Chapter 10.2, Problem 61E

To determine

To find: The time when particles collide.

Explanation of Solution

Given information:

The position vector of two particles,

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

Calculation:

If two particles intersect then the position of both particles at same point.

  $\begin{array}{c}{r}_1=r_2\\ 〈t-3,{\left(t-3\right)}^2〉=〈\frac{3t}{2}-4,\frac{3t}{2}-2〉\\ t-3=\frac{3t}{2}-4\\ \frac{3t}{2}-t=4-3\\ t=2\end{array}$

Hence, the particles collides at $t=2$ .

To determine

To find: The direction of the motion of each particle at $t=2$ .

Answer to Problem 61E

  $117°,45°$

Explanation of Solution

Given information:

The position vector of two particles,

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

Calculation:

The motion of the particles,

  $\begin{array}{ccc}{r}_1\left(t\right)=〈t-3,{\left(t-3\right)}^2〉& \text{and}& r_2\left(t\right)=〈\frac{3t}{2}-4,\frac{3t}{2}-2〉\\ r{\text{'}}_1\left(t\right)=〈1,2\left(t-3\right)〉& \text{and}& r{\text{'}}_2\left(t\right)=〈\frac{3}{2},\frac{3}{2}〉\\ r{\text{'}}_1\left(2\right)=〈1,-2〉& \text{and}& r{\text{'}}_2\left(2\right)=〈\frac{3}{2},\frac{3}{2}〉\end{array}$

Now calculate the direction of the motion.

The direction of first particles is

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-2}{1}\right)\\ =117°\end{array}$

The direction of second particles is

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{3/2}{3/2}\right)\\ =45°\end{array}$

Hence, the direction of first particle is $117°$ and the direction of second particles is $45°$ .