**Answer:**

To find the acceleration of the particle at t = 2s, we need to differentiate the linear momentum equation with respect to time.

Given: p = 2t^2

**Differentiation of the Linear Momentum Equation:**

Differentiating p = 2t^2 with respect to t, we get:

dp/dt = d/dt(2t^2)

dp/dt = 4t

Therefore, the derivative of the linear momentum equation is dp/dt = 4t.

**Calculating Acceleration at t = 2s:**

To find the acceleration at t = 2s, we substitute t = 2 into the expression for dp/dt:

dp/dt = 4t

dp/dt = 4(2)

dp/dt = 8 m/s^2

Therefore, the acceleration of the particle at t = 2s is 8 m/s^2.

**Explanation:**

The linear momentum of a particle is given by the product of its mass and velocity. In this problem, the linear momentum (p) is represented as a function of time (t) by the equation p = 2t^2, where p is in kg m/s and t is in seconds.

To find the acceleration of the particle at a specific time, we need to differentiate the linear momentum equation with respect to time. This is because acceleration is the rate of change of velocity, and velocity is the derivative of position with respect to time.

By differentiating p = 2t^2 with respect to t, we obtain dp/dt = 4t. This represents the rate of change of the linear momentum with respect to time.

Substituting t = 2 into the expression for dp/dt, we find that the acceleration at t = 2s is 8 m/s^2.

This means that at t = 2s, the particle is experiencing an acceleration of 8 m/s^2, indicating that its velocity is increasing at a constant rate of 8 m/s^2.