If `X = 3 - 4t^(2) + t^(3)`, then work done in first 4s. will be (Mass of the particle is 3 gram) :

Options

A. 1. 384 mJ
B. 2. 168 mJ
C. 3. 192 mJ
D. 4. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the position function
Given the position function:
\[ X(t) = 3 - 4t^2 + t^3 \]

### Step 2: Find the velocity function
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dX}{dt} = \frac{d}{dt}(3 - 4t^2 + t^3) \]
Calculating the derivative:
\[ v(t) = 0 - 8t + 3t^2 = 3t^2 - 8t \]

### Step 3: Calculate the initial and final velocities
Now, we will calculate the velocities at \( t = 0 \) seconds and \( t = 4 \) seconds.

- At \( t = 0 \):
\[ v(0) = 3(0)^2 - 8(0) = 0 \, \text{m/s} \]

- At \( t = 4 \):
\[ v(4) = 3(4)^2 - 8(4) = 3(16) - 32 = 48 - 32 = 16 \, \text{m/s} \]

### Step 4: Calculate the change in kinetic energy
Using the work-energy theorem, the work done \( W \) is equal to the change in kinetic energy \( \Delta KE \):
\[ W = \Delta KE = KE_f - KE_i \]
Where:
- \( KE_f = \frac{1}{2} m v_f^2 \)
- \( KE_i = \frac{1}{2} m v_i^2 \)

Given that the mass \( m = 3 \) grams \( = 3 \times 10^{-3} \) kg, we can calculate:
- \( KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (3 \times 10^{-3}) (0)^2 = 0 \)
- \( KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} (3 \times 10^{-3}) (16)^2 \)

Calculating \( KE_f \):
\[ KE_f = \frac{1}{2} (3 \times 10^{-3}) (256) = \frac{3 \times 256}{2 \times 1000} = \frac{768}{2000} = 0.384 \, \text{J} \]

### Step 5: Convert to millijoules
To convert joules to millijoules:
\[ 0.384 \, \text{J} = 384 \, \text{mJ} \]

### Final Answer
Thus, the work done in the first 4 seconds is:
\[ \boxed{384 \, \text{millijoules}} \]

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