Question

(a) find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 8t i + sin t j + cos 2t k, v(0) = i, r(0) = j

Answer 1

`\mathbf a(t)=8t\,\mathbf i+\sin t\,\mathbf j+\cos2t\,\mathbf k`


`\mathbf v(t)=\displaystyle\int\mathbf a(t)\,\mathrm dt`

`\mathbf v(t)=(4t^2+C_1)\,\mathbf i+(-\cos t+C_2)\,\mathbf j+\left(\frac12\sin2t+C_3\right)\,\mathbf k`


`\mathbf v(0)=\mathbf i`

`\implies\begin{cases}4(0)^2+C_1=1\\-\cos0+C_2=0\\\frac12\sin2(0)+C_3=0\end{cases}\implies C_1=1,C_2=1,C_3=0`


`\mathbf v(t)=(4t^2+1)\,\mathbf i+(1-\cos t)\,\mathbf j+\frac12\sin2t\,\mathbf k`


`\mathbf r(t)=\displaystyle\int\mathbf v(t)\,\mathrm dt`

`\mathbf r(t)=\left(\frac43t^3+t+C_4\right)\,\mathbf i+(t-\sin t+C_5)\,\mathbf j+\left(-\frac12\cos2t+C_6\right)\,\mathbf k`


`\mathbf r(0)=\mathbf j`

`\implies\begin{cases}\frac43(0)^2+0+C_4=0\\0-\sin0+C_5=1\\-\frac12\cos2(0)+C_6=0\end{cases}\implies C_4=0,C_5=1,C_6=\frac12`


`\mathbf r(t)=\left(\frac43t^3+t\right)\,\mathbf i+(t-\sin t+1)\,\mathbf j+\left(\frac12-\frac12\cos2t\right)\,\mathbf k`