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 start with the given equation that describes the relationship between displacement \( x \) and time \( t \):

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

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

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

4. **Finding Velocity**:
The velocity \( v \) is the rate of change of displacement with respect to time, which can be expressed as:
\[ v = \frac{dx}{dt} \]
To find \( \frac{dx}{dt} \), we differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t^2 + 18t + 81) = 2t + 18 \]

5. **Finding Acceleration**:
The acceleration \( a \) is the rate of change of velocity with respect to time, which can be expressed as:
\[ a = \frac{dv}{dt} \]
We differentiate \( v \) with respect to \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt}(2t + 18) = 2 \]

6. **Conclusion**:
From our calculations, we find that:
- The velocity \( v = 2t + 18 \) is not constant and depends on time.
- The acceleration \( a = 2 \, \text{m/s}^2 \) is constant.

Thus, the correct option from the given choices is that the acceleration is constant and equal to \( 2 \, \text{m/s}^2 \).