A particle confined to motion along an `x` axis moves with constant acceleration from `x=2.0m ` to `x=8.0` m during a 2.5 s time interval. The velocity of the particle at `x=8.0m` is `2.8` m/s. What is the constant acceleration during this time interval ?

Answer

To find the constant acceleration of the particle moving from \(x = 2.0 \, \text{m}\) to \(x = 8.0 \, \text{m}\) over a time interval of \(2.5 \, \text{s}\) with a final velocity of \(2.8 \, \text{m/s}\), we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify Known Values**:
- Initial position, \(x_0 = 2.0 \, \text{m}\)
- Final position, \(x = 8.0 \, \text{m}\)
- Time interval, \(t = 2.5 \, \text{s}\)
- Final velocity, \(v = 2.8 \, \text{m/s}\)

2. **Calculate Displacement**:
\[ \Delta x = x - x_0 = 8.0 \, \text{m} - 2.0 \, \text{m} = 6.0 \, \text{m} \]

3. **Use the First Equation of Motion**:
The first equation of motion relates final velocity, initial velocity, acceleration, and time:
\[ v = u + at \]
where \(u\) is the initial velocity and \(a\) is the acceleration. Rearranging gives:
\[ a = \frac{v - u}{t} \]

4. **Use the Second Equation of Motion**:
The second equation of motion relates displacement, initial velocity, acceleration, and time:
\[ \Delta x = ut + \frac{1}{2} a t^2 \]
Substituting for \(\Delta x\):
\[ 6.0 = ut + \frac{1}{2} a t^2 \]
This can be rewritten as:
\[ 6.0 = u(2.5) + \frac{1}{2} a (2.5)^2 \]
Simplifying gives:
\[ 6.0 = 2.5u + 3.125a \quad \text{(Equation 1)} \]

5. **Substituting for \(u\)**:
From the first equation of motion, we can express \(u\):
\[ u = v - at \]
Substituting \(v = 2.8 \, \text{m/s}\):
\[ u = 2.8 - 2.5a \quad \text{(Equation 2)} \]

6. **Substituting Equation 2 into Equation 1**:
Replace \(u\) in Equation 1:
\[ 6.0 = 2.5(2.8 - 2.5a) + 3.125a \]
Expanding gives:
\[ 6.0 = 7.0 - 6.25a + 3.125a \]
Combining like terms:
\[ 6.0 = 7.0 - 3.125a \]
Rearranging gives:
\[ 3.125a = 7.0 - 6.0 \]
\[ 3.125a = 1.0 \]
\[ a = \frac{1.0}{3.125} \approx 0.32 \, \text{m/s}^2 \]

### Final Answer:
The constant acceleration of the particle is approximately \(0.32 \, \text{m/s}^2\).