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 start with the given velocity-displacement equation:

\[ v = 5 \sqrt{1 + s} \]

### Step 1: Square both sides

To eliminate the square root, we square both sides of the equation:

\[ v^2 = (5 \sqrt{1 + s})^2 \]

This simplifies to:

\[ v^2 = 25(1 + s) \]

### Step 2: Expand the equation

Expanding the right-hand side gives us:

\[ v^2 = 25 + 25s \]

### Step 3: Compare with the standard equation

We know from kinematics that the equation relating velocity, initial velocity, acceleration, and displacement is:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( s \) is the displacement.

### Step 4: Identify terms

By comparing the two equations:

1. From \( v^2 = 25 + 25s \), we can see that:
- \( u^2 = 25 \) (which implies \( u = 5 \) or \( u = -5 \))
- \( 2a = 25 \) (which gives \( a = \frac{25}{2} = 12.5 \, \text{m/s}^2 \))

### Step 5: Determine initial conditions

Since the problem states that the particle is moving in a straight line, we consider the positive value for the initial velocity:

- Initial velocity \( u = 5 \, \text{m/s} \)
- Acceleration \( a = 12.5 \, \text{m/s}^2 \)

### Conclusion

Based on the analysis, we conclude that:

- The initial velocity of the particle is \( 5 \, \text{m/s} \).
- The particle has a constant acceleration of \( 12.5 \, \text{m/s}^2 \).

Thus, the correct option is:

**Option 2: Initial velocity of the particle is 5 meters per second and particle has constant acceleration of 12.5 meters per second square.**

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