Power applied to a particle varices with time as `P =(3t^(2)-2t + 1)` watt, where t is in second. Find the change in its kinetic energy between time `t=2s` and `t = 4 s` .

Options

A. 32 J
B. 46 J
C. 61 J
D. 102 J

Explanation

To find the change in kinetic energy of a particle when the power applied to it varies with time, we can use the relationship between power, work done, and kinetic energy. The power \( P \) is given as:

\[ P(t) = 3t^2 - 2t + 1 \quad \text{(in watts)} \]

We need to find the change in kinetic energy \( \Delta K \) between \( t = 2 \) seconds and \( t = 4 \) seconds. The change in kinetic energy is equal to the work done over that time interval, which can be calculated using the integral of power with respect to time:

\[ \Delta K = W = \int_{t_1}^{t_2} P(t) \, dt \]

where \( t_1 = 2 \) s and \( t_2 = 4 \) s.

### Step-by-Step Solution:

1. **Set up the integral for work done**:
\[ \Delta K = \int_{2}^{4} (3t^2 - 2t + 1) \, dt \]

2. **Integrate each term**:
- The integral of \( 3t^2 \) is \( t^3 \).
- The integral of \( -2t \) is \( -t^2 \).
- The integral of \( 1 \) is \( t \).

Therefore, we can write:
\[ \Delta K = \left[ t^3 - t^2 + t \right]_{2}^{4} \]

3. **Evaluate the integral at the limits**:
- First, substitute \( t = 4 \):
\[ (4^3 - 4^2 + 4) = (64 - 16 + 4) = 52 \]
- Next, substitute \( t = 2 \):
\[ (2^3 - 2^2 + 2) = (8 - 4 + 2) = 6 \]

4. **Calculate the change in kinetic energy**:
\[ \Delta K = 52 - 6 = 46 \, \text{Joules} \]

Thus, the change in kinetic energy between \( t = 2 \) seconds and \( t = 4 \) seconds is:

\[ \Delta K = 46 \, \text{Joules} \]