A particle moving with uniform acceleration has velocity of `6 ms^(-1)` at a distance 5 m from the initial position. After moving another 7 m the velocity becomes `8 ms^(-1)` The initial velocity and acceleration of the particle are 

Options

A. `2 ms^(-1) , 4m s ^(-2)`
B. `4 ms^(-1) , 2 m s ^(-2)`
C. `4 ms^(-1) , 4m s ^(-2)`
D. `6 ms^(-1) , 1m s ^(-2)`

Explanation

To solve the problem, we need to find the initial velocity (U) and the acceleration (A) of a particle that moves with uniform acceleration. We are given two conditions:

1. The particle has a velocity of 6 m/s at a distance of 5 m from the initial position.
2. After moving another 7 m (total distance of 12 m from the initial position), the velocity becomes 8 m/s.

### Step-by-Step Solution:

1. **Identify the known values:**
- Initial velocity (U) = ?
- First velocity (V1) = 6 m/s at distance (s1) = 5 m
- Second velocity (V2) = 8 m/s at distance (s2) = 12 m (5 m + 7 m)
- Distance traveled between the two velocities (s) = 7 m

2. **Use the kinematic equation to find acceleration (A):**
We can use the equation:
\[ V^2 = U^2 + 2As \]
For the second part of the motion (from 5 m to 12 m):
\[ V2^2 = V1^2 + 2A \cdot s \]
Substituting the known values:
\[ 8^2 = 6^2 + 2A \cdot 7 \]
This simplifies to:
\[ 64 = 36 + 14A \]
Rearranging gives:
\[ 64 - 36 = 14A \]
\[ 28 = 14A \]
\[ A = \frac{28}{14} = 2 \, \text{m/s}^2 \]

3. **Now, find the initial velocity (U):**
We will use the same kinematic equation for the first part of the motion (from 0 m to 5 m):
\[ V1^2 = U^2 + 2As_1 \]
Substituting the known values:
\[ 6^2 = U^2 + 2 \cdot 2 \cdot 5 \]
This simplifies to:
\[ 36 = U^2 + 20 \]
Rearranging gives:
\[ U^2 = 36 - 20 \]
\[ U^2 = 16 \]
Taking the square root:
\[ U = 4 \, \text{m/s} \]

### Final Answers:
- Initial velocity (U) = 4 m/s
- Acceleration (A) = 2 m/s²