Question

Use the given acceleration vectors $a\left(t\right)=v\left(t\right)=r\left(t\right)$and initial conditions in Exercises to find the position function $r\left(t\right).$

localid="1650040775983" $\mathrm{a}\left(\mathrm{t}\right)=-32\mathrm{j},\mathrm{v}\left(0\right)=5\mathrm{i}+5\mathrm{j},\mathrm{r}\left(0\right)=26\mathrm{j}$

Short answer

$a\left(t\right)=-32j,v\left(0\right)=5i+5j,r\left(0\right)=26j$

$\mathrm{r}\left(\mathrm{t}\right)=5\mathrm{ti}+1\left(-16{\mathrm{t}}^2+5\mathrm{t}+26\right)\mathrm{j}$

Step-by-step solution

Given Information

Consider, $\mathrm{a}\left(\mathrm{t}\right)=-32\mathrm{j},\mathrm{v}\left(0\right)=5\mathrm{i}+5\mathrm{j},\mathrm{r}\left(0\right)=26\mathrm{j}$

The objective is to find the position function r(t)

since

$$\begin{gathered} \mathrm{a}\left(\mathrm{t}\right)={\mathrm{v}}^2\left(\mathrm{t}\right)={\mathrm{r}}^2\left(\mathrm{t}\right), \\ \mathrm{We}\mathrm{have} \\ \mathrm{v}\left(\mathrm{t}\right)=\int \mathrm{a}\left(\mathrm{t}\right)\mathrm{dt}\mathrm{and}\mathrm{r}\left(\mathrm{t}\right)=\int \mathrm{v}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{now} \\ \mathrm{v}\left(\mathrm{t}\right)=\int \mathrm{a}\left(\mathrm{t}\right)\mathrm{dt} \\ =\int -32\mathrm{jdt} \\ \mathrm{v}\left(0\right)={\mathrm{c}}_1\mathrm{i}+{\mathrm{c}}_2\mathrm{j}............\left(1\right) \\ \mathrm{v}\left(0\right)=5\mathrm{i}+5\mathrm{j}............\left(2\right) \end{gathered}$$

Calculation

Comparing(1) and (2) equations,

$c1=5,c2=5$

so,

$$\begin{gathered} v\left(t\right)=5i+\left(-32t+5\right)j \\ r\left(t\right)=\int v\left(t\right)dt \\ =\int \left(5\mathrm{i}+\left(-32\mathrm{i}+5\right)\mathrm{j}\right)\mathrm{dt} \end{gathered}$$

Where $C$is a vector constant

$=\left(5\mathrm{i}+{\mathrm{c}}_3\right)\mathrm{i}+\left(-16{\mathrm{t}}^2+5\mathrm{t}+{\mathrm{c}}_4\right)\mathrm{j}$

Where $C_3$and $C_4$are scalars

$$\begin{gathered} r\left(0\right)=26j \\ thus, \\ r\left(t\right)=5ti+1\left(-16t^2+5t+26\right)j \end{gathered}$$