Answer:
Explanation:
Given
Acceleration [tex]a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex][/tex]
and [tex]v(0)=\hat{i}[/tex]
[tex]r(0)=\hat{j}[/tex]
we know [tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]\int dv=\int adt[/tex]
[tex]v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt[/tex]
[tex]v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c[/tex]
at [tex]t=0[/tex]
[tex]v(0)=0-1\cdot \hat{j}+0+c[/tex]
[tex]c=\hat{i}+\hat{j}[/tex]
[tex]v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}[/tex]
and [tex]\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)[/tex]
[tex]\int dr=\int vdt[/tex]
[tex]r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt[/tex]
[tex]r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2[/tex]
at [tex]t=0[/tex]
[tex]r(0)=\hat{j}[/tex]
[tex]r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}[/tex]
