A particle of mass `2 kg` starts moving in a straight line with an initial velocity of `2 m//s` at a constant acceleration of `2 m//s^(2)`. Then rate of change of kinetic energy.

Options

A. is four times the velocity at any moment
B. is two times the displacement at any moment
C. is four times the rate of change of velocity at any moment
D. is constant throughout

Answer

To solve the problem of finding the rate of change of kinetic energy for a particle with given mass, initial velocity, and constant acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Initial velocity, \( u = 2 \, \text{m/s} \)
- Constant acceleration, \( a = 2 \, \text{m/s}^2 \)

2. **Determine the Velocity at Any Time \( t \):**
Using the kinematic equation:
\[ V = u + at \]
Substitute the known values:
\[ V = 2 + 2t \]

3. **Calculate the Kinetic Energy \( K \):**
The formula for kinetic energy is:
\[ K = \frac{1}{2} mv^2 \]
Substitute \( m \) and \( V \):
\[ K = \frac{1}{2} \times 2 \times (2 + 2t)^2 \]
Simplifying this:
\[ K = 1 \times (2 + 2t)^2 \]
Expanding \( (2 + 2t)^2 \):
\[ K = (2^2 + 2 \cdot 2 \cdot 2t + (2t)^2) = 4 + 8t + 4t^2 \]

4. **Differentiate Kinetic Energy with Respect to Time \( t \):**
We need to find \( \frac{dK}{dt} \):
\[ \frac{dK}{dt} = \frac{d}{dt}(4 + 8t + 4t^2) \]
Differentiating term by term:
\[ \frac{dK}{dt} = 0 + 8 + 8t \]
Thus,
\[ \frac{dK}{dt} = 8 + 8t \]

5. **Express the Result in Terms of Velocity:**
We know from our earlier calculation that:
\[ V = 2 + 2t \]
Therefore, we can express \( \frac{dK}{dt} \) as:
\[ \frac{dK}{dt} = 8 + 8t = 4(2 + 2t) = 4V \]

### Final Result:
The rate of change of kinetic energy is:
\[ \frac{dK}{dt} = 4V \]