If the displacement of a particle varies with time as ` sqrt(x) = t + 7 ` . The -

Options

A. velocity of the particle is inversely proportional to t
B. velocity of the particle is proportional to t
C. velocity of the particle is proportional to `sqrt(t)`
D. the particle moves with a constant acceleration

Answer

To solve the problem, we start with the given equation for the displacement of a particle:

\[ \sqrt{x} = t + 7 \]

### Step 1: Square both sides
To eliminate the square root, we square both sides of the equation:

\[ x = (t + 7)^2 \]

### Step 2: Differentiate to find velocity
The velocity \( v \) is the rate of change of displacement with respect to time, which can be expressed as:

\[ v = \frac{dx}{dt} \]

Now, we differentiate \( x \) with respect to \( t \):

\[ \frac{dx}{dt} = \frac{d}{dt}[(t + 7)^2] \]

Using the chain rule, we get:

\[ \frac{dx}{dt} = 2(t + 7) \cdot \frac{d}{dt}(t + 7) = 2(t + 7) \cdot 1 = 2(t + 7) \]

Thus, the velocity \( v \) is:

\[ v = 2(t + 7) \]

### Step 3: Analyze the velocity
From the expression for velocity, we can see that:

\[ v = 2t + 14 \]

This shows that the velocity is a linear function of time \( t \), which means it is proportional to \( t \).

### Step 4: Find acceleration
To find the acceleration \( a \), we differentiate the velocity with respect to time:

\[ a = \frac{dv}{dt} \]

Differentiating \( v = 2t + 14 \):

\[ \frac{dv}{dt} = 2 \]

This indicates that the acceleration is constant, specifically:

\[ a = 2 \, \text{m/s}^2 \]

### Conclusion
Based on the analysis:
1. The velocity of the particle is proportional to \( t \).
2. The particle moves with a constant acceleration of \( 2 \, \text{m/s}^2 \).

### Final Answer
- The velocity of the particle is proportional to \( t \).
- The particle moves with constant acceleration.

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