A particle moving in a straight line has velocity-displacement equation as `v = 5 sqrt(1 + s).` Here v is in `ms^-1` and s in metres. Select the correct alternative.

Options

A. Particle is initially at rest
B. Initially velocity of the particle is `5 m//s` and the particle has a constant acceleration of `12.5 ms^-2`
C. Particle moves with a uniform velocity
D. None of the above

Explanation

To solve the problem, we need to analyze the given velocity-displacement equation and find the initial velocity and acceleration of the particle.

### Step-by-Step Solution:

1. **Understanding the Given Equation:**
The velocity \( v \) of the particle is given by the equation:
\[ v = 5 \sqrt{1 + s} \]
where \( v \) is in meters per second (m/s) and \( s \) is in meters (m).

2. **Finding the Initial Velocity:**
To find the initial velocity, we need to evaluate the velocity when the displacement \( s \) is zero:
\[ v = 5 \sqrt{1 + 0} = 5 \sqrt{1} = 5 \, \text{m/s} \]
Therefore, the initial velocity \( v_0 \) is:
\[ v_0 = 5 \, \text{m/s} \]

3. **Finding the Acceleration:**
Acceleration \( a \) can be expressed in terms of velocity and displacement using the formula:
\[ a = v \frac{dv}{ds} \]
We need to differentiate \( v \) with respect to \( s \).

4. **Differentiating Velocity:**
First, we differentiate the velocity function:
\[ v = 5 \sqrt{1 + s} \]
Using the chain rule:
\[ \frac{dv}{ds} = 5 \cdot \frac{1}{2\sqrt{1 + s}} \cdot \frac{d(1 + s)}{ds} = 5 \cdot \frac{1}{2\sqrt{1 + s}} \cdot 1 = \frac{5}{2\sqrt{1 + s}} \]

5. **Substituting into the Acceleration Formula:**
Now, substituting \( v \) and \( \frac{dv}{ds} \) into the acceleration formula:
\[ a = v \cdot \frac{dv}{ds} = 5 \sqrt{1 + s} \cdot \frac{5}{2\sqrt{1 + s}} = \frac{25}{2} \, \text{m/s}^2 \]

6. **Final Results:**
The initial velocity is:
\[ v_0 = 5 \, \text{m/s} \]
The acceleration at any time \( t \) is:
\[ a = \frac{25}{2} \, \text{m/s}^2 \]

### Summary:
- Initial Velocity: \( 5 \, \text{m/s} \)
- Acceleration: \( \frac{25}{2} \, \text{m/s}^2 \)