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="1650737775645" $\mathrm{a}\left(\mathrm{t}\right)=2\mathrm{t},3{\mathrm{t}}^2,\mathrm{v}\left(0\right)=1,-2,\mathrm{r}\left(0\right)=2,-3$

Short answer

$r\left(t\right)=\left(\frac{r^2}{3}+t+2\frac{t^2}{4}-2i-3\right)$

$r\left(t\right)=\left(\frac{t^3}{3}+t+2,\frac{t^4}{4}-2t-3\right)$

Step-by-step solution

Given Information

Consider $\mathrm{a}\left(\mathrm{t}\right)=\left(2\mathrm{t},3{\mathrm{t}}^2\right),\mathrm{v}\left(0\right)=\left(1,-2\right),\mathrm{r}\left(0\right)=\left(2,-3\right)$

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

Since $a\left(t\right)=v^t\left(t\right)=r^n\left(t\right)$, we have $\mathrm{v}\left(\mathrm{t}\right)=\int \mathrm{a}\left(\mathrm{t}\right)\mathrm{dt}\mathrm{and}\mathrm{r}\left(\mathrm{t}\right)=\int \mathrm{v}\left(\mathrm{t}\right)\mathrm{dt}$

Now,

But the given initial position $\mathrm{v}\left(0\right)=\left(1,-2\right)=\mathrm{i}-2\mathrm{j}..............\left(2\right)$

Comparing (1) and (2) equations,$C_1=1,{\mathrm{C}}_2=-2$

Calculation

so

$\mathrm{v}\left(\mathrm{t}\right)=\left({\mathrm{t}}^2+1\right)\mathrm{i}+\left({\mathrm{t}}^3-2\right)\mathrm{j}$

Now,

$$\begin{gathered} r\left(t\right)=\int v(t0dt \\ =\int ((t^2+1)i+(t^2-2)j)dt \end{gathered}$$

$=\left(\frac{t^3}{3}+t\right)i+\left(\frac{t^4}{4}-2t\right)j+c$

Where c is a vector constant

$=\left(\frac{t^3}{3}+t+c_3\right)i+\left(\frac{t^4}{4}-2t+c_4\right)j$

where $c_3$and $c_4$are scalars.

$$\begin{gathered} r\left(0\right)=\left(\frac{0^3}{3}+0+c_3\right)i+\left(\frac{0^4}{4}-2\left(0\right)+c_4\right)j \\ r\left(0\right)=c_{3}i+c_{4}j..........\left(3\right) \end{gathered}$$

But as per problem, $r\left(0\right)=\left(2,-3\right)=2i-3j...........\left(4\right)$

Comparing $\left(3\right)\&\left(4\right)$equations, $c_3=2,c_4=-3$

Thus

$$\begin{gathered} r\left(t\right)=\left(\frac{t^3}{3}+t+2\right)i+\left(\frac{t^4}{4}-2t-3\right)j \\ =\left(\frac{t^3}{3}+t+2,\frac{t^4}{4}-2t-3\right) \end{gathered}$$