A time varying force, `F=2t` is acting on a particle of mass `2kg` moving along x-axis. velocity of the particle is `4m//s` along negative x-axis at time `t=0`. Find the velocity of the particle at the end of 4s.

Explanation

To solve the problem, we need to find the velocity of a particle under the influence of a time-varying force. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Initial velocity, \( v_0 = -4 \, \text{m/s} \) (along the negative x-axis)
- Time-varying force, \( F(t) = 2t \)
- Time duration, \( t = 4 \, \text{s} \)

### Step 2: Relate force to momentum
The force acting on the particle can be expressed in terms of the rate of change of momentum:
\[ F = \frac{dp}{dt} \]
Since momentum \( p = mv \), we can write:
\[ F = m \frac{dv}{dt} \]

### Step 3: Substitute the force into the equation
Substituting the expression for force:
\[ 2t = m \frac{dv}{dt} \]
Now, substituting \( m = 2 \, \text{kg} \):
\[ 2t = 2 \frac{dv}{dt} \]
Dividing both sides by 2 gives:
\[ t = \frac{dv}{dt} \]

### Step 4: Rearranging and integrating
Rearranging the equation:
\[ dv = t \, dt \]
Now, we integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[ \int_{v_0}^{v} dv = \int_{0}^{4} t \, dt \]

### Step 5: Perform the integration
The left side integrates to:
\[ v - v_0 \]
The right side integrates to:
\[ \frac{t^2}{2} \bigg|_0^4 = \frac{4^2}{2} - 0 = \frac{16}{2} = 8 \]
Thus, we have:
\[ v - (-4) = 8 \]
This simplifies to:
\[ v + 4 = 8 \]

### Step 6: Solve for final velocity
Now, solving for \( v \):
\[ v = 8 - 4 = 4 \, \text{m/s} \]

### Conclusion
The velocity of the particle at the end of 4 seconds is:
\[ \boxed{4 \, \text{m/s}} \quad \text{(along the positive x-axis)} \]