A particle moves in a straight line and its position x and time t are related as follows: `x=(2+t)^(1//2)`
Acceleration of the particle is given by

Options

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

Explanation

To find the acceleration of the particle given the position function \( x = (2 + t)^{1/2} \), we will follow these steps:

### Step 1: Differentiate the position function to find velocity
The velocity \( v \) is defined as the rate of change of position with respect to time, which is given by the derivative \( \frac{dx}{dt} \).

Given:
\[ x = (2 + t)^{1/2} \]

We differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{1}{2}(2 + t)^{-1/2} \cdot \frac{d(2 + t)}{dt} \]
Since \( \frac{d(2 + t)}{dt} = 1 \), we have:
\[ v = \frac{1}{2}(2 + t)^{-1/2} \]

### Step 2: Differentiate the velocity function to find acceleration
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which is given by the derivative \( \frac{dv}{dt} \).

Now we differentiate \( v \):
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{1}{2}(2 + t)^{-1/2} \right) \]

Using the chain rule:
\[ a = \frac{1}{2} \cdot (-\frac{1}{2})(2 + t)^{-3/2} \cdot \frac{d(2 + t)}{dt} \]
Again, since \( \frac{d(2 + t)}{dt} = 1 \), we have:
\[ a = -\frac{1}{4}(2 + t)^{-3/2} \]

### Step 3: Substitute \( x \) in terms of \( a \)
From the original position equation, we know:
\[ x = (2 + t)^{1/2} \]
We can express \( (2 + t) \) in terms of \( x \):
\[ (2 + t) = x^2 \]
Thus, we can substitute this into our acceleration equation:
\[ a = -\frac{1}{4}(x^2)^{-3/2} \]
This simplifies to:
\[ a = -\frac{1}{4} \cdot \frac{1}{x^3} \]

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

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