The acceleration of a particle moving in straight line is defined by the relation `a= -4x(-1+(1)/(4)x^(2))`, where a is acceleration in `m//s^(2)` and `x` is position in meter. If velocity `v = 17 m//s` when `x = 0`, then the velocity of the particle when `x = 4` meter is :

Options

A. `12 m//s`
B. `15 m//s`
C. `20 m//s`
D. `25 m//s`

Explanation

To solve the problem, we need to find the velocity of a particle when its position \( x = 4 \) meters, given the acceleration function and the initial conditions.

### Step-by-Step Solution

1. **Understand the given acceleration equation**:
The acceleration \( a \) is given by:
\[ a = -4x \left(-1 + \frac{1}{4} x^2\right) \]
This can be rewritten as:
\[ a = -4x + x^3 \]

2. **Relate acceleration to velocity**:
We know that acceleration \( a \) can also be expressed as:
\[ a = v \frac{dv}{dx} \]
where \( v \) is the velocity of the particle. Therefore, we can set up the equation:
\[ v \frac{dv}{dx} = -4x + x^3 \]

3. **Separate variables and integrate**:
Rearranging gives:
\[ v \, dv = (-4x + x^3) \, dx \]
Now, we integrate both sides. The left side integrates to:
\[ \int v \, dv = \frac{1}{2} v^2 \]
The right side integrates as follows:
\[ \int (-4x + x^3) \, dx = -2x^2 + \frac{1}{4} x^4 + C \]

4. **Set up the equation after integration**:
After integration, we have:
\[ \frac{1}{2} v^2 = -2x^2 + \frac{1}{4} x^4 + C \]

5. **Use initial conditions to find \( C \)**:
We know that when \( x = 0 \), \( v = 17 \, \text{m/s} \):
\[ \frac{1}{2} (17)^2 = -2(0)^2 + \frac{1}{4}(0)^4 + C \]
This simplifies to:
\[ \frac{1}{2} \times 289 = C \implies C = 144.5 \]

6. **Substitute \( C \) back into the equation**:
Now our equation becomes:
\[ \frac{1}{2} v^2 = -2x^2 + \frac{1}{4} x^4 + 144.5 \]

7. **Find the velocity when \( x = 4 \)**:
Substitute \( x = 4 \):
\[ \frac{1}{2} v^2 = -2(4)^2 + \frac{1}{4}(4)^4 + 144.5 \]
Calculate the right side:
\[ = -2(16) + \frac{1}{4}(256) + 144.5 = -32 + 64 + 144.5 = 176.5 \]
So,
\[ \frac{1}{2} v^2 = 176.5 \]
Multiplying both sides by 2 gives:
\[ v^2 = 353 \]
Therefore, taking the square root:
\[ v = \sqrt{353} \approx 18.79 \, \text{m/s} \]

### Final Answer
The velocity of the particle when \( x = 4 \) meters is approximately \( 18.79 \, \text{m/s} \).