A body at rest is moved in a stright line by supplying constant power to it . Distance moved by the body in time t is found to be proportional to `t^(n)` Value of n is .

Options

A. `3//2`
B. `2//3`
C. `1/2`
D. `3//4`

Explanation

To solve the problem, we need to analyze the motion of a body being moved by a constant power. The key steps are as follows:

### Step 1: Understand the relationship between power, force, and velocity.
Power (P) is defined as the product of force (F) and velocity (v):
\[ P = F \cdot v \]

### Step 2: Relate 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 of the body and \( a \) is its acceleration.

### Step 3: Express acceleration in terms of velocity.
Acceleration \( a \) can be expressed as the rate of change of velocity:
\[ a = \frac{dv}{dt} \]
Substituting this into the force equation gives:
\[ F = m \cdot \frac{dv}{dt} \]

### Step 4: Substitute force into the power equation.
Substituting \( F \) into the power equation, we have:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

### Step 5: Rearrange the equation for integration.
Rearranging gives:
\[ P \cdot dt = m \cdot v \cdot dv \]
Now we can integrate both sides. Since power is constant, we can take it out of the integral.

### Step 6: Integrate both sides.
Integrating from 0 to \( t \) on the left side and from 0 to \( v \) on the right side:
\[ \int_0^t P \, dt = \int_0^v m \cdot v \, dv \]
This results in:
\[ P \cdot t = \frac{m}{2} v^2 \]

### Step 7: Solve for velocity.
Rearranging gives:
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root:
\[ v = \sqrt{\frac{2Pt}{m}} \]

### Step 8: Relate velocity to distance.
Now, we know that velocity \( v \) is also the rate of change of distance \( x \):
\[ v = \frac{dx}{dt} \]
Substituting for \( v \):
\[ \frac{dx}{dt} = \sqrt{\frac{2Pt}{m}} \]

### Step 9: Separate variables and integrate.
Rearranging gives:
\[ dx = \sqrt{\frac{2P}{m}} \cdot t^{1/2} \, dt \]
Integrating both sides from 0 to \( x \) and 0 to \( t \):
\[ \int_0^x dx = \sqrt{\frac{2P}{m}} \int_0^t t^{1/2} \, dt \]
The integral on the right side evaluates to:
\[ x = \sqrt{\frac{2P}{m}} \cdot \left( \frac{2}{3} t^{3/2} \right) \]

### Step 10: Final relationship.
Thus, we find:
\[ x = \frac{2\sqrt{2P}}{3\sqrt{m}} t^{3/2} \]
This shows that distance \( x \) is proportional to \( t^{3/2} \).

### Conclusion:
The value of \( n \) is \( \frac{3}{2} \).