The equation `sqrt x = t + 9` gives the variation of displacement with time. Which of the following is correct ?

Options

A. Velocity is proporational to the square of time
B. Velocity is inversely proporational to time
C. Acceleration depends upon time
D. Acceleration is constant

Answer

To solve the problem, we need to analyze the given equation and derive the necessary quantities step by step.

### Step-by-Step Solution:

1. **Given Equation**:
The equation provided is:
\[ \sqrt{x} = t + 9 \]

2. **Square Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ x = (t + 9)^2 \]

3. **Expand the Right Side**:
Now, we expand the right side:
\[ x = t^2 + 18t + 81 \]

4. **Find Velocity**:
Velocity \( v \) is defined as the rate of change of displacement with respect to time. We can find velocity by differentiating \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 18t + 81) \]
Differentiating gives:
\[ v = 2t + 18 \]

5. **Find Acceleration**:
Acceleration \( a \) is defined as the rate of change of velocity with respect to time. We differentiate \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(2t + 18) \]
Differentiating gives:
\[ a = 2 \]

6. **Conclusion**:
From the calculations, we find that:
- The velocity \( v \) is a linear function of time \( t \) (specifically \( v = 2t + 18 \)).
- The acceleration \( a \) is constant and equal to \( 2 \, \text{m/s}^2 \).

### Final Answer:
The correct interpretation from the options provided is that the acceleration is constant.