Chapter 11.2, Problem 43E

(a)

To determine

To find: The position of the particle at time $t=3$ .

Answer to Problem 43E

The required position vector is $x\left(t\right)=t^3-t^2+2$ .

Explanation of Solution

Given Data:

The given equation for the velocity is $v\left(t\right)=\langle 3t^2-2t,1+2\cos \pi t\rangle ;\left(2,6\right)$ .

Calculation:

Consider the velocity vector is $v\left(t\right)=v_1\langle t\rangle +v_2\langle t\rangle$ and it is equal to the first derivative of the position vector $r\left(t\right)=x\left(t\right),y\left(t\right)$ that givens $\frac{dx\left(t\right)}{dt}=v_1\left(t\right)$ and $\frac{dy\left(t\right)}{dt}=v_2\left(t\right)$ .

Then, from the above explanation.

  $\begin{array}{l}\frac{dx\left(t\right)}{dt}=3t^2-2t\\ x\left(t\right)={\displaystyle \int \left(3t^2-2t\right)dt}\\ x\left(t\right)=t^3-t^2+C\end{array}$

Substitute the position of the particle as,

  $\begin{array}{l}2=0-2+C\\ C=2\end{array}$

Thus, the position particle is,

  $x\left(t\right)=t^3-t^2+2$

(b)

To determine

To find: The distance that the particle travels from $t=0$ to $t=3$ .

Answer to Problem 43E

The distance that the particle travels is shown in Figure 1

Explanation of Solution

Consider the equation,

  $\begin{array}{l}\frac{dy\left(t\right)}{dt}=1+\cos \pi t\\ y\left(t\right)={\displaystyle \int \left(1+\cos \pi t\right)dt}\\ y\left(t\right)=1+\frac{1}{\pi }\sin \pi t+C\end{array}$

Consider the position of the particle at $y=6$ and $t=0$ as,

  $\begin{array}{l}6=0+0+C\\ C=6\end{array}$

The required position of the particle is,

  $y\left(t\right)=t+\frac{1}{\pi }\sin \pi t+6$

Thus, the required position vector is,

  $r\left(t\right)=\langle t^3-t^2,t+\frac{1}{\pi }\sin \pi t+6\rangle$

The graph to show the distance travelled by the particle is shown in Figure 1

  

Figure 1