If for a particle position `x prop t^(2)` then :-

Options

A. velocity is constant
B. acceleration is constant
C. acceleration is variable
D. None of these

Answer

To solve the problem, we need to analyze the relationship between the position \( x \) of a particle and time \( t \) given that \( x \) is directly proportional to \( t^2 \). We will derive expressions for velocity and acceleration and determine their nature.

### Step-by-step Solution:

1. **Express the Position Function**:
Since \( x \) is directly proportional to \( t^2 \), we can write:
\[ x = k t^2 \]
where \( k \) is a constant of proportionality.

**Hint**: Remember that "directly proportional" means we can express the relationship using a constant multiplier.

2. **Differentiate to Find Velocity**:
The velocity \( v \) is the rate of change of position with respect to time, which can be expressed as:
\[ v = \frac{dx}{dt} \]
Differentiating \( x = k t^2 \):
\[ v = \frac{d}{dt}(k t^2) = k \cdot \frac{d}{dt}(t^2) = k \cdot 2t = 2kt \]

**Hint**: Use the power rule for differentiation, which states that \(\frac{d}{dt}(t^n) = n t^{n-1}\).

3. **Differentiate to Find Acceleration**:
The acceleration \( a \) is the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Differentiating \( v = 2kt \):
\[ a = \frac{d}{dt}(2kt) = 2k \cdot \frac{d}{dt}(t) = 2k \cdot 1 = 2k \]

**Hint**: Remember that the derivative of a constant multiplied by a variable is just the constant times the derivative of the variable.

4. **Analyze the Results**:
- The velocity \( v = 2kt \) is dependent on \( t \), which means it is **variable**.
- The acceleration \( a = 2k \) is a constant since \( k \) is a constant.

5. **Conclusion**:
Based on the analysis:
- Velocity is variable.
- Acceleration is constant.

Therefore, the correct option is **B**: Velocity is variable and acceleration is constant.