Step 1: Read the problem carefully.
The acceleration is not constant here. It keeps changing with time and is given by $a = 6t$. The particle starts from rest, which means at the start its velocity is zero. We want the velocity after $2$ seconds.
Step 2: Why normal formulas will not work.
The simple equations like $v = u + at$ only work when acceleration stays the same. Here the acceleration grows as time passes, so we must use a method that handles changing values. That method is integration.
Step 3: Write the basic link between velocity and acceleration.
Acceleration is just how fast velocity changes with time: \[ a = \frac{dv}{dt} \] So we can write a small change in velocity as $dv = a\,dt$.
Step 4: Put in the given acceleration.
Replace $a$ with $6t$: \[ dv = 6t\,dt \] Now we add up all these tiny changes from time $0$ up to time $2$ seconds.
Step 5: Do the integration.
The velocity at rest is zero, so we integrate from $0$ to the final velocity on the left side, and from $0$ to $2$ on the right side: \[ \int_0^{v} dv = \int_0^{2} 6t\,dt \] This gives \[ v = 6 \times \frac{t^2}{2} \Big|_0^{2} = 3t^2 \Big|_0^{2} \]
Step 6: Put the time value in.
Substitute $t = 2$: \[ v = 3 \times (2)^2 = 3 \times 4 = 12 \] So the velocity after $2$ seconds is twelve metres per second. \[ \boxed{12\text{ ms}^{-1}} \]