Question

Use the given acceleration vectors $a\left(t\right)=v\text{'}\left(t\right)=r"\left(t\right)$and initial conditions in Exercises to find the position function $r\left(t\right)$.

Short answer

Step-by-step solution

Given Information

Consider,

The objective is to find the position function $r\left(t\right)$.

Since$a\left(t\right)=v\text{'}\left(t\right)=r"\left(t\right)$, we have$v\left(t\right)=\int a\left(t\right)dt$and$r\left(t\right)=\int v\left(t\right)dt$

Now

Where c is a constant

$v\left(t\right)=\left(\frac{-\cos 3t}{3}+c_1\right)i+\left(\frac{t^3}{3}+c_2\right)j+\left(\frac{\sin 3t}{3}+c_3\right)k$

Where $c_1,c_2$and $c_3$are scalars

$v\left(0\right)=\left(\frac{-1}{3}+c_1\right)i+c_{2}j+c_{3}k..............\left(1\right)$

Expression

But as per problem,$v\left(0\right)=\left(0,0,0\right)=0i+0j+0k............\left(2\right)$

Comparing (1) and (2) equations

$$\begin{gathered} \frac{-1}{3}+c_1=0,c_2=0,c_3=0 \\ {c}_1=\frac{1}{3} \\ sov\left(t\right)=\left(\frac{-\cos 3t+1}{3}\right)i+\frac{t^2}{3}j+\frac{\sin 3t}{3}k \end{gathered}$$

Now

$$\begin{gathered} r\left(t\right)=\int v\left(t\right)dt \\ =\int \left(\frac{-\cos 3t+1}{3}\right)i+\frac{t^2}{3}j+\frac{\sin 3t}{3}kdt \\ r\left(0\right)=c_{4}i+c_{5}j+\left(\frac{-\cos 0}{9}+c_6\right)k \\ r\left(0\right)=c_{4}i+c_{5}j+\left(\frac{-1}{9}+c_6\right)k............\left(3\right) \\ r\left(0\right)=\left(2,-5,3\right)=2i-5j+3k.............\left(4\right) \end{gathered}$$

Comparing (3) and (4) equations,

$c_4=2,c_5=-5\frac{-1}{9}+c_6=3$

$$\begin{gathered} C_6=3+\frac{1}{9} \\ {C}_6=\frac{28}{9} \end{gathered}$$

$r\left(t\right)=\left(\frac{-\sin 3t}{9}+\frac{1}{3}+2\frac{t^3}{12}5,\frac{-\cos 3t}{9}+\frac{28}{9}\right)$