1. **State the problem:** We are given the acceleration function of a particle as $a(t) = 36t^2 + 12t$. We need to find the velocity at time $t=2$ given that the velocity at $t=1$ is 1 and the position at $t=1$ is 1. 2. **Recall the relationship:** Acceleration is the derivative of velocity, so $$a(t) = \frac{dv}{dt}$$ To find velocity, integrate acceleration: $$v(t) = \int a(t) \, dt + C$$ 3. **Integrate acceleration:** $$v(t) = \int (36t^2 + 12t) \, dt + C = 36 \int t^2 \, dt + 12 \int t \, dt + C$$ $$= 36 \cdot \frac{t^3}{3} + 12 \cdot \frac{t^2}{2} + C = 12t^3 + 6t^2 + C$$ 4. **Use initial velocity to find $C$:** Given $v(1) = 1$, $$1 = 12(1)^3 + 6(1)^2 + C = 12 + 6 + C = 18 + C$$ $$\Rightarrow C = 1 - 18 = -17$$ 5. **Velocity function:** $$v(t) = 12t^3 + 6t^2 - 17$$ 6. **Find velocity at $t=2$:** $$v(2) = 12(2)^3 + 6(2)^2 - 17 = 12 \cdot 8 + 6 \cdot 4 - 17 = 96 + 24 - 17 = 103$$ **Final answer:** The velocity at $t=2$ is $\boxed{103}$.