A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :

Options

A. `2pi mk^(2)r^(2)t`
B. `mk^(2)r^(2)t`
C. `(mk^(4)r^(2)t^(5))//3`
D. Zero

Answer

To solve the problem, we need to analyze the motion of the particle and calculate the power delivered to it by the forces acting on it.

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

### Step 2: Relating Centripetal Acceleration to Speed
According to the problem, the centripetal acceleration is given as:
\[ a_c = k^2 r t^2 \]
Setting the two expressions for centripetal acceleration equal gives us:
\[ \frac{v^2}{r} = k^2 r t^2 \]

### Step 3: Solving for Speed \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = k^2 r^2 t^2 \]
Taking the square root of both sides, we find:
\[ v = k r t \]

### Step 4: Finding Kinetic Energy
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the expression for \( v \):
\[ KE = \frac{1}{2} m (k^2 r^2 t^2) = \frac{1}{2} m k^2 r^2 t^2 \]

### Step 5: Calculating Power
Power \( P \) is defined as the rate of change of kinetic energy with respect to time:
\[ P = \frac{d(KE)}{dt} \]
Now, substituting the expression for \( KE \):
\[ P = \frac{d}{dt} \left( \frac{1}{2} m k^2 r^2 t^2 \right) \]
Using the derivative:
\[ P = \frac{1}{2} m k^2 r^2 \cdot 2t = m k^2 r^2 t \]

### Final Result
Thus, the power delivered to the particle by the forces acting on it is:
\[ P = m k^2 r^2 t \]