A body travels such that square of time is proportional to the displacement. Its acceleration is:

Options

A. zero
B. infinite
C. constant
D. variable

Explanation

To solve the problem, we need to analyze the relationship between time and displacement given in the question.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We are given that the square of time (t²) is proportional to the displacement (s). This can be expressed mathematically as:
\[ s \propto t^2 \]
This means:
\[ s = k t^2 \]
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. Mathematically, this is given by:
\[ v = \frac{ds}{dt} \]
To find \( v \), we differentiate \( s = k t^2 \) with respect to \( t \):
\[ v = \frac{d}{dt}(k t^2) = k \cdot \frac{d}{dt}(t^2) = k \cdot 2t = 2kt \]

3. **Finding Acceleration**:
Acceleration (a) is defined as the rate of change of velocity with respect to time. This can be expressed as:
\[ a = \frac{dv}{dt} \]
Now, we differentiate \( v = 2kt \) with respect to \( t \):
\[ a = \frac{d}{dt}(2kt) = 2k \cdot \frac{d}{dt}(t) = 2k \cdot 1 = 2k \]

4. **Conclusion**:
The acceleration \( a \) is a constant value, specifically \( 2k \). Since \( k \) is a constant, the acceleration does not change over time.

### Final Answer:
The acceleration of the body is a constant value \( 2k \).

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