Power of the only force acting on a particle of mass m=1 kg moving in straight line depends on its velocity as `P=v^(2)` where v is in m/s and P is in watt. If initial velocity of the particle is 1m/s, then the displacement of the particle in `ln2` second will be :

Options

A. `(ln2-1)m`
B. `(ln2)^(2)m`
C. `1m`
D. `2m`

Answer

To solve the problem step by step, we will analyze the relationship between power, force, and displacement.

### Step 1: Understand the relationship between power and force
The power \( P \) is given by the equation:
\[ P = F \cdot v \]
where \( F \) is the force acting on the particle and \( v \) is its velocity. From the problem, we know that:
\[ P = v^2 \]
Thus, we can equate the two expressions for power:
\[ F \cdot v = v^2 \]

### Step 2: Solve for force
From the equation \( F \cdot v = v^2 \), we can solve for the force \( F \):
\[ F = \frac{v^2}{v} = v \]
This indicates that the force acting on the particle is equal to its velocity.

### Step 3: Apply Newton's second law
Using Newton's second law, we have:
\[ F = m \cdot a \]
where \( m \) is the mass of the particle and \( a \) is its acceleration. Given that the mass \( m = 1 \, \text{kg} \), we can write:
\[ v = m \cdot a \]
Since \( m = 1 \):
\[ v = a \]

### Step 4: Relate acceleration to velocity
The acceleration \( a \) can also be expressed as:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ v = \frac{dv}{dt} \]

### Step 5: Rearranging and integrating
Rearranging gives:
\[ dt = \frac{dv}{v} \]
Now we can integrate both sides. We need to find the displacement \( x \) in terms of velocity \( v \):
\[ dx = v \cdot dt \]
Substituting for \( dt \):
\[ dx = v \cdot \frac{dv}{v} = dv \]

### Step 6: Set up the integral
Now we need to integrate from the initial velocity \( v_0 = 1 \, \text{m/s} \) to some final velocity \( v \), and from \( x = 0 \) to \( x \):
\[ \int_{0}^{x} dx = \int_{1}^{v} dv \]
This simplifies to:
\[ x = v - 1 \]

### Step 7: Express velocity in terms of time
From the earlier relationship, we have:
\[ v = x + 1 \]
Now we need to find the relationship with time. Since we have:
\[ \frac{dx}{dt} = x + 1 \]
Rearranging gives:
\[ \frac{dx}{x + 1} = dt \]

### Step 8: Integrate with limits
Integrating both sides:
\[ \int_{0}^{x} \frac{dx}{x + 1} = \int_{0}^{t} dt \]
This gives:
\[ \ln(x + 1) \bigg|_{0}^{x} = t \bigg|_{0}^{t} \]
Thus:
\[ \ln(x + 1) - \ln(1) = t \]
So:
\[ \ln(x + 1) = t \]

### Step 9: Solve for displacement at \( t = \ln 2 \)
Substituting \( t = \ln 2 \):
\[ \ln(x + 1) = \ln 2 \]
Exponentiating both sides gives:
\[ x + 1 = 2 \]
Thus:
\[ x = 2 - 1 = 1 \, \text{m} \]

### Final Answer
The displacement of the particle in \( \ln 2 \) seconds is:
\[ \boxed{1 \, \text{m}} \]