Chapter 11.2, Problem 61E

(a)

To determine

To find: The exact times at which the particle collide.

Answer to Problem 61E

The particle collide at $t=2$ .

Explanation of Solution

Given Data:

The path of the two particle for the time $t\ge 0$ are given by the position vectors,

  $\begin{array}{l}{r}_1\left(t\right)=\langle t-3,{\left(t-3\right)}^2\rangle \\ r_2\left(t\right)=\langle \frac{3t}{2}-4,\frac{3t}{2}-2\rangle \end{array}$

Calculation:

Consider that the particle collide when the components of the position vectors are equal then by equating the $x$ component then,

  $\begin{array}{l}t-3=\frac{3t}{2}-4\\ t-\frac{3t}{2}=-1\\ t=2\end{array}$

Thus, the particle collide at $t=2$ .

(b)

To determine

To find: The find the direction of the motion of each particle at the time of the collision.

Answer to Problem 61E

The direction of the unit vector is $\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$

Explanation of Solution

Consider that at time $t=2$ the velocity vector is,

  $\begin{array}{c}{v}_1\left(t\right)=\frac{d\left(r_1\left(t\right)\right)}{dt}\\ =\frac{d}{d}\langle t-3,{\left(t-3\right)}^2\rangle \\ =\langle \frac{d}{dt}\left(t-3\right),\frac{d}{dt}{\left(t-3\right)}^2\rangle \\ =\langle 1,2\left(t-3\right)\rangle \end{array}$

Consider that at $t=2$ , $v_1\left(t\right)=\langle 1,-2\rangle$ and the magnitude of the vector is $\langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\rangle$ .

For the second particle the velocity vector is,

  $\begin{array}{c}{v}_2\left(t\right)=\frac{d\left(r_2\left(t\right)\right)}{dt}\\ =\langle \frac{d}{dt}\left(\frac{3t}{2}-4\right),\frac{d}{dt}\left(\frac{3t}{2}-1\right)\rangle \\ =\langle \frac{3}{2},\frac{3}{2}\rangle \end{array}$

Consider that at $t=2$ , $v_2\left(2\right)=\left(\frac{3}{2},\frac{3}{2}\right)$ and the magnitude of the vector is $\langle \frac{1}{\sqrt{5}},\frac{-2}{\sqrt{5}}\rangle$ .

Then, the magnitude is $\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2}=\frac{3}{\sqrt{2}}$

Thus, the direction of the unit vector is $\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle$