The displacement of a particle is proportional to the cube of time. Then magnitude of its acceleration

Options

A. increases with time.
B. decreases with time.
C. constant but non-zero.
D. zero.

Answer

To solve the problem, we need to analyze the relationship between displacement, velocity, and acceleration when displacement is proportional to the cube of time.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The problem states that the displacement \( x \) of a particle is proportional to the cube of time \( t \). This can be expressed mathematically as:
\[ x = k t^3 \]
where \( k \) is a constant of proportionality.

2. **Finding Velocity**:
Velocity \( v \) is defined as the rate of change of displacement with respect to time. Therefore, we differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = \frac{d(kt^3)}{dt} \]
Using the power rule of differentiation:
\[ v = 3kt^2 \]

3. **Finding Acceleration**:
Acceleration \( a \) is defined as the rate of change of velocity with respect to time. Thus, we differentiate \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d(3kt^2)}{dt} \]
Again, applying the power rule:
\[ a = 6kt \]

4. **Magnitude of Acceleration**:
The magnitude of acceleration is given by \( |a| \). Since \( k \) is a constant, the magnitude of acceleration can be expressed as:
\[ |a| = 6k|t| \]
This shows that the magnitude of acceleration is directly proportional to time \( t \).

### Conclusion:
The magnitude of acceleration increases linearly with time, which means as time increases, the magnitude of acceleration also increases.