12.11-bo'lim : Tezlik va tezlanish
1. Ob'ektlarning tezlanishi \(\vec a = 3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k\) tomonidan berilgan. Ob'ektlarning boshlang'ich tezligi \(\vec v\left( 0 \right) = \vec j - 3\vec k\) va ob'ektlarning boshlang'ich pozitsiyasi \(\vec r\left( 0 \right) = - 5\vec i + 2\vec j - 3\vec k\). Jismlarning tezligi va joylashuv funksiyalarini aniqlang.
Yechim
Tezlik funksiyasini aniqlash uchun biz qilishimiz kerak bo'lgan narsa tezlashtirish funksiyasini integrallashdir.
\[\vec v\left( t \right) = \int{{3t\,\vec i - 4{{\bf{e}}^{ - t}}\,\vec j + 12{t^2}\vec k\,dt}} = \frac{3}{2}{t^2}\,\vec i + 4{{\bf{e}}^{ - t}}\,\vec j + 4{t^3}\vec k + \vec c\]Integratsiyaning "doimiy" ni unutmang, bu holda aslida vektor \(\vec c = {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\) . Integratsiya konstantasini aniqlash uchun bizga muammo bayonida berilgan \(\vec v\left( 0 \right)\) qiymatidan foydalanish kifoya.
\[\vec j - 3\vec k = \vec v\left( 0 \right) = 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\,\]\({c_1}\), \({c_2}\) va \({c_3}\) qiymatlarini aniqlash uchun biz qilishimiz kerak bo'lgan yagona narsa turli komponentlarni tenglashtirishdir.
\[\begin{align*}&{\vec i:0 = {c_1}}\\&{\vec j:1 = 4 + {c_2}}\\&{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.5in}{c_1} = 0,\,\,\,\,{c_2} = - 3,\,\,\,\,{c_3} = - 3\]Tezlik u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec v\left( t \right) = \frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k}}\]2-qadam
Joylashuv funktsiyasi oddiygina oldingi bosqichda topilgan tezlik funktsiyasining ajralmas qismidir.
\[\vec r\left( t \right) = \int{{\frac{3}{2}{t^2}\,\vec i + \left( {4{{\bf{e}}^{ - t}} - 3} \right)\,\vec j + \left( {4{t^3} - 3} \right)\vec k\,dt}} = \frac{1}{2}{t^3}\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t} \right)\,\vec j + \left( {{t^4} - 3t} \right)\vec k + \vec c\] Integratsiya konstantasining qiymatini aniqlash uchun muammo bayonotidagi \(\vec r\left( 0 \right)\) qiymatidan foydalanamiz. \[ - 5\vec i + 2\vec j - 3\vec k = \vec r\left( 0 \right) = - 4\,\vec j + {c_1}\vec i + {c_2}\vec j + {c_3}\vec k\] \[\begin{align*} &{\vec i: - 5 = {c_1}}\\& {\vec j:2 = - 4 + {c_2}}\\ &{\vec k: - 3 = {c_3}}\end{align*}\hspace{0.5in} \Rightarrow \hspace{0.25in}{c_1} = - 5,\,\,\,\,{c_2} = 6,\,\,\,\,{c_3} = - 3\]Pozitsiya funktsiyasi u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left( {\frac{1}{2}{t^3} - 5} \right)\,\vec i + \left( { - 4{{\bf{e}}^{ - t}} - 3t + 6} \right)\,\vec j + \left( {{t^4} - 3t - 3} \right)\vec k}}\]