To find the expression for the car's velocity $v(t)$, we need to take the derivative of the position function $x(t)$ with respect to time $t$.

The position function is given as: $x(t) = \frac{2}{3}t^3 - 8t^2 + 30t + 36$

Now, let's differentiate each term:

1. The derivative of $\frac{2}{3}t^3$ is: $\frac{d}{dt} \left(\frac{2}{3}t^3\right) = 2t^2$
2. The derivative of $-8t^2$ is: $\frac{d}{dt} (-8t^2) = -16t$
3. The derivative of $30t$ is: $\frac{d}{dt} (30t) = 30$
4. The derivative of the constant $36$ is: $\frac{d}{dt} (36) = 0$

Thus, the velocity function $v(t)$ is: $v(t) = 2t^2 - 16t + 30$

Would you like more details on this differentiation process or further clarifications?

Related questions:

1. What is the car's acceleration $a(t)$ based on the velocity function?
2. At what time does the car reach its maximum velocity?
3. How can we find the time when the car comes to a stop?
4. How would the behavior of the car change if the $t^3$ term was absent in the position function?
5. What is the physical meaning of each term in the velocity function?

Tip: Remember that velocity is the first derivative of position, while acceleration is the derivative of velocity.