Displacement x of a particle varies with time as `sqrt(x)=t+5`, where x is in metres and time t is in seconds, Select the correct option.

Options

A. Acceleration of the particle is constant.
B. Velocity of the particle at t=0 is 10m/s.
C. Particle never reverses its direction of motion for `t gt 0`
D. All of the above.

Explanation

To solve the problem where the displacement \( x \) of a particle varies with time as \( \sqrt{x} = t + 5 \), we will follow these steps:

### Step 1: Express \( x \) in terms of \( t \)
Starting from the equation given:
\[ \sqrt{x} = t + 5 \]
We can square both sides to eliminate the square root:
\[ x = (t + 5)^2 \]

### Step 2: Differentiate \( x \) with respect to \( t \) to find velocity
To find the velocity \( v \), we differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}[(t + 5)^2] \]
Using the chain rule:
\[ v = 2(t + 5) \cdot \frac{d(t + 5)}{dt} = 2(t + 5) \cdot 1 = 2(t + 5) \]

### Step 3: Simplify the expression for velocity
Now we can simplify the expression for velocity:
\[ v = 2t + 10 \]

### Step 4: Calculate the velocity at \( t = 0 \)
To find the velocity at \( t = 0 \):
\[ v(0) = 2(0) + 10 = 10 \, \text{m/s} \]

### Step 5: Differentiate \( v \) with respect to \( t \) to find acceleration
Next, we differentiate \( v \) with respect to \( t \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}[2t + 10] = 2 \]

### Step 6: Analyze the results
1. The velocity at \( t = 0 \) is \( 10 \, \text{m/s} \).
2. The acceleration is constant and equal to \( 2 \, \text{m/s}^2 \).

### Step 7: Determine if the particle ever reverses direction
Since the velocity \( v = 2t + 10 \) is always positive for \( t \geq 0 \), the particle never reverses its direction of motion.

### Conclusion
Based on the calculations:
- The velocity at \( t = 0 \) is \( 10 \, \text{m/s} \).
- The acceleration is constant at \( 2 \, \text{m/s}^2 \).
- The particle does not reverse its direction of motion.

Thus, all statements in the options provided are correct.

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