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 step by step, we will use the impulse-momentum theorem, which states that the impulse applied to an object is equal to the change in its momentum.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Time-varying force: \( F(t) = 2t \)
- Mass of the particle: \( m = 2 \, \text{kg} \)
- Initial velocity at \( t = 0 \): \( v_0 = -4 \, \text{m/s} \) (negative indicates direction along the negative x-axis)
- Time interval: \( t = 4 \, \text{s} \)

2. **Calculate the Impulse:**
- Impulse \( J \) is given by the integral of force over time:
\[ J = \int_{0}^{4} F(t) \, dt = \int_{0}^{4} 2t \, dt \]
- Evaluating the integral:
\[ J = 2 \int_{0}^{4} t \, dt = 2 \left[ \frac{t^2}{2} \right]_{0}^{4} = 2 \left[ \frac{4^2}{2} - \frac{0^2}{2} \right] = 2 \left[ \frac{16}{2} \right] = 2 \times 8 = 16 \, \text{Ns} \]

3. **Relate Impulse to Change in Momentum:**
- The change in momentum \( \Delta p \) is given by:
\[ J = \Delta p = m(v_f - v_0) \]
- Substituting the known values:
\[ 16 = 2(v_f - (-4)) \]
- Simplifying:
\[ 16 = 2(v_f + 4) \]
\[ 8 = v_f + 4 \]
\[ v_f = 8 - 4 = 4 \, \text{m/s} \]

4. **Final Result:**
- The final velocity of the particle at \( t = 4 \, \text{s} \) is \( v_f = 4 \, \text{m/s} \).