Acceleration of a particle moving in `x-y` plane varies with time t as. `vec(a) = (ti + 3t^(2)j)`.
Here a is in `m//s^(2)` and t in sec. At time `t = 0` particle is at rest origin. Mass of the particles is `1 kg`. Find the net work done on the particle in first `2 sec`.

Options

A. 40 J
B. 34 J
C. 16 J
D. 48 J

Answer

To solve the problem, we need to find the net work done on a particle moving in the x-y plane with a given acceleration vector over a time interval of 2 seconds. Here are the steps to arrive at the solution:

### Step 1: Understand the Given Information
The acceleration vector is given as:
\[ \vec{a} = t \hat{i} + 3t^2 \hat{j} \]
At time \( t = 0 \), the particle is at rest at the origin. The mass of the particle is \( 1 \, \text{kg} \).

### Step 2: Use the Work-Energy Theorem
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
Since the particle starts from rest, the initial kinetic energy \( KE_i = 0 \).

### Step 3: Find the Final Velocity
To find the final velocity, we need to integrate the acceleration vector with respect to time. We can express acceleration as the derivative of velocity:
\[ \vec{a} = \frac{d\vec{v}}{dt} \]
Thus,
\[ d\vec{v} = (t \hat{i} + 3t^2 \hat{j}) dt \]
Integrating both sides from \( t = 0 \) to \( t = 2 \) seconds:
\[ \vec{v} = \int_0^2 (t \hat{i} + 3t^2 \hat{j}) dt \]

### Step 4: Perform the Integration
We can split the integral:
\[ \vec{v} = \int_0^2 t \hat{i} dt + \int_0^2 3t^2 \hat{j} dt \]
Calculating each integral:
1. For the \( \hat{i} \) component:
\[ \int_0^2 t \, dt = \left[\frac{t^2}{2}\right]_0^2 = \frac{2^2}{2} - 0 = 2 \]
So, the \( \hat{i} \) component is \( 2 \hat{i} \).

2. For the \( \hat{j} \) component:
\[ \int_0^2 3t^2 \, dt = 3 \left[\frac{t^3}{3}\right]_0^2 = \left[t^3\right]_0^2 = 2^3 - 0 = 8 \]
So, the \( \hat{j} \) component is \( 8 \hat{j} \).

Combining both components, we find:
\[ \vec{v} = 2 \hat{i} + 8 \hat{j} \]

### Step 5: Calculate the Magnitude of Velocity
The magnitude of the velocity vector is:
\[ |\vec{v}| = \sqrt{(2)^2 + (8)^2} = \sqrt{4 + 64} = \sqrt{68} \]

### Step 6: Calculate the Final Kinetic Energy
The kinetic energy is given by:
\[ KE_f = \frac{1}{2} mv^2 \]
Substituting the mass \( m = 1 \, \text{kg} \) and \( v = \sqrt{68} \):
\[ KE_f = \frac{1}{2} \cdot 1 \cdot (68) = \frac{68}{2} = 34 \, \text{J} \]

### Step 7: Calculate the Work Done
Since the initial kinetic energy \( KE_i = 0 \):
\[ W = KE_f - KE_i = 34 - 0 = 34 \, \text{J} \]

### Final Answer
The net work done on the particle in the first 2 seconds is:
\[ \boxed{34 \, \text{J}} \]