The velocity `upsilon` of a particle as a function of its position (x) is expressed as `upsilon = sqrt(c_(1)-c_(2)x)`, where `c_(1)` and `c_(2)` are positive constants. The acceleration of the particle is

Options

A. `c_(2)`
B. `-(c_(2))/(2)`
C. `c_(1)-c_(2)`
D. `(c_(1)+c_(2))/(2)`

Explanation

To find the acceleration of the particle given the velocity as a function of position, we start with the equation:

\[ v = \sqrt{c_1 - c_2 x} \]

### Step 1: Square the velocity equation
To eliminate the square root, we square both sides:

\[ v^2 = c_1 - c_2 x \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ c_2 x = c_1 - v^2 \]

### Step 3: Differentiate with respect to time
To find acceleration, we need to express it in terms of velocity and position. We know that acceleration \( a \) can be expressed as:

\[ a = \frac{dv}{dt} \]

Using the chain rule, we can write:

\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]

### Step 4: Differentiate \( v \) with respect to \( x \)
Now we need to find \( \frac{dv}{dx} \). From the equation \( v^2 = c_1 - c_2 x \), we can differentiate both sides with respect to \( x \):

\[ 2v \frac{dv}{dx} = -c_2 \]

### Step 5: Solve for \( \frac{dv}{dx} \)
Rearranging gives us:

\[ \frac{dv}{dx} = -\frac{c_2}{2v} \]

### Step 6: Substitute \( \frac{dv}{dx} \) into the acceleration equation
Now substituting \( \frac{dv}{dx} \) back into the equation for acceleration:

\[ a = \left(-\frac{c_2}{2v}\right) v = -\frac{c_2}{2} \]

### Conclusion
Thus, the acceleration of the particle is:

\[ a = -\frac{c_2}{2} \]