The velocity v and displacement r of a body are related as `v^(2) =kr`, where k is a constant. The acceleration of the body is

Options

A. `sqrt(kr)`
B. `kr^(3//2)`
C. k/2
D. `sqrt(k) xx r`

Explanation

To find the acceleration of a body whose velocity \( v \) and displacement \( r \) are related by the equation \( v^2 = kr \), where \( k \) is a constant, we can follow these steps:

### Step 1: Understand the relationship
We start with the equation:
\[ v^2 = kr \]
This implies that the velocity \( v \) can be expressed in terms of displacement \( r \).

### Step 2: Express velocity in terms of displacement
From the equation \( v^2 = kr \), we can take the square root to find \( v \):
\[ v = \sqrt{kr} \]
Note that we can consider both the positive and negative roots, but for the purpose of finding acceleration, we will use the positive root.

### Step 3: Use the relationship between acceleration, velocity, and displacement
The acceleration \( a \) can be expressed in terms of velocity and displacement using the formula:
\[ a = v \frac{dv}{dr} \]
This means we need to find \( \frac{dv}{dr} \).

### Step 4: Differentiate velocity with respect to displacement
We have \( v = \sqrt{kr} \). Now, we differentiate \( v \) with respect to \( r \):
\[ \frac{dv}{dr} = \frac{d}{dr}(\sqrt{kr}) = \frac{1}{2\sqrt{kr}} \cdot k = \frac{k}{2\sqrt{kr}} \]

### Step 5: Substitute \( v \) and \( \frac{dv}{dr} \) into the acceleration formula
Now we substitute \( v \) and \( \frac{dv}{dr} \) back into the acceleration formula:
\[ a = v \frac{dv}{dr} = \sqrt{kr} \cdot \frac{k}{2\sqrt{kr}} \]

### Step 6: Simplify the expression
Simplifying the expression, we have:
\[ a = \frac{k\sqrt{kr}}{2\sqrt{kr}} = \frac{k}{2} \]

### Final Result
Thus, the acceleration \( a \) of the body is:
\[ a = \frac{k}{2} \]