A body is moving under the expense of a constant power. If .X. is the displacement in time. `t_1`, then .x. is proportional to

Options

A. `t^(2//3)`
B. `t^(3//2)`
C. `t^(2)`
D. `t^(1//2)`

Answer

To solve the problem, we need to analyze the relationship between displacement \( x \) and time \( t \) when a body is moving under the influence of a constant power.

### Step-by-Step Solution:

1. **Understanding Power**:
Power \( P \) is defined as the rate at which work is done. Mathematically, it can be expressed as:
\[ P = F \cdot v \]
where \( F \) is the force and \( v \) is the velocity of the body.

2. **Expressing Power in Terms of Velocity**:
Since the power is constant, we can denote it as \( k \):
\[ F \cdot v = k \]

3. **Relating Force to Mass and Acceleration**:
According to Newton's second law, force can also be expressed as:
\[ F = m \cdot a \]
where \( m \) is the mass and \( a \) is the acceleration. Since acceleration \( a \) can be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]

4. **Substituting for Force**:
Substituting the expression for force into the power equation gives:
\[ m \cdot a \cdot v = k \]
or
\[ m \cdot v \cdot \frac{dv}{dt} = k \]

5. **Rearranging the Equation**:
Rearranging this equation leads to:
\[ m \cdot v \cdot dv = k \cdot dt \]

6. **Integrating Both Sides**:
We can integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( t \):
\[ \int m \cdot v \, dv = \int k \, dt \]
This results in:
\[ \frac{m}{2} v^2 = k \cdot t + C \]
where \( C \) is the constant of integration.

7. **Finding Velocity**:
Solving for \( v \):
\[ v^2 = \frac{2k}{m} t + \frac{2C}{m} \]
For simplicity, we can assume \( C = 0 \) (starting from rest), leading to:
\[ v^2 = \frac{2k}{m} t \]
Thus,
\[ v = \sqrt{\frac{2k}{m} t} \]

8. **Relating Displacement to Velocity**:
Since velocity \( v \) is also defined as the derivative of displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
We can substitute for \( v \):
\[ \frac{dx}{dt} = \sqrt{\frac{2k}{m} t} \]

9. **Integrating to Find Displacement**:
Now we integrate to find \( x \):
\[ dx = \sqrt{\frac{2k}{m}} \sqrt{t} \, dt \]
Integrating both sides gives:
\[ x = \sqrt{\frac{2k}{m}} \cdot \frac{2}{3} t^{3/2} + C \]
Again, assuming \( C = 0 \), we have:
\[ x = \frac{2}{3} \sqrt{\frac{2k}{m}} t^{3/2} \]

10. **Final Relationship**:
Thus, we conclude that:
\[ x \propto t^{3/2} \]

### Conclusion:
The displacement \( x \) is proportional to \( t^{3/2} \).