Chapter 11, Problem 46RE

To determine

To calculate: The position vector.

Answer to Problem 46RE

The required position vector is $\overrightarrow{r}\left(t\right)=\langle -t^2+6t-2,-t^2+2t+2\rangle$ .

Explanation of Solution

Given information:

The vectors: $\overrightarrow{a}\left(t\right)=\langle -2,-2\rangle$ , $\overrightarrow{v}\left(1\right)=\langle 4,0\rangle$ and $\overrightarrow{r}\left(1\right)=\langle 3,3\rangle$

Calculation:

The given vectors are $\overrightarrow{a}\left(t\right)=\langle -2,-2\rangle$ , $\overrightarrow{v}\left(1\right)=\langle 4,0\rangle$ and $\overrightarrow{r}\left(1\right)=\langle 3,3\rangle$ .

It is known that the first derivative of the velocity vector $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ is the acceleration vector $\overrightarrow{a}\left(t\right)=\langle {\overrightarrow{a}}_1\left(t\right),{\overrightarrow{a}}_2\left(t\right)\rangle$ .

So, $\frac{d{\overrightarrow{v}}_1\left(t\right)}{dt}={\overrightarrow{a}}_1\left(t\right)$ , $\frac{d{\overrightarrow{v}}_2\left(t\right)}{dt}={\overrightarrow{a}}_2\left(t\right)$

Now, use the above theory and write:

  $\begin{array}{c}\frac{d{\overrightarrow{v}}_1\left(t\right)}{dt}=-2\\ {\overrightarrow{v}}_1\left(t\right)={\displaystyle \int \left(-2\right)dt}\\ {\overrightarrow{v}}_1\left(t\right)=-2t+C\end{array}$

The velocity component of the particle is ${\overrightarrow{v}}_1=4$ at $t=1$ .

Substitute 4 for ${\overrightarrow{v}}_1\left(t\right)$ and 1 for t in the above expression.

  $\begin{array}{c}4=-2\left(1\right)+C\\ C-2=4\\ C=4+2\\ C=6\end{array}$

Substitute 6 for C in the expression ${\overrightarrow{v}}_1\left(t\right)=-2t+C$ .

  ${\overrightarrow{v}}_1\left(t\right)=-2t+6$

Thus, the velocity component of the particle is ${\overrightarrow{v}}_1\left(t\right)=-2t+C$ .

Also, $\begin{array}{c}\frac{d{\overrightarrow{v}}_2\left(t\right)}{dt}=-2\\ {\overrightarrow{v}}_2\left(t\right)={\displaystyle \int \left(-2\right)dt}\\ {\overrightarrow{v}}_2\left(t\right)=-2t+C\end{array}$

The velocity component of the particle is ${\overrightarrow{v}}_2=0$ at $t=0$ .

Substitute 0 for ${\overrightarrow{v}}_2\left(t\right)$ and 1 for t in the above expression.

  $\begin{array}{c}0=-2\left(1\right)+C\\ C-2=0\\ C=2\end{array}$

Substitute 2 for C in the expression ${\overrightarrow{v}}_1\left(t\right)=-2t+C$ .

  ${\overrightarrow{v}}_2\left(t\right)=-2t+2$

Thus, the velocity vector is $\overrightarrow{v}\left(t\right)=\left(-2t+6,-2t+2\right)$ .

It is known that the first derivative of the position vector $\overrightarrow{r}\left(t\right)=\langle \overrightarrow{x}\left(t\right),\overrightarrow{y}\left(t\right)\rangle$ is the velocity vector $\overrightarrow{v}\left(t\right)=\langle {\overrightarrow{v}}_1\left(t\right),{\overrightarrow{v}}_2\left(t\right)\rangle$ .

So, $\frac{d\overrightarrow{x}\left(t\right)}{dt}={\overrightarrow{v}}_1\left(t\right)$ , $\frac{d\overrightarrow{y}\left(t\right)}{dt}={\overrightarrow{v}}_2\left(t\right)$

Now, use the above theory and write:

  $\begin{array}{c}\frac{d\overrightarrow{x}\left(t\right)}{dt}=-2t+6\\ \overrightarrow{x}\left(t\right)={\displaystyle \int \left(-2t+6\right)dt}\\ \overrightarrow{x}\left(t\right)=-t^2+6t+C\end{array}$

The position of the particle is at $\overrightarrow{x}=3$ at $t=1$ .

Substitute 3 for $\overrightarrow{x}\left(t\right)$ and 1 for t in the above expression.

  $\begin{array}{c}3=-1+6+C\\ C+5=3\\ C=3-5\\ C=-2\end{array}$

Substitute $-2$ for C in the expression $\overrightarrow{x}\left(t\right)=-t^2+6t+C$ .

  $\overrightarrow{x}\left(t\right)=-t^2+6t-2$

Also, $\begin{array}{c}\frac{d\overrightarrow{y}\left(t\right)}{dt}=-2t+2\\ \overrightarrow{y}\left(t\right)={\displaystyle \int \left(-2t+2\right)dt}\\ \overrightarrow{y}\left(t\right)=-t^2+2t+C\end{array}$

The position of the particle is at $\overrightarrow{y}=3$ at $t=1$ .

Substitute 3 for $\overrightarrow{y}\left(t\right)$ in the above expression.

  $\begin{array}{c}3=-1+2+C\\ C+1=3\\ C=3-1\\ C=2\end{array}$

Substitute 2 for C in the expression $\overrightarrow{y}\left(t\right)=-t^2+2t+C$ .

  $\overrightarrow{y}\left(t\right)=-t^2+2t+2$

Thus, the velocity component of the particle is $\overrightarrow{y}\left(t\right)=-t^2+2t+2$ .

Hence, the required position vector is $\overrightarrow{r}\left(t\right)=\langle -t^2+6t-2,-t^2+2t+2\rangle$ .