Step 1: Understanding the Concept:
Power is the rate of change of work done, and according to the Work-Energy Theorem, the work done on a particle is equal to the change in its kinetic energy. Since the particle starts from rest, the work done up to time $t$ equals its kinetic energy at that time.
Step 2: Key Formula or Approach:
1. Work Done \( W = \int P \, dt \).
2. Work-Energy Theorem: \( W = \Delta K.E. = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \).
Step 3: Detailed Explanation:
Given \( P = \frac{3}{2}t^2 \). Calculate work done from \( t = 0 \) to \( t = 2 \):
\[ W = \int_{0}^{2} \frac{3}{2}t^2 \, dt = \frac{3}{2} \left[ \frac{t^3}{3} \right]_0^2 = \frac{1}{2} [2^3 - 0^3] = 4 \text{ J} \]
Using the Work-Energy Theorem (\( u = 0 \)):
\[ W = \frac{1}{2}mv^2 \implies 4 = \frac{1}{2}(2)v^2 \]
\[ 4 = v^2 \implies v = 2 \text{ m/s} \]
Step 4: Final Answer:
The velocity at \( t = 2 \) s is 2 m/s.