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 find the acceleration of the particle whose position \( x \) at time \( t \) is given by the equation \( x^2 = 2 + t \), we will follow these steps:

### Step 1: Express \( x \) in terms of \( t \)
We start with the equation:
\[ x^2 = 2 + t \]
Taking the square root of both sides, we get:
\[ x = \sqrt{2 + t} \]

### Step 2: Differentiate \( x \) with respect to \( t \) to find velocity \( v \)
To find the velocity \( v \), we differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(\sqrt{2 + t}) \]
Using the chain rule, we have:
\[ \frac{dx}{dt} = \frac{1}{2\sqrt{2 + t}} \cdot \frac{d}{dt}(2 + t) = \frac{1}{2\sqrt{2 + t}} \cdot 1 = \frac{1}{2\sqrt{2 + t}} \]
Thus, the velocity \( v \) is:
\[ v = \frac{1}{2\sqrt{2 + t}} \]

### Step 3: Differentiate \( v \) with respect to \( t \) to find acceleration \( a \)
Next, we differentiate \( v \) with respect to \( t \) to find the acceleration \( a \):
\[ \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{2\sqrt{2 + t}}\right) \]
Using the quotient rule or chain rule, we have:
\[ \frac{dv}{dt} = -\frac{1}{2} \cdot \frac{1}{(2 + t)^{3/2}} \cdot \frac{d}{dt}(2 + t) = -\frac{1}{2} \cdot \frac{1}{(2 + t)^{3/2}} \cdot 1 = -\frac{1}{2(2 + t)^{3/2}} \]
Thus, the acceleration \( a \) is:
\[ a = -\frac{1}{2(2 + t)^{3/2}} \]

### Step 4: Express acceleration in terms of \( x \)
Since \( x = \sqrt{2 + t} \), we can express \( 2 + t \) as \( x^2 \):
\[ a = -\frac{1}{2(x^2)^{3/2}} = -\frac{1}{2x^3} \]

### Final Answer
Thus, the acceleration of the particle is given by:
\[ a = -\frac{1}{2x^3} \]

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