**Given:**
Acceleration of the particle, a = (2ti + 3t^2j) m/s^2
Initial velocity, u = (2i + 3j) m/s
Time, t = 4 s

**To find:**
Velocity of the particle at t = 4 s

**Solution:**
To find the velocity of the particle at t = 4 s, we need to integrate the acceleration function with respect to time.

1. **Integration of acceleration with respect to time:**
The integration of acceleration function with respect to time gives the velocity function. Since the acceleration function is given as a function of time, we can directly integrate it with respect to time.

Given, a = (2ti + 3t^2j)
To find the velocity, v, we integrate the acceleration function with respect to time:
v = ∫ a dt

Integrating each component separately:
∫ (2ti) dt = t^2i + C1
∫ (3t^2j) dt = t^3j + C2

Therefore, the velocity function is given by:
v = t^2i + C1 + t^3j + C2

2. **Finding the constants of integration:**
To find the constants of integration, we can use the given initial velocity, u.

Given, u = (2i + 3j)

Substituting t = 0 in the velocity function, we get:
v(0) = (0^2i + C1) + (0^3j + C2)
u = C1 + C2

Therefore, the constants of integration are C1 = 2i and C2 = 3j.

3. **Substituting the values and finding the velocity at t = 4 s:**
Substituting the constants of integration and the given time, t = 4 s, in the velocity function, we get:
v = t^2i + C1 + t^3j + C2
v = (4^2i + 2i) + (4^3j + 3j)
v = 16i + 2i + 64j + 3j
v = 18i + 67j

Therefore, the velocity of the particle at t = 4 s is given by:
v = 18i + 67j m/s.

Hence, the velocity of the particle at t = 4 s is 18i + 67j m/s.

Is the answer is ( 18i + 16j ) m/s...