Constant power P is supplied to a particle having mass `m` and initially at rest. choose correct graph.

Options

A.
B.
C.
D.

Answer

To solve the problem of determining the correct graph for a particle under constant power, we can follow these steps:

### Step 1: Understand the relationship between power, work, and kinetic energy
Given that constant power \( P \) is supplied to a particle of mass \( m \) that starts from rest, we can relate power to kinetic energy. The work done by the power over time \( t \) is equal to the change in kinetic energy.

### Step 2: Relate power to kinetic energy
The work done \( W \) by the power over time \( t \) can be expressed as:
\[ W = P \cdot t \]
This work is equal to the kinetic energy \( KE \) of the particle:
\[ W = \frac{1}{2} m v^2 \]
Setting these two equations equal gives:
\[ P \cdot t = \frac{1}{2} m v^2 \]

### Step 3: Solve for velocity \( v \)
Rearranging the equation for \( v \):
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2P}{m} t} \]

### Step 4: Find the relationship between distance \( s \) and time \( t \)
The distance \( s \) traveled by the particle can be found by integrating the velocity over time:
\[ s = \int_0^t v \, dt \]
Substituting the expression for \( v \):
\[ s = \int_0^t \sqrt{\frac{2P}{m} t'} \, dt' \]

### Step 5: Perform the integration
The integral can be computed as follows:
\[ s = \sqrt{\frac{2P}{m}} \int_0^t (t')^{1/2} \, dt' \]
The integral of \( (t')^{1/2} \) is:
\[ \int (t')^{1/2} \, dt' = \frac{2}{3} (t')^{3/2} \]
Evaluating this from 0 to \( t \) gives:
\[ s = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ s = \frac{2}{3} \sqrt{\frac{2P}{m}} t^{3/2} \]

### Step 6: Determine the relationship between \( s \) and \( t \)
From the final expression, we can see that:
\[ s \propto t^{3/2} \]
This indicates that the graph of \( s \) versus \( t \) will be a curve that is concave upward, resembling a parabola.

### Conclusion
The correct graph representing the relationship between distance \( s \) and time \( t \) is a concave upward parabola.

---