661634507[s]2.2[s] \begin{aligned} t&=\frac{1}{4} \left(3+\sqrt{33}\right)\ut{s}(\because t>0)\\ &\approx2.186140661634507\ut{s}\\ &\approx2.2\ut{s}\\ \end{aligned}