(a) find the position vector of a particle that has the given acceleration and the specified initial…

Explanation:

Step1: Integrate acceleration to get velocity

Given (a(t)=5t\mathbf{i}+e^{t}\mathbf{j}+e^{-t}\mathbf{k}). Integrate each component: (v(t)=\int a(t)dt=\int5t\mathbf{i}dt+\int e^{t}\mathbf{j}dt+\int e^{-t}\mathbf{k}dt) (v(t)=\left(\frac{5t^{2}}{2}+C_{1}\right)\mathbf{i}+(e^{t}+C_{2})\mathbf{j}+(-e^{-t}+C_{3})\mathbf{k}) Use (v(0)=\mathbf{k}), so (C_{1} = 0), (C_{2}=0), (C_{3}=2) (v(t)=\frac{5t^{2}}{2}\mathbf{i}+e^{t}\mathbf{j}+(-e^{-t}+2)\mathbf{k})

Step2: Integrate velocity to get position

(r(t)=\int v(t)dt=\int\frac{5t^{2}}{2}\mathbf{i}dt+\int e^{t}\mathbf{j}dt+\int(-e^{-t}+2)\mathbf{k}dt) (r(t)=\left(\frac{5t^{3}}{6}+D_{1}\right)\mathbf{i}+(e^{t}+D_{2})\mathbf{j}+(e^{-t}+2t+D_{3})\mathbf{k}) Use (r(0)=\mathbf{j}+\mathbf{k}), so (D_{1} = 0), (D_{2}=0), (D_{3}=0) (r(t)=\frac{5t^{3}}{6}\mathbf{i}+e^{t}\mathbf{j}+(e^{-t}+2t)\mathbf{k})

Answer:

(\frac{5t^{3}}{6}\mathbf{i}+e^{t}\mathbf{j}+(e^{-t}+2t)\mathbf{k})