If a particle is moving such that the velocity acquired is proportional to the square root of the distance covered, then its acceleration is

Options

A. a constant
B. `prop s^(2)`
C. `prop(1)/(s^(2))`
D. `prop s`

Explanation

To solve the problem, we need to find the acceleration of a particle whose velocity is proportional to the square root of the distance covered. Let's break down the solution step by step.

### Step 1: Define the relationship between velocity and distance
We are given that the velocity \( v \) of the particle is proportional to the square root of the distance \( s \) covered. This can be expressed mathematically as:

\[ v = k \sqrt{s} \]

where \( k \) is a proportionality constant.

**Hint for Step 1:** Remember that "proportional to" means you can express the relationship using a constant multiplier.

### Step 2: Differentiate velocity with respect to time
To find the acceleration \( a \), we need to differentiate the velocity \( v \) with respect to time \( t \):

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

Using the chain rule, we can express this as:

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

### Step 3: Differentiate the velocity function
Now, we need to differentiate \( v = k \sqrt{s} \) with respect to \( s \):

\[ \frac{dv}{ds} = k \cdot \frac{1}{2} s^{-1/2} = \frac{k}{2\sqrt{s}} \]

### Step 4: Substitute \( \frac{ds}{dt} \) (which is the velocity)
From our earlier definition, we know that \( \frac{ds}{dt} = v = k \sqrt{s} \). We can substitute this into our expression for acceleration:

\[ a = \frac{dv}{ds} \cdot \frac{ds}{dt} = \left(\frac{k}{2\sqrt{s}}\right) \cdot (k\sqrt{s}) \]

### Step 5: Simplify the expression for acceleration
Now, we simplify the expression:

\[ a = \frac{k}{2\sqrt{s}} \cdot k\sqrt{s} = \frac{k^2}{2} \]

### Conclusion
Thus, the acceleration \( a \) is a constant value given by:

\[ a = \frac{k^2}{2} \]

### Final Answer
The acceleration of the particle is constant.

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