A body is moving unidirectionally under the influence of a sources of constant power. The square of its displacement in time t is proportional to

Options

A. `t^(3)`
B. `t^(2)`
C. t
D. `sqrt(t)`

Explanation

To solve the problem, we need to analyze the relationship between the displacement of a body moving under constant power and time. The key points to consider are the definitions of power, work, and kinetic energy.

### Step-by-Step Solution:

1. **Understanding Power**:
Power (P) is defined as the rate at which work is done. Mathematically, it is given by:
\[ P = \frac{W}{t} \]
where \(W\) is the work done and \(t\) is the time taken.

2. **Work Done**:
The work done by the body can be expressed in terms of kinetic energy. According to the work-energy theorem:
\[ W = \Delta KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the body and \(v\) is its velocity.

3. **Relating Work and Power**:
Since power is constant, we can express work done over time as:
\[ W = P \cdot t \]
Setting the two expressions for work equal gives:
\[ P \cdot t = \frac{1}{2} mv^2 \]

4. **Solving for Velocity**:
Rearranging the equation to find \(v^2\):
\[ v^2 = \frac{2Pt}{m} \]

5. **Relating Velocity to Displacement**:
Velocity is also defined as the rate of change of displacement:
\[ v = \frac{ds}{dt} \]
Substituting for \(v\) in terms of displacement gives:
\[ \frac{ds}{dt} = \sqrt{\frac{2Pt}{m}} \]

6. **Separating Variables**:
To find displacement \(s\), we can separate variables:
\[ ds = \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]

7. **Integrating**:
Integrating both sides:
\[ s = \int \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]
The integral of \(t^{1/2}\) is:
\[ \int t^{1/2} dt = \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} + C \]
where \(C\) is the constant of integration.

8. **Finding the Square of Displacement**:
The square of displacement \(s^2\) is then proportional to \(t^3\):
\[ s^2 \propto t^3 \]

### Conclusion:
The square of the displacement of the body in time \(t\) is proportional to \(t^3\).