A particle of mass m is moving on a circular path of radius r . Centripetal acceleration of the particle or radius r . Centripetal acceleration of the particle depends on time t according to relation `a_(c ) = kt^(2)` . What power is delievered to the particle ?

Options

A. mrkt
B. `mrkt^(2)`
C. `mr^(2)kt`
D. zero

Explanation

To solve the problem, we need to determine the power delivered to a particle of mass \( m \) moving in a circular path of radius \( r \) with centripetal acceleration given by the relation \( a_c = kt^2 \). Here are the steps to derive the power delivered to the particle:

### Step 1: Understand Centripetal Acceleration
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the tangential velocity of the particle and \( r \) is the radius of the circular path.

### Step 2: Relate Centripetal Acceleration to Velocity
From the given relation \( a_c = kt^2 \), we can equate this to the formula for centripetal acceleration:
\[ \frac{v^2}{r} = kt^2 \]
Multiplying both sides by \( r \) gives:
\[ v^2 = krt^2 \]

### Step 3: Solve for Velocity
Taking the square root of both sides, we find the velocity \( v \):
\[ v = \sqrt{krt^2} = \sqrt{kr} \cdot t \]

### Step 4: Find Tangential Acceleration
The tangential acceleration \( a_t \) is related to the change in velocity with respect to time:
\[ a_t = \frac{dv}{dt} \]
Substituting \( v = \sqrt{kr} \cdot t \) into the derivative:
\[ \frac{dv}{dt} = \sqrt{kr} \]

### Step 5: Calculate Tangential Force
The tangential force \( F_t \) acting on the particle can be expressed using Newton's second law:
\[ F_t = m \cdot a_t = m \cdot \frac{dv}{dt} = m \cdot \sqrt{kr} \]

### Step 6: Calculate Power Delivered to the Particle
Power \( P \) delivered to the particle is given by the dot product of the force and velocity:
\[ P = F_t \cdot v \]
Since both \( F_t \) and \( v \) are in the same direction, we can simplify this to:
\[ P = F_t \cdot v = (m \cdot \sqrt{kr}) \cdot (\sqrt{kr} \cdot t) \]
This simplifies to:
\[ P = m \cdot kr \cdot t \]

### Final Answer
Thus, the power delivered to the particle is:
\[ P = m \cdot kr \cdot t \]