The velocity of a particle moving in the positive direction of X-axis varies as `v=5sqrt(x)`. Assuming that at t = 0, particle was at x = 0. What is the acceleration of the particle ?

Options

A. `12.5 ms^(-2)`
B. `7.5 ms^(-2)`
C. `5 ms^(-2)`
D. `2.5 ms^(-2)`

Explanation

To solve the problem, we need to find the acceleration of a particle whose velocity varies with position according to the equation \( v = 5\sqrt{x} \).

### Step 1: Understanding the relationship between velocity and position
The given equation for velocity is:
\[ v = 5\sqrt{x} \]
We know that acceleration \( a \) can be expressed in terms of velocity \( v \) and position \( x \) using the chain rule:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]
This means we need to find \( \frac{dv}{dx} \).

### Step 2: Differentiate the velocity with respect to position
To find \( \frac{dv}{dx} \), we differentiate \( v = 5\sqrt{x} \):
\[ \frac{dv}{dx} = \frac{d}{dx}(5\sqrt{x}) = 5 \cdot \frac{1}{2\sqrt{x}} = \frac{5}{2\sqrt{x}} \]

### Step 3: Substitute \( v \) into the acceleration formula
Now we can substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula:
\[ a = \frac{dv}{dx} \cdot v = \left(\frac{5}{2\sqrt{x}}\right) \cdot (5\sqrt{x}) = \frac{25\sqrt{x}}{2\sqrt{x}} = \frac{25}{2} \]

### Step 4: Conclusion
The acceleration of the particle is:
\[ a = 12.5 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle is \( 12.5 \, \text{m/s}^2 \).

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