A particle is moving along x-axis. At time `t=0,` Its x-coordinate is `x=-4 m.` Its velocity-time equation is `v=8-2t` where, v is in m//s and t in seconds.
(a) At how many times, particle is at a distance of `8m` from the origin?
(b) Find those times.

Explanation

To solve the problem step by step, we will analyze the motion of the particle based on the given information.

### Given:
- Initial position of the particle at \( t = 0 \): \( x_0 = -4 \, \text{m} \)
- Velocity-time equation: \( v = 8 - 2t \)

### Part (a): At how many times is the particle at a distance of 8 m from the origin?

1. **Determine the positions where the particle is 8 m from the origin:**
The particle can be at two positions:
- \( x = 8 \, \text{m} \) (to the right of the origin)
- \( x = -8 \, \text{m} \) (to the left of the origin)

2. **Calculate the total displacement required to reach these positions:**
- For \( x = 8 \, \text{m} \):
\[ \text{Displacement} = x_{\text{final}} - x_{\text{initial}} = 8 - (-4) = 12 \, \text{m} \]
- For \( x = -8 \, \text{m} \):
\[ \text{Displacement} = x_{\text{final}} - x_{\text{initial}} = -8 - (-4) = -4 \, \text{m} \]

3. **Determine the number of times the particle crosses these positions:**
The particle will cross \( x = 8 \, \text{m} \) twice (once while moving away from the origin and once while returning) and \( x = -8 \, \text{m} \) once. Thus, the particle is at a distance of 8 m from the origin a total of **three times**.

### Part (b): Find those times.

1. **Finding times for \( x = 8 \, \text{m} \):**
Use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = 12 \, \text{m} \)
- \( u = 8 \, \text{m/s} \) (initial velocity)
- \( a = -2 \, \text{m/s}^2 \) (acceleration)

Plugging in the values:
\[ 12 = 8t - \frac{1}{2} \cdot 2t^2 \]
Simplifying:
\[ 12 = 8t - t^2 \]
Rearranging:
\[ t^2 - 8t + 12 = 0 \]
Factoring:
\[ (t - 2)(t - 6) = 0 \]
Thus, \( t_1 = 2 \, \text{s} \) and \( t_2 = 6 \, \text{s} \).

2. **Finding time for \( x = -8 \, \text{m} \):**
Use the same equation of motion:
\[ s = -4 \, \text{m} \]
Plugging in the values:
\[ -4 = 8t - \frac{1}{2} \cdot 2t^2 \]
Simplifying:
\[ -4 = 8t - t^2 \]
Rearranging:
\[ t^2 - 8t - 4 = 0 \]
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 + 16}}{2} = \frac{8 \pm \sqrt{80}}{2} = \frac{8 \pm 4\sqrt{5}}{2} \]
Thus:
\[ t_3 = 4 \pm 2\sqrt{5} \approx 8.47 \, \text{s} \]

### Final Times:
- The particle is at a distance of 8 m from the origin at:
- \( t_1 = 2 \, \text{s} \)
- \( t_2 = 6 \, \text{s} \)
- \( t_3 \approx 8.47 \, \text{s} \)

### Summary:
(a) The particle is at a distance of 8 m from the origin **three times**.
(b) The times are approximately \( t = 2 \, \text{s}, 6 \, \text{s}, \) and \( 8.47 \, \text{s} \).