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 a particle with mass `m` initially at rest and subjected to a constant power `P`, we need to derive the relationship between distance `x` and time `t`. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between power, work, and energy
Power is defined as the rate at which work is done or energy is transferred. The relationship can be expressed as:
\[ P = \frac{dW}{dt} \]
where \( W \) is the work done.

### Step 2: Relate power to kinetic energy
The work done on the particle is converted into kinetic energy. The kinetic energy \( K \) of a particle is given by:
\[ K = \frac{1}{2} mv^2 \]
If we supply power \( P \) for a time \( t \), the total work done (or energy supplied) is:
\[ W = P \cdot t \]

### Step 3: Set up the equation for kinetic energy
Since the work done is equal to the change in kinetic energy, we can write:
\[ P \cdot t = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity \( v \)
Rearranging the equation gives us:
\[ v^2 = \frac{2Pt}{m} \]
Taking the square root, we find:
\[ v = \sqrt{\frac{2Pt}{m}} \]

### Step 5: Relate velocity to distance
Velocity is defined as the rate of change of distance with respect to time:
\[ v = \frac{dx}{dt} \]
Substituting our expression for \( v \):
\[ \sqrt{\frac{2Pt}{m}} = \frac{dx}{dt} \]

### Step 6: Separate variables and integrate
Rearranging gives:
\[ dx = \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]
Now we integrate both sides. The left side integrates to \( x \), and the right side integrates as follows:
\[ \int dx = \int \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]

The integral of \( t^{1/2} \) is:
\[ \int t^{1/2} dt = \frac{2}{3} t^{3/2} \]
Thus, we have:
\[ x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} + C \]
Since the particle starts from rest at \( t = 0 \), \( C = 0 \). Therefore:
\[ x = \frac{2\sqrt{2P}}{3\sqrt{m}} t^{3/2} \]

### Step 7: Analyze the relationship between \( x \) and \( t \)
From the derived equation, we see that:
\[ x \propto t^{3/2} \]
This indicates that the distance \( x \) varies with the time \( t \) in a non-linear fashion, specifically as a power function.

### Step 8: Identify the correct graph
Since \( x \) varies as \( t^{3/2} \), the graph of \( x \) versus \( t \) will be a curve that starts at the origin and increases steeply as \( t \) increases. The correct graph will show this non-linear relationship.

### Conclusion
The correct graph representing the relationship between distance and time for a particle under constant power is one that curves upwards, indicating that distance increases with the time raised to the power of \( 3/2 \).

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