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 find the acceleration of the particle given that its velocity varies as \( v = 5\sqrt{x} \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and displacement
Acceleration (\( a \)) can be expressed as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
However, since velocity is also a function of displacement (\( x \)), we can use the chain rule to express acceleration in terms of displacement:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} \) is the velocity (\( v \)) itself.

### Step 2: Substitute the expression for velocity
From the problem, we know:
\[ v = 5\sqrt{x} \]
Thus, we can rewrite the acceleration as:
\[ a = v \cdot \frac{dv}{dx} \]

### Step 3: Differentiate the velocity with respect to displacement
Now, we need to find \( \frac{dv}{dx} \):
\[ v = 5\sqrt{x} = 5x^{1/2} \]
Differentiating \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = 5 \cdot \frac{1}{2} x^{-1/2} = \frac{5}{2\sqrt{x}} \]

### Step 4: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now we can substitute \( v \) and \( \frac{dv}{dx} \) back into the acceleration equation:
\[ a = v \cdot \frac{dv}{dx} = (5\sqrt{x}) \cdot \left(\frac{5}{2\sqrt{x}}\right) \]

### Step 5: Simplify the expression
Simplifying the expression, we have:
\[ a = 5\sqrt{x} \cdot \frac{5}{2\sqrt{x}} = \frac{25}{2} = 12.5 \, \text{m/s}^2 \]

### Final Result
Thus, the acceleration of the particle is:
\[ \boxed{12.5 \, \text{m/s}^2} \]

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