Chapter 11, Problem 52RE

(a)

To determine

To find: The velocity vector for each particle at time $t=3$ .

Answer to Problem 52RE

The velocity vectors for each particle at time $t=3$ are $\langle 1,2\rangle$ , and $\langle \frac{3}{2},\frac{3}{2}\rangle$ .

Explanation of Solution

Given:

The position of the particle $A$ at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t-2\\ y\left(t\right)={\left(t-2\right)}^2\end{array}$

The position of the particle $B$ at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=\frac{3}{2}t-4\\ y\left(t\right)=\frac{3}{2}t-2\end{array}$

Calculation:

Find the velocity vector of the particle $A$ as follows.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle \left(t-2\right),{\left(t-2\right)}^2\rangle \\ =\langle 1,2\left(t-2\right)\rangle \end{array}$

Find the speed of the particle when $t=3$ .

  $\begin{array}{c}v{\left(t\right)}_{t=3}=\langle 1,2\left(3-2\right)\rangle \\ =\langle 1,2\rangle \end{array}$

Find the velocity vector of the particle $B$ as follows.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle \frac{3}{2}t-4,\frac{3}{2}t-2\rangle \\ =\langle \frac{3}{2},\frac{3}{2}\rangle \end{array}$

Therefore, the velocity vectors for each particle at time $t=3$ are $\langle 1,2\rangle$ , and $\langle \frac{3}{2},\frac{3}{2}\rangle$ .

(b)

To determine

To find: The distance traveled by particle $A$ from $t=0$ to $t=3$ .

Answer to Problem 52RE

The distance traveled by particle $A$ from $t=0$ to $t=3$ is $6.126$ .

Explanation of Solution

Given:

The position of the particle $A$ at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t-2\\ y\left(t\right)={\left(t-2\right)}^2\end{array}$

Calculation:

Find the total distance traveled by the particle from $t=0$ to $t=3$ as follows.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left|v\left(t\right)\right|}={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{1+{\left(2t-4\right)}^2}dt}\\ =6.126\end{array}$

Therefore, the distance traveled by particle $A$ from $t=0$ to $t=3$ is $6.126$ .

(c)

To determine

To find: The exact time when the particles collide.

Answer to Problem 52RE

The particles collide at $t=4$ .

Explanation of Solution

Given:

The position of the particle $A$ at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=t-2\\ y\left(t\right)={\left(t-2\right)}^2\end{array}$

The position of the particle $B$ at time $t\ge 0$ is given by,

  $\begin{array}{c}x\left(t\right)=\frac{3}{2}t-4\\ y\left(t\right)=\frac{3}{2}t-2\end{array}$

Calculation:

The particles collide when components of position vectors are equal. So, equate the $x$ component.

  $\begin{array}{c}t-2=\frac{3}{2}t-4\\ t-\frac{3}{2}t=-2\\ t=4\end{array}$

Therefore, the particles collide at $t=4$ .