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 between time \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) given the power function \( P(t) = 3t^2 - 2t + 1 \, \text{W} \), we can follow these steps:

### Step 1: Understand the relationship between power and work
Power is defined as the rate of doing work, which can be expressed mathematically as:
\[ P = \frac{dW}{dt} \]
where \( W \) is the work done. The work done over a time interval can be found by integrating the power function over that interval.

### Step 2: Set up the integral for work done
To find the work done (which equals the change in kinetic energy) from \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \), we need to integrate the power function:
\[ W = \int_{2}^{4} P(t) \, dt = \int_{2}^{4} (3t^2 - 2t + 1) \, dt \]

### Step 3: Calculate the integral
We will calculate the integral:
\[ W = \int (3t^2 - 2t + 1) \, dt \]
Calculating the integral term by term:
- The integral of \( 3t^2 \) is \( t^3 \)
- The integral of \( -2t \) is \( -t^2 \)
- The integral of \( 1 \) is \( t \)

Thus, we have:
\[ W = \left[ t^3 - t^2 + t \right]_{2}^{4} \]

### Step 4: Evaluate the definite integral
Now we will evaluate this from \( t = 2 \) to \( t = 4 \):
\[ W = \left(4^3 - 4^2 + 4\right) - \left(2^3 - 2^2 + 2\right) \]
Calculating the upper limit:
\[ 4^3 = 64, \quad 4^2 = 16 \quad \Rightarrow \quad 64 - 16 + 4 = 52 \]
Calculating the lower limit:
\[ 2^3 = 8, \quad 2^2 = 4 \quad \Rightarrow \quad 8 - 4 + 2 = 6 \]
Now substituting back:
\[ W = 52 - 6 = 46 \, \text{J} \]

### Step 5: Conclusion
The change in kinetic energy between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) is:
\[ \Delta KE = W = 46 \, \text{J} \]

Thus, the final answer is:
\[ \Delta KE = 46 \, \text{J} \]

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