A particle moves in a straight line and its position x at time t is given by `x^(2)=2+t`. Its acceleration is given by :-

Options

A. `(-2)/(x^(3))`
B. `-(1)/(4x^(3))`
C. `-(1)/(4x^(2))`
D. `(1)/(x^(2))`

Answer

To solve the problem, we need to find the acceleration of a particle whose position \( x \) at time \( t \) is given by the equation \( x^2 = 2 + t \).

### Step-by-Step Solution:

1. **Express \( x \) in terms of \( t \)**:
\[ x^2 = 2 + t \implies x = \sqrt{2 + t} \]

**Hint**: Remember that \( x \) is the square root of the expression on the right side.

2. **Find the velocity \( v \)**:
The velocity \( v \) is defined as the derivative of position with respect to time:
\[ v = \frac{dx}{dt} = \frac{d}{dt}(\sqrt{2 + t}) \]
Using the chain rule:
\[ v = \frac{1}{2\sqrt{2 + t}} \cdot \frac{d}{dt}(2 + t) = \frac{1}{2\sqrt{2 + t}} \cdot 1 = \frac{1}{2\sqrt{2 + t}} \]

**Hint**: Use the chain rule for differentiation when dealing with composite functions.

3. **Find the acceleration \( a \)**:
Acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{2\sqrt{2 + t}}\right) \]
Again, using the chain rule:
\[ a = \frac{1}{2} \cdot \frac{d}{dt}((2 + t)^{-1/2}) = \frac{1}{2} \cdot \left(-\frac{1}{2}(2 + t)^{-3/2}\right) \cdot \frac{d}{dt}(2 + t) \]
Since \(\frac{d}{dt}(2 + t) = 1\):
\[ a = -\frac{1}{4}(2 + t)^{-3/2} \]

**Hint**: When differentiating a power function, remember to multiply by the derivative of the inside function.

4. **Substituting back for \( x \)**:
Since \( x = \sqrt{2 + t} \), we can express \( 2 + t \) as \( x^2 \):
\[ a = -\frac{1}{4}(x^2)^{-3/2} = -\frac{1}{4} \cdot \frac{1}{(x^2)^{3/2}} = -\frac{1}{4} \cdot \frac{1}{x^3} \]

**Hint**: Be careful with the exponents when substituting back for \( x \).

5. **Final expression for acceleration**:
\[ a = -\frac{1}{4}x^3 \]

### Conclusion:
The acceleration of the particle is given by:
\[ a = -\frac{1}{4}x^3 \]

### Answer:
The correct option is option number 2.