1. **State the problem:** We are given the displacement function $s(t) = 4t^3 + 3t^2$ and need to find the instantaneous rate of change of displacement at $t=3$ seconds. This rate of change is the velocity at that time. 2. **Formula used:** The instantaneous rate of change of a function at a point is given by the derivative of the function evaluated at that point. So velocity $v(t) = s'(t) = \frac{ds}{dt}$. 3. **Find the derivative:** Differentiate $s(t)$ with respect to $t$: $$s'(t) = \frac{d}{dt}(4t^3 + 3t^2) = 4 \cdot 3t^{3-1} + 3 \cdot 2t^{2-1} = 12t^2 + 6t$$ 4. **Evaluate at $t=3$:** Substitute $t=3$ into $s'(t)$: $$v(3) = 12(3)^2 + 6(3) = 12 \cdot 9 + 18 = 108 + 18 = 126$$ 5. **Interpretation:** The instantaneous velocity of the car at 3 seconds is $126$ meters per second. **Final answer:** The velocity at $t=3$ seconds is $126$ m/s.